MC LCLI NI U ................................................................................................. 3Phn th nht: 10 PHNG PHP GII NHANH BI TP TRC NGHIM HA HC ..............................................4Phng php 1: p dng nh lut bo ton khi lng ........................ 4Phng php 2: Bo ton mol nguyn t ............................................... 13Phng php 3: Bo ton mol electron ................................................... 22Phng php 4: S dng phng trnh ion - electron ............................36Phng php 5: S dng cc gi tr trung bnh ...................................... 49Phng php 6: Tng gim khi lng .................................................. 60Phng php 7: Qui i hn hp nhiu cht v s lng cht t hn ..... 71Phng php 8: S ng cho ........................................................77Phng php 9: Cc i lng dng khi qut .................................... 85Phng php 10: T chn lng cht ..................................................... 97Phn th hai: 25 THI TH TUYN SINH I HC, CAO NG. . 108 s 01 ................................................................................................... 108 s 02 ................................................................................................... 115 s 03 ................................................................................................... 122 s 04 ................................................................................................... 129 s 05 ................................................................................................... 136 s 06 ................................................................................................... 143 s 07 ................................................................................................... 150 s 08 ................................................................................................... 157 s 09 ................................................................................................... 163 s 10 ................................................................................................... 170 s 11 ................................................................................................... 177 s 12 ................................................................................................... 185 s 13 ................................................................................................... 193 s 14 ................................................................................................... 201 s 15 ................................................................................................... 209 s 16 ................................................................................................... 216 s 17 ................................................................................................... 223 s 18 ................................................................................................... 231 s 19 ................................................................................................... 238 s 20 ................................................................................................... 247 s 21 ................................................................................................... 254 s 22 ................................................................................................... 262 s 23 ................................................................................................... 270 s 24 ................................................................................................... 2771 s 25 ................................................................................................... 284Phn th ba: P N 25 THI TH TUYN SINH I HC, CAO NG ..............................................................................291p n 01 ............................................................................................ 291p n 02 ............................................................................................ 291p n 03 ............................................................................................ 291p n 04 ............................................................................................ 292p n 05 ............................................................................................ 292p n 06 ............................................................................................ 292p n 07 ............................................................................................ 292p n 08 ............................................................................................ 293p n 09 ............................................................................................ 293p n 10 ............................................................................................ 293p n 11 ............................................................................................ 293p n 12 ............................................................................................ 294p n 13 ............................................................................................ 294p n 14 ............................................................................................ 294p n 15 ............................................................................................ 294p n 16 ............................................................................................ 295p n 17 ............................................................................................ 295p n 18 ............................................................................................ 295p n 19 ............................................................................................ 295p n 20 ............................................................................................ 296p n 21 ............................................................................................ 296p n 22 ............................................................................................ 296p n 23 ............................................................................................ 296p n 24 ............................................................................................ 297p n 25 ............................................................................................ 297LI NI U gip cho Gio vin v hc sinh n tp, luyn tp v vn dng cc kin thc vo vic gii cc bi tp trc nghim mn ha hc v c bit khi gii nhng bi tp cn phi tnh tonmt cch nhanh nht, thun li nht ng thi p ng cho k thi tuyn sinh i hc v cao ng. Chng ti xin trn trng gii thiu cun : 10 phng php gii nhanh trc nghim ha hc v 25 thi th tuyn sinh i hc v cao ng.2Cu trc ca cun sch gm 3 phn:Phn I: 10 phng php gii nhanh trc nghim ha hc. mi phng php gii nhanh trc nghim ha hc chng ti u trnh by phn hng dn gii mu chi tit nhng bi tp trc nghim kh, gip hc sinh c cch nhn nhn mi v phng php gii bi tp trc nghim tht ngn gn trong thi gian nhanh nht, bo m tnh chnh xc cao. gii bi tp trc nghim nhanhtrongvngt1-2pht chngtaphi bit phnloivnmchccc phng php suy lun. Vic gii bi tp trc nghim khng nht thit phi theo ng qui trnh cc bc gii, khng nht thit phis dng ht cc d kin u bi v i khi khng cnvit v cn bng tt c cc phng trnh phn ng. Phn II: 25 thi th tuyn sinh i hc, cao ng. Cc thi c xy dng vi ni dung a dng phong ph vi hm lng kin thc hon ton nm trong chng trnh ha hc THPT theo qui nh ca B Gio dc v o to. B thi c kh tng ng hoc cao hn cc c s dng trong cc k thi tuyn sinh i hc v cao ng gn y. Phn III: p n ca b 25 thi gii thiu phn II.Chng ti hi vng cun sch ny s l mt ti liu tham kho b ch cho gio vin v hc sinh THPT.Chng ti xin chn thnh cm n nhng kin ng gp xy dng ca Qu Thy,C gio, cc ng nghip v bn c. Cc tc gi. H Ni thng 1 nm 2008Phn th nht10 PHNG PHP GII NHANH BI TP TRC NGHIM HA HC3Phng php 1P DNG NH LUT BO TON KHI LNGNguyn tc ca phng php ny kh n gin, da vo nh lut bo ton khi lng: Tng khi lng cc cht tham gia phn ng bng tng khi lng cc cht to thnh trong phn ng. Cn lu l: khng tnh khi lng ca phn khng tham gia phn ng cng nh phn cht c sn, v d nc c sn trong dung dch.Khi c cn dung dch th khi lng mui thu c bng tng khi lng cc cation kim loi v anion gc axit.V d 1:Hn hp X gm Fe, FeO v Fe2O3. Cho mt lung CO i qua ng s ngmgamhn hp X nungnng. Saukhiktthcth nghimthu c 64 gam cht rn A trong ng s v 11,2 lt kh B (ktc) c t khi so vi H2 l 20,4. Tnh gi tr m.A. 105,6 gam. B. 35,2 gam. C. 70,4 gam. D. 140,8 gam.Hng dn giiCc phn ng kh st oxit c th c:3Fe2O3 + COot2Fe3O4 + CO2(1)Fe3O4 + COot3FeO + CO2(2)FeO + COotFe + CO2(3)Nh vy cht rn A c th gm 3 cht Fe, FeO, Fe3O4hoc t hn, iu khng quan trng v vic cn bng cc phng trnh trn cng khng cn thit, quan trng l s mol CO phn ng bao gi cng bng s mol CO2 to thnh.B11, 2n 0,522,5 mol.Gi x l s mol ca CO2 ta c phng trnh v khi lng ca B:44x + 28(0,5 x) = 0,5 20,4 2 = 20,4nhn c x = 0,4 mol v cng chnh l s mol CO tham gia phn ng.Theo LBTKL ta c: mX + mCO = mA + 2COm4 m = 64 + 0,4 44 0,4 28 = 70,4 gam. (p n C)V d 2: un 132,8 gam hn hp 3 ru no, n chc vi H2SO4 c 140oC thu c hn hp cc ete c s mol bng nhau v c khi lng l 111,2 gam. S mol ca mi ete trong hn hp l bao nhiu?A. 0,1 mol. B. 0,15 mol.C. 0,4 mol. D. 0,2 mol.Hng dn giiTa bit rng c 3 loi ru tch nc iu kin H2SO4c, 140oC th to thnh 6 loi ete v tch ra 6 phn t H2O.Theo LBTKL ta c2H O etem m m 132,8 11, 2 21,6 r u gam2H O21,6n 1,218 mol.Mt khc c hai phn t ru th to ra mt phn t ete v mt phn t H2O do s mol H2O lun bng s mol ete, suy ra s mol mi ete l 1,20,26 mol. (p n D)Nhn xt:Chng ta khng cn vit 6 phng trnh phn ng t ru tch nc to thnh 6 ete, cng khng cn tm CTPT ca cc ru v cc ete trn. Nu cc bn xa vo vic vit phng trnh phn ng v t n s mol cc ete tnh ton th khng nhng khng gii c m cn tn qu nhiu thi gian.V d 3: Cho 12 gam hn hp hai kim loi Fe, Cu tc dng va vi dung dch HNO3 63%. Sau phn ng thu c dung dch A v 11,2 lt kh NO2 duy nht (ktc). Tnh nng % cc cht c trong dung dch A.A. 36,66% v 28,48%. B. 27,19% v 21,12%.C. 27,19% v 72,81%. D. 78,88% v 21,12%.Hng dn giiFe+6HNO3 Fe(NO3)3+3NO2+3H2OCu+4HNO3 Cu(NO3)2 +2NO2+2H2O2NOn 0,5 mol 3 2HNO NOn 2n 1 mol.5p dng nh lut bo ton khi lng ta c:223NOd HNOm m m m1 63 10012 46 0,5 89 gam.63 + + 2 2d mui h k.loitnFe = x mol, nCu = y mol ta c:56x 64y 123x 2y 0,5+ '+ x 0,1y 0,1'3 3Fe( NO )0,1 242 100%m 27,19%89 3 2Cu( NO )0,1 188 100%m 21,12%.89 (p n B)V d 4:Ho tan hon ton 23,8 gam hn hp mt mui cacbonat ca cc kim loi ho tr (I) v mui cacbonat ca kim loi ho tr (II) trong dung dch HCl. Sau phn ng thu c 4,48 lt kh (ktc). em c cn dung dch thu c bao nhiu gam mui khan?A. 13 gam. B. 15 gam.C. 26 gam. D. 30 gam.Hng dn giiM2CO3+2HCl 2MCl+CO2+H2OR2CO3+2HCl 2MCl2+CO2+H2O2CO4,88n 0, 222, 4 mol Tng nHCl = 0,4 mol v 2H On 0, 2 mol. p dng nh lut bo ton khi lng ta c:23,8 + 0,4 36,5 = mmui + 0,2 44 + 0,2 18 mmui = 26 gam. (p n C)V d 5: Hn hp A gm KClO3, Ca(ClO2)2, Ca(ClO3)2, CaCl2 v KCl nng 83,68 gam. Nhit phn hon ton A ta thu c cht rn B gm CaCl2, KCl v 17,472 lt kh ( ktc). Cho cht rn B tc dng vi 360 ml dung dch K2CO30,5M (va ) thu c kt ta C v dung dch D. Lng KCl 6trong dung dch D nhiu gp 22/3 ln lng KCl c trong A. % khi lng KClO3 c trong A lA. 47,83%. B. 56,72%. C. 54,67%. D. 58,55%.Hng dn giiooo2t3 2t3 2 2 2t2 2 2 22 2( A) (A)h B3KClO KCl O (1)2Ca(ClO ) CaCl 3O (2)83,68 gam A Ca(ClO ) CaCl 2O (3)CaCl CaClKCl KCl + + +'1 2 32On 0,78 mol. p dng nh lut bo ton khi lng ta c: mA = mB + 2Om mB = 83,68 32 0,78 = 58,72 gam.Cho cht rn B tc dng vi 0,18 mol K2CO3Hn hp B 2 2 33( B) ( B)CaCl K CO CaCO 2KCl (4)0,18 0,18 0,36 molKCl KCl + + ' ' hn hp D( B) 2KCl B CaCl ( B)m m m58,72 0,18 111 38,74 gam ( D)KCl KCl (B) KCl ( pt 4)m m m38,74 0,36 74,5 65,56 gam + + ( A) ( D)KCl KCl3 3m m 65,56 8,94 gam22 22 (B) (A)KClpt (1) KCl KClm =m m 38,74 8,94 29,8 gam. Theo phn ng (1):73KClO29,8m 122,5 49 gam.74,5 3KClO ( A)49 100%m 58,55%.83,68 (p n D)V d 6: t chy hon ton 1,88 gam cht hu c A (cha C, H, O) cn 1,904 lt O2 (ktc) thu c CO2 v hi nc theo t l th tch 4:3. Hy xc nh cng thc phn t ca A. Bit t khi ca A so vi khng kh nh hn 7.A. C8H12O5. B. C4H8O2. C. C8H12O3. D. C6H12O6.Hng dn gii1,88 gam A + 0,085 mol O2 4a mol CO2 + 3a mol H2O.p dng nh lut bo ton khi lng ta c:2 2CO H Om m 1,88 0,085 32 46 gam + + Ta c: 44 4a + 18 3a = 46 a = 0,02 mol.Trong cht A c: nC = 4a = 0,08 molnH = 3a 2 = 0,12 molnO = 4a 2 + 3a 0,085 2 = 0,05 mol nC : nH : no=0,08 : 0,12 : 0,05=8 : 12 : 5Vy cng thc ca cht hu c A l C8H12O5 c MA < 203. (p n A)V d 7: Cho 0,1 mol este to bi 2 ln axit v ru mt ln ru tc dng hon tonvi NaOHthuc6,4gamruvmt lngmi ckhi lngnhiuhnlngestel13,56%(sovilngeste). Xcnh cng thc cu to ca este.A. CH3COOCH3.B. CH3OCOCOOCH3.C. CH3COOCOOCH3.D. CH3COOCH2COOCH3.Hng dn giiR(COOR )2+2NaOH R(COONa)2+2R OH8 0,10,2 0,1 0,2 molR OH6, 4M 320, 2 Ru CH3OH.p dng nh lut bo ton khi lng ta c:meste + mNaOH=mmui + mru mmui meste=0,2 40 64=1,6 gam.m mmui meste=13,56100meste meste = 1,6 10011,8 gam13,56 Meste=118 vCR + (44 + 15) 2=118 R = 0.Vy cng thc cu to ca este l CH3OCOCOOCH3. (p n B)V d 8: Thu phn hon ton 11,44 gam hn hp 2 este n chc l ng phn ca nhau bng dung dch NaOH thu c 11,08 gam hn hp mui v 5,56 gam hn hp ru. Xc nh cng thc cu to ca 2 este.A. HCOOCH3 vC2H5COOCH3,B. C2H5COOCH3 v CH3COOC2H5.C. HCOOC3H7 vC2H5COOCH3.D. C B, C u ng.Hng dn giit cngthctrungbnhtngqut cahai estenchcngphnl RCOOR .RCOOR+NaOH RCOONa+ R OH 11,4411,085,56 gamp dng nh lut bo ton khi lng ta c:MNaOH= 11,08 + 5,56 11,44=5,2 gamNaOH5, 2n 0,13 mol40 RCOONa11,08M 85,230,13 R 18,23 9R OH5,56M 42,770,13 R 25,77 RCOOR11, 44M 880,13 CTPT ca este l C4H8O2Vy cng thc cu to 2 este ng phn l:HCOOC3H7 vC2H5COOCH3 hocC2H5COOCH3 v CH3COOC2H5. (p n D)V d 9: Chia hn hp gm hai anehit no n chc lm hai phn bng nhau:- Phn 1: em t chy hon ton thu c 1,08 gam H2O.- Phn 2: Tc dng vi H2 d (Ni, to) th thu c hn hp A. em t chy hon ton th th tch kh CO2 (ktc) thu c lA. 1,434 lt. B. 1,443 lt. C. 1,344 lt. D. 0,672 lt.Hng dn giiPhn 1: V anehit no n chc nn 2 2CO H On n = 0,06 mol.2CO Cn n 0,06(phn2) (phn2) mol.Theo bo ton nguyn t v bo ton khi lng ta c:C C ( A)n n 0,06(phn2) mol.2CO ( A)n= 0,06 mol2COV= 22,4 0,06 = 1,344 lt. (p n C)V d 10: Cho mt lung CO i qua ng s ng 0,04 mol hn hp A gm FeO v Fe2O3 t nng. Sau khi kt thc th nghim thu c B gm 4 cht nng4,784gam. Kh irakhi ngschohpthvodungdch Ba(OH)2dth thuc9,062gamkt ta. Phntrmkhi lng Fe2O3 trong hn hp A lA. 86,96%. B. 16,04%. C. 13,04%. D.6,01%.Hng dn gii0,04 mol hn hp A (FeO v Fe2O3) + CO 4,784 gam hn hp B + CO2.10CO2+Ba(OH)2 d BaCO3 +H2O2 3CO BaCOn n 0,046 mol v2CO( ) COn n 0,046 molp. p dng nh lut bo ton khi lng ta c:mA + mCO=mB + 2COm mA=4,784 + 0,046 44 0,046 28 = 5,52 gam.t nFeO = x mol, 2Fe O3n y mol trong hn hp B ta c:x y 0,0472x 160y 5,52+ '+ x 0,01 moly 0,03 mol' %mFeO= 0,01 72 10113,04%5,52 %Fe2O3 = 86,96%. (p n A)MT S BI TP VN DNG GII THEO PHNG PHP S DNG NH LUT BO TON KHI LNG01. Ha tan 9,14 gam hp kim Cu, Mg, Al bng mt lng va dung dch HCl thu c 7,84 lt kh X (ktc) v 2,54 gam cht rn Y v dung dch Z. Lc b cht rn Y, c cn cn thn dung dch Z thu c lng mui khan lA. 31,45 gam. B. 33,99 gam. C. 19,025 gam. D. 56,3 gam.02. Cho 15 gam hn hp 3 amin n chc, bc mt tc dng va vi dung dch HCl 1,2 M th thu c 18,504 gam mui. Th tch dung dch HCl phi dng l A. 0,8 lt. B. 0,08 lt.C. 0,4 lt.D. 0,04 lt.03. Trn 8,1 gam bt Al vi 48 gam bt Fe2O3 ri cho tin hnh phn ng nhit nhm trong iu kin khng c khng kh, kt thc th nghim lng cht rn thu c lA. 61,5 gam. B. 56,1 gam. C. 65,1 gam. D. 51,6 gam.04. Ha tan hon ton 10,0 gam hn hp X gm hai kim loi (ng trc H trong dy in ha) bng dung dch HCl d thu c 2,24 lt kh H2 (ktc). C cn dung dch sau phn ng thu c lng mui khan l11A. 1,71 gam. B. 17,1 gam. C. 13,55 gam. D. 34,2 gam.05. Nhit phn hon ton m gam hn hp X gm CaCO3 v Na2CO3 thu c 11,6 gam cht rn v 2,24 lt kh (ktc). Hm lng % CaCO3 trong X lA. 6,25%. B. 8,62%. C. 50,2%. D. 62,5%.06. Cho 4,4 gam hn hp hai kim loi nhm IA hai chu k lin tip tc dng vi dung dch HCl d thu c 4,48 lt H2 (ktc) v dung dch cha m gam mui tan. Tn hai kim loi v khi lng m lA. 11 gam; Li v Na. B. 18,6 gam; Li v Na.C. 18,6 gam; Na v K. D. 12,7 gam; Na v K.07. t chy hon ton 18 gam FeS2 v cho ton b lng SO2 vo 2 lt dung dch Ba(OH)2 0,125M. Khi lng mui to thnh lA. 57,40 gam. B. 56,35 gam. C. 59,17 gam. D.58,35 gam.08. Ha tan 33,75 gam mt kim loi M trong dung dch HNO3 long, d thu c 16,8 lt kh X (ktc) gm hai kh khng mu ha nu trong khng kh c t khi hi so vi hiro bng 17,8.a) Kim loi lA. Cu. B. Zn. C. Fe. D. Al. b) Nu dng dung dch HNO3 2M v ly d 25% th th tch dung dch cn ly lA. 3,15 lt. B. 3,00 lt. C. 3,35 lt. D. 3,45 lt.09. Ho tan hon ton 15,9 gam hn hp gm 3 kim loi Al, Mg v Cu bng dung dch HNO3 thu c 6,72 lt kh NO v dung dch X. em c cn dung dch X thu c bao nhiu gam mui khan?A. 77,1 gam. B. 71,7 gam. C. 17,7 gam. D. 53,1 gam.10. Ha tan hon ton 2,81 gam hn hp gm Fe2O3, MgO, ZnO trong 500 ml axit H2SO4 0,1M (va ). Sau phn ng, hn hp mui sunfat khan thu c khi c cn dung dch c khi lng lA. 6,81 gam.B. 4,81 gam.C. 3,81 gam.D. 5,81 gam.p n cc bi tp vn dng:1. A 2. B 3. B 4. B 5. D6. B 7. D 8. a-D, b-B 9. B 10. APhng php 2BO TON MOL NGUYN T12C rt nhiu phng php gii ton ha hc khc nhau nhng phng php bo ton nguyn t v phng php bo ton s mol electron cho php chng ta gpnhiuphngtrnhphnngli lmmt,quign vic tnhtonv nhm nhanh p s. Rt ph hp vi vic gii cc dng bi ton ha hc trc nghim. Cch thc gp nhng phng trnh lm mt v cch lp phng trnh theo phng php bo ton nguyn t s c gii thiu trong mt s v d sau y.V d 1: kh hon ton 3,04 gam hn hp X gm FeO, Fe3O4, Fe2O3 cn 0,05 mol H2. Mt khc ha tan hon ton 3,04 gam hn hp X trong dung dch H2SO4c thu c th tch kh SO2(sn phm kh duy nht) iu kin tiu chun lA. 448 ml. B. 224 ml. C. 336 ml. D. 112 ml.Hng dn giiThc cht phn ng kh cc oxit trn lH2+O H2O0,05 0,05 molt s mol hn hp X gm FeO, Fe3O4, Fe2O3 ln lt l x, y, z. Ta c:nO=x + 4y + 3z=0,05 mol (1)Fe3,04 0,05 16n 0,04 mol56 x + 3y + 2z= 0,04 mol (2)Nhn hai v ca (2) vi 3 ri tr (1) ta c:x + y=0,02 mol.Mt khc:2FeO + 4H2SO4 Fe2(SO4)3 + SO2 + 4H2Ox x/22Fe3O4 + 10H2SO4 3Fe2(SO4)3 + SO2 + 10H2O y y/2tng:SO2x y 0,2n 0,01 mol2 2+ Vy:2SOV 224 ml. (p n B)13V d 2: Thi t t V lt hn hp kh (ktc) gm CO v H2 i qua mt ng ng 16,8 gam hn hp 3 oxit: CuO, Fe3O4, Al2O3 nung nng, phn ng hon ton. Sau phn ng thu c m gam cht rn v mt hn hp kh v hi nng hn khi lng ca hn hp V l 0,32 gam. Tnh V v m.A. 0,224 lt v 14,48 gam. B. 0,448 lt v 18,46 gam.C. 0,112 lt v 12,28 gam.D. 0,448 lt v 16,48 gam.Hng dn giiThc cht phn ng kh cc oxit trn lCO+O CO2 H2+O H2O.Khi lng hn hp kh to thnh nng hn hn hp kh ban u chnh l khi lng ca nguyn t Oxi trong cc oxit tham gia phn ng. Do vy:mO = 0,32 gam.O0,32n 0,02 mol16 ( )2CO Hn n 0,02 mol + .p dng nh lut bo ton khi lng ta c:moxit = mcht rn + 0,32 16,8 = m + 0,32 m = 16,48 gam.2hh (CO H )V 0,02 22,4 0,448+ lt. (p n D)V d 3: Thi rt chm 2,24 lt (ktc) mt hn hp kh gm CO v H2qua mt ng s ng hn hp Al2O3, CuO, Fe3O4, Fe2O3 c khi lng l 24 gam d ang c un nng. Sau khi kt thc phn ng khi lng cht rn cn li trong ng s lA. 22,4 gam. B. 11,2 gam. C. 20,8 gam. D. 16,8 gam.Hng dn gii2hh(CO H )2, 24n 0,1 mol22, 4+ 14Thc cht phn ng kh cc oxit l:CO+O CO2H2+O H2O.Vy:2O CO Hn n n 0,1 mol + . mO = 1,6 gam.Khi lng cht rn cn li trong ng s l: 24 1,6 = 22,4 gam. (p n A)V d 4: Cho m gam mt ancol (ru) no, n chc X qua bnh ng CuO (d), nung nng. Sau khi phn ng hon ton, khi lng cht rn trong bnh gim 0,32 gam. Hn hp hi thu c c t khi i vi hiro l 15,5. Gi tr ca m lA. 0,92 gam. B. 0,32 gam. C. 0,64 gam. D. 0,46 gam.Hng dn giiCnH2n+1CH2OH + CuO ot CnH2n+1CHO + Cu + H2OKhi lng cht rn trong bnh gim chnh l s gam nguyn t O trong CuO phn ng. Do nhn c:mO = 0,32 gam O0,32n 0,02 mol16 Hn hp hi gm:n 2n 12C H CHO : 0,02 molH O : 0,02 mol.+'Vy hn hp hi c tng s mol l 0,04 mol.CM= 31 mhh hi = 31 0,04 = 1,24 gam.mancol + 0,32=mhh himancol=1,24 0,32=0,92 gam. (p n A)Ch : Vi ru bc (I) hoc ru bc (II) u tha mn u bi.V d 5: t chy hon ton 4,04 gam mt hn hp bt kim loi gm Al, Fe, Cu trong khng kh thu c 5,96 gam hn hp 3 oxit. Ha tan ht hn hp 3 oxit bng dung dch HCl 2M. Tnh th tch dung dch HCl cn dng.15A. 0,5 lt. B. 0,7 lt. C. 0,12 lt. D. 1 lt.Hng dn giimO=moxit mkl=5,96 4,04=1,92 gam.O1,92n 0,12 mol16 .Ha tan ht hn hp ba oxit bng dung dch HCl to thnh H2O nh sau:2H+ +O2 H2O 0,24 0,12 molHCl0, 24V 0,122 lt. (p n C)V d 6: t chy hon ton 0,1 mol mt axit cacbonxylic n chc cn va V lt O2 ( ktc), thu c 0,3 mol CO2 v 0,2 mol H2O. Gi tr ca V lA. 8,96 lt. B. 11,2 lt. C. 6,72 lt. D. 4,48 lt.Hng dn giiAxit cacbonxylic n chc c 2 nguyn t Oxi nn c th t l RO2. Vy:2 2 2 2O ( RO ) O (CO ) O (CO ) O ( H O)n n n n + +0,1 2 + nO (p.)=0,3 2 + 0,2 1 nO (p.)=0,6 mol2On 0,3mol 2OV 6,72 lt. (p n C)V d 7: (Cu 46 - M 231 - TSC Khi A 2007)Cho 4,48 lt CO ( ktc) t t i qua ng s nung nng ng 8 gam mt oxit st n khi phn ng xy ra hon ton. Kh thu c sau phn ng c t khi so vi hiro bng 20. Cngthc ca oxit st v phn trm th tch ca kh CO2 trong hn hp kh sau phn ng lA. FeO; 75%. B. Fe2O3; 75%.C. Fe2O3; 65%. D. Fe3O4; 65%.Hng dn giiFexOy+yCO xFe + yCO216Kh thu c cM 40 gm 2 kh CO2 v CO d2COCOn3n 1 2CO%V 75% .Mt khc:2CO ( ) CO75n n 0,2 0,15100p. mol nCO d = 0,05 mol.Thc cht phn ng kh oxit st l doCO+O (trong oxit st) CO2 nCO = nO = 0,15 mol mO = 0,15 16 = 2,4 gam mFe = 8 2,4 = 5,6 gam nFe = 0,1 mol.Theo phng trnh phn ng ta c:2FeCOn x 0,1 2n y 0,15 3 Fe2O3. (p n B)V d 8:Cho hn hp A gm Al, Zn, Mg. em oxi ho hon ton 28,6 gam A bngoxidthu c 44,6 gam hn hp oxit B. Ho tan ht B trong dung dch HCl thu c dung dch D. C cn dung dch D c hn hp mui khan lA. 99,6 gam. B. 49,8 gam.C. 74,7 gam.D. 100,8 gam.Hng dn giiGi M l kim loi i din cho ba kim loi trn vi ho tr l n.M+n2O2 M2On(1)M2On+2nHCl 2MCln+nH2O (2)Theo phng trnh (1) (2)2HCl On 4.n .p dng nh lut bo ton khi lng 2Om 44,6 28,6 16 gam172COCOn 44 1240n 28 42On 0,5 mol nHCl = 4 0,5 = 2 molCln 2 mol mmui = mhhkl + Clm = 28,6 + 2 35,5 = 99,6 gam. (p n A)V d 9: Cho mt lung kh CO i qua ng ng 0,01 mol FeO v 0,03 mol Fe2O3 (hn hp A) t nng. Sau khi kt thc th nghim thu c 4,784 gam cht rn B gm 4 cht. Ho tan cht rn B bng dung dch HCl d thy thot ra 0,6272 lt H2 ( ktc). Tnh s mol oxit st t trong hn hp B. Bit rng trong B s mol oxit st t bng 1/3 tng s mol st (II) oxit v st (III) oxit.A. 0,006. B. 0,008. C. 0,01. D. 0,012.Hng dn giiHnhpA 2 3FeO : 0,01 molFe O : 0,03 mol'+ CO 4,784gamB (Fe,Fe2O3,FeO, Fe3O4) tng ng vi s mol l: a, b, c, d (mol).Ho tan B bng dung dch HCl d thu c 2Hn 0,028 mol.Fe+2HCl FeCl2 +H2 a = 0,028 mol.(1)Theo u bi:( )3 4 2 3Fe O FeO Fe O1n n n3 + ( )1d b c3 + (2)Tng mB l:(56.a + 160.b + 72.c + 232.d) = 4,78 gam. (3)S mol nguyn t Fe trong hn hp A bng s mol nguyn t Fe trong hn hp B. Ta c: nFe (A) = 0,01 + 0,03 2 = 0,07 mol nFe (B) = a + 2b + c + 3d a + 2b + c + 3d = 0,07 (4)T (1, 2, 3, 4) b = 0,006 molc = 0,012 mold = 0,006 mol. (p n A)18V d 10: Kh hon ton 24 gam hn hp CuO v FexOy bng H2 d nhit cao thu c 17,6 gam hn hp 2 kim loi. Khi lng H2O to thnh lA. 1,8 gam. B. 5,4 gam. C. 7,2 gam.D. 3,6 gam.Hng dn giimO (trong oxit)=moxit mkloi = 24 17,6 = 6,4 gam. ( )2O H Om 6,4 gam ;2H O6, 4n 0, 416 mol.2H Om 0, 4 18 7, 2 gam. (p n C)V d 11: Kh ht m gam Fe3O4 bng CO thu c hn hp A gm FeO v Fe. A tan va trong 0,3 lt dung dch H2SO4 1M cho ra 4,48 lt kh (ktc). Tnh m?A. 23,2 gam. B. 46,4 gam. C. 11,2 gam. D. 16,04 gam.Hng dn giiFe3O4 (FeO, Fe) 3Fe2+ n mol( )244Fe trongFeSOSOn n 0,3 molp dng nh lut bo ton nguyn t Fe:( ) ( )4 3 4Fe FeSO Fe Fe On n 3n = 0,3 n = 0,13 4Fe Om 23, 2 gam (p n A)V d 12: un hai ru n chc vi H2SO4 c, 140oC c hn hp ba ete. Ly 0,72 gam mt trong ba ete em t chy hon ton thu c 1,76 gam CO2 v 0,72 gam H2O. Hai ru l A. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.C. C2H5OH v C4H9OH.D. CH3OH v C3H5OH.Hng dn giit cng thc tng qut ca mt trong ba ete l CxHyO, ta c: 19C0,72m 12 0,4844 gam ;H0,72m 2 0,0818 gam mO = 0,72 0,48 0,08 = 0,16 gam.0, 48 0,08 0,16x : y :1 : :12 1 16=4 : 8 : 1.Cng thc phn t ca mt trong ba ete l C4H8O. Cng thc cu to l CH3OCH2CH=CH2.Vy hai ancol l CH3OH v CH2=CHCH2OH. (p n D)MT S BI TP VN DNG GII THEO PHNG PHP BO TON MOL NGUYN T01.Ha tan hon ton hn hp X gm 0,4 mol FeO v 0,1mol Fe2O3vo dung dch HNO3long, d thu c dung dch A v kh B khng mu, ha nu trong khng kh. Dung dch A cho tc dng vi dung dch NaOH d thu c kt ta. Ly ton b kt ta nung trong khng kh n khi lng khng i thu c cht rn c khi lng lA. 23,0 gam. B. 32,0 gam. C. 16,0 gam. D. 48,0 gam.02.Cho kh CO i qua ng s cha 16 gam Fe2O3un nng, sau phn ng thu c hn hp rn X gm Fe, FeO, Fe3O4, Fe2O3. Ha tan hon ton X bng H2SO4c, nngthuc dung dch Y. C cn dung dch Y, lng mui khan thu c lA. 20 gam. B. 32 gam. C. 40 gam. D. 48 gam.03. Kh hon ton 17,6 gam hn hp X gm Fe, FeO, Fe2O3 cn 2,24 lt CO ( ktc). Khi lng st thu c lA. 5,6 gam. B. 6,72 gam. C. 16,0 gam. D. 11,2 gam.04.tchy hn hphirocacbon X thu c 2,24 lt CO2(ktc) v 2,7 gam H2O. Th tch O2 tham gia phn ng chy (ktc) lA. 5,6 lt. B. 2,8 lt. C. 4,48 lt. D. 3,92 lt.05. Ho tan hon ton a gam hn hp X gm Fe v Fe2O3 trong dung dch HCl thu c 2,24 lt kh H2 ktc v dung dch B. Cho dung dch B tc dng dung 20dch NaOH d, lc ly kt ta, nung trong khng kh n khi lng khng i thu c 24 gam cht rn. Gi tr ca a lA. 13,6 gam. B. 17,6 gam. C. 21,6 gam. D. 29,6 gam.06. Hn hp X gm Mg v Al2O3. Cho 3 gam X tc dng vi dung dch HCl d gii phng V lt kh (ktc). Dung dch thu c cho tc dng vi dung dch NH3 d, lc v nung kt ta c 4,12 gam bt oxit. V c gi tr l:A. 1,12 lt. B. 1,344 lt. C. 1,568 lt. D. 2,016 lt.07. Hn hp A gm Mg, Al, Fe, Zn. Cho 2 gam A tc dng vi dung dch HCl d gii phng 0,1 gam kh. Cho 2 gam A tc dng vi kh clo d thu c 5,763 gam hn hp mui. Phn trm khi lng ca Fe trong A lA. 8,4%. B. 16,8%. C. 19,2%. D. 22,4%.08. (Cu 2 - M 231 - TSC - Khi A 2007)tchyhontonmtthtchkhthinnhingmmetan, etan, propan bng oxi khng kh (trong khng kh Oxi chim 20% th tch), thu c 7,84 lt kh CO2 (ktc) v 9,9 gam H2O. Th tch khng kh (ktc) nh nht cn dng t chy hon ton lng kh thin nhin trn lA. 70,0 lt. B. 78,4 lt. C. 84,0 lt. D. 56,0 lt.09.Ho tan hon ton 5 gam hn hp 2 kim loi X v Y bng dung dch HCl thu c dung dch A v kh H2. C cn dung dch A thu c 5,71 gam mui khan. Hy tnh th tch kh H2 thu c ktc.A. 0,56 lt. B. 0,112 lt.C. 0,224 lt D. 0,448 lt10. t chy hon ton m gam hn hp Y gm C2H6, C3H4 v C4H8 th thu c 12,98 gam CO2 v 5,76 gam H2O. Vy m c gi tr lA. 1,48 gam.B. 8,14 gam. C. 4,18 gam.D. 16,04 gam.p n cc bi tp vn dng:1. D 2. C 3. C 4. D 5. C6. C 7. B 8. A 9. C 10. C21Phng php 3BO TON MOL ELECTRONTrc ht cn nhn mnh y khng phi l phng php cn bng phn ng oxi ha - kh, mc d phng php thng bng electron dng cn bng phn ng oxi ha - kh cng da trn s bo ton electron.Nguyn tc ca phng php nh sau: khi c nhiu cht oxi ha, cht kh trong mt hn hp phn ng (nhiu phn ng hoc phn ng qua nhiu giai on) th tng s electron ca cc cht kh cho phi bng tng s electron m cc cht oxi ha nhn. Ta ch cn nhn nh ng trng thi u v trng thi cui ca cc cht oxi ha hoc cht kh, thm ch khng cn quan tm n vic cn bng cc phng trnh phn ng. Phng php ny c bit l th i vi cc bi ton cn phi bin lun nhiu trng hp c th xy ra.Sau y l mt s v d in hnh.V d 1: Oxi ha hon ton 0,728 gam bt Fe ta thu c 1,016 gam hn hp hai oxit st (hn hp A).1. Ha tan hn hp A bng dung dch axit nitric long d. Tnh th tch kh NO duy nht bay ra ( ktc).A. 2,24 ml. B. 22,4 ml.C. 33,6 ml. D. 44,8 ml.2. Cng hn hp A trn trn vi 5,4 gam bt Al ri tin hnh phn ng nhit nhm (hiu sut 100%). Ha tan hn hp thu c sau phn ng bng dung dch HCl d. Tnh th tch bay ra ( ktc).A. 6,608 lt. B. 0,6608 lt.C. 3,304 lt. D. 33,04. ltHng dn gii1. Cc phn ng c th c:2Fe + O2

ot2FeO (1)2Fe + 1,5O2

otFe2O3(2)223Fe + 2O2

otFe3O4(3)Cc phn ng ha tan c th c:3FeO + 10HNO3 3Fe(NO3)3 + NO + 5H2O (4)Fe2O3 + 6HNO3 2Fe(NO3)3 + 3H2O (5)3Fe3O4 + 28HNO3 9Fe(NO3)3 + NO + 14H2O (6)Ta nhn thy tt c Fe t Fe0 b oxi ha thnh Fe+3, cn N+5 b kh thnh N+2, O20 b kh thnh 2O2 nn phng trnh bo ton electron l:0,7283n 0,009 4 3 0,03956+ mol.trong , n l s mol NO thot ra. Ta d dng rt ran = 0,001 mol;VNO = 0,001 22,4 = 0,0224 lt = 22,4 ml. (p n B)2. Cc phn ng c th c:2Al + 3FeOot3Fe + Al2O3(7)2Al + Fe2O3

ot2Fe + Al2O3(8)8Al + 3Fe3O4

ot9Fe + 4Al2O3(9)Fe + 2HCl FeCl2 + H2(10)2Al + 6HCl 2AlCl3 + 3H2(11)Xt cc phn ng (1, 2, 3, 7, 8, 9, 10, 11) ta thy Fe0cui cng thnh Fe+2, Al0thnh Al+3, O20thnh 2O2v 2H+thnh H2nn ta c phng trnh bo ton electron nh sau:5, 4 30,013 2 0,009 4 n 227 + + Fe0 Fe+2 Al0 Al+3O20 2O2 2H+ H2 n = 0,295 mol2HV 0, 295 22, 4 6,608 lt. (p n A)23Nhn xt: Trong bi ton trn cc bn khng cn phi bn khon l to thnh hai oxit st (hn hp A) gm nhng oxit no v cng khng cn phi cn bng 11 phng trnh nh trn m ch cn quan tm ti trng thi u v trng thi cui ca cc cht oxi ha v cht kh ri p dng lut bo ton electron tnh lc bt c cc giai on trung gian ta s tnh nhm nhanh c bi ton.V d 2: Trn 0,81 gam bt nhm vi bt Fe2O3 v CuO ri t nng tin hnh phn ng nhit nhm thu c hn hp A. Ho tan hon ton A trong dung dch HNO3un nng thu c V lt kh NO (sn phm kh duy nht) ktc. Gi tr ca V lA. 0,224 lt. B. 0,672 lt. C. 2,24 lt. D. 6,72 lt.Hng dn giiTm tt theo s :o2 3 tNOFe O0,81 gam Al V ?CuO3hatan hon tondung dch HNOhnh pA+ 'Thccht trongbi tonnych cqutrnhchovnhnelectronca nguyn t Al v N.Al Al+3 + 3e 0,8127 0,09 molv N+5 +3eN+2 0,09 mol 0,03 mol VNO = 0,03 22,4 = 0,672 lt. (p n D)Nhn xt: Phn ng nhit nhm cha bit l hon ton hay khng hon ton do hn hp A khng xc nh c chnh xc gm nhng cht no nn vic vit phng trnh ha hc v cn bng phng trnh phc tp. Khi ha tan hon ton hn hp A trong axit HNO3 th Al0 to thnh Al+3, nguyn t Fe v Cu c bo ton ha tr.C bn s thc mc lng kh NO cn c to bi kim loi Fe v Cu trong hn hp A. Thc cht lng Al phn ng b li lng Fe v Cu to thnh.V d 3: Cho 8,3 gam hn hp X gm Al, Fe (nAl = nFe) vo 100 ml dung dch Y gm Cu(NO3)2 v AgNO3. Sau khi phn ng kt thc thu c cht rn 24A gm 3 kim loi. Ha tan hon ton cht rn A vo dung dch HCl d thy c 1,12 lt kh thot ra (ktc) v cn li 28 gam cht rn khng tan B. Nng CM ca Cu(NO3)2 v ca AgNO3 ln lt lA. 2M v 1M. B. 1M v 2M.C. 0,2M v 0,1M. D. kt qu khc.Tm tt s :Al Fe8,3 gam hn h p X(n =n )AlFe' + 100 ml dung dch Y 33 2AgNO : x molCu(NO ) :y mol' Cht rn A(3 kim loi) 2HCl d1,12lt H2,8gamcht rn khngtanB+Z]Hng dn giiTa c: nAl = nFe = 8,30,1mol.83 t 3AgNOn x mol v 3 2Cu( NO )n y mol X + Y Cht rn A gm 3 kim loi.Al ht, Fe cha phn ng hoc cn d. Hn hp hai mui ht. Qu trnh oxi ha:Al Al3+ + 3e Fe Fe2+ + 2e0,10,3 0,10,2Tng s mol e nhng bng 0,5 mol.Qu trnh kh:Ag+ + 1e Ag Cu2+ + 2e Cu 2H+ +2e H2 x xx y 2y y 0,1 0,05Tng s e mol nhn bng (x + 2y + 0,1).Theo nh lut bo ton electron, ta c phng trnh:x + 2y + 0,1=0,5 hay x + 2y=0,4 (1)Mt khc, cht rn B khng tan l:Ag: x mol ;Cu: y mol.25 108x + 64y =28 (2)Gii h (1), (2) ta c:x = 0,2 mol ;y = 0,1 mol.3MAgNO0, 2C0,1 = 2M;3 2MCu( NO )0,1C0,1 = 1M. (p n B)V d 4: Ha tan 15 gam hn hp X gm hai kim loi Mg v Al vo dung dch Y gm HNO3 v H2SO4 c thu c 0,1 mol mi kh SO2, NO, NO2, N2O. Phn trm khi lng ca Al v Mg trong X ln lt lA. 63% v 37%. B. 36% v 64%.C. 50% v 50%. D. 46% v 54%.Hng dn giit nMg = x mol ;nAl = y mol. Ta c: 24x + 27y = 15.(1)Qu trnh oxi ha:Mg Mg2+ + 2e AlAl3+ + 3ex2x y3yTng s mol e nhng bng (2x + 3y).Qu trnh kh:N+5 + 3e N+22N+5 + 24e 2N+10,3 0,1 0,80,2N+5 + 1e N+4S+6 + 2e S+40,1 0,10,2 0,1Tng s mol e nhn bng 1,4 mol.Theo nh lut bo ton electron:2x + 3y=1,4 (2)Gii h (1), (2) ta c: x = 0,4 mol ;y = 0,2 mol.27 0,2%Al 100% 36%.15 26%Mg = 100% 36% = 64%. (p n B)V d 5: Trn 60 gam bt Fe vi 30 gam bt lu hunh ri un nng (khng c khng kh) thu c cht rn A. Ho tan A bng dung dch axit HCl d c dung dch B v kh C. t chy C cn V lt O2(ktc). Bit cc phn ng xy ra hon ton. V c gi tr lA. 11,2 lt. B. 21 lt. C. 33 lt. D. 49 lt.Hng dn giiVFe S30n n32> nn Fe d v S ht.Kh C l hn hp H2S v H2. t C thu c SO2 v H2O. Kt qu cui cng ca qu trnh phn ng l Fe v S nhng e, cn O2 thu e.Nhng e: Fe Fe2+ + 2e 60mol56 60256 molS S+4 +4e

30mol32 30432molThu e: Gi s mol O2 l x mol.O2 +4e 2O-2 x mol 4xTa c: 60 304x 2 456 32 + gii ra x = 1,4732 mol.2OV 22, 4 1, 4732 33 lt. (p n C)V d 6: Hn hp A gm 2 kim loi R1, R2 c ho tr x, y khng i (R1, R2 khng tc dng vi nc v ng trc Cu trong dy hot ng ha hc ca kim loi). Cho hn hp A phn ng hon ton vi dung dch HNO3 d thu c 1,12 lt kh NO duy nht ktc.Nu cho lng hn hp A trn phn ng hon ton vi dung dch HNO3 th thu c bao nhiu lt N2. Cc th tch kh o ktc.A. 0,224 lt. B. 0,336 lt. C. 0,448 lt. D. 0,672 lt.27Hng dn giiTrong bi ton ny c 2 th nghim:TN1: R1 v R2 nhng e cho Cu2+ chuyn thnh Cu sau Cu li nhng e cho 5N+ thnh 2N+(NO). S mol e do R1 v R2 nhng ra l5N+ + 3e 2N+ 0,15 05 , 04 , 2212 , 1 TN2: R1 v R2 trc tip nhng e cho 5N+ to ra N2. Gi x l s mol N2, th s mol e thu vo l25N+ + 10e02N10x x molTa c:10x = 0,15 x = 0,0152NV= 22,4.0,015 = 0,336 lt. (p n B)Vd 7:Cho1,35 gam hn hp gm Cu, Mg, Al tc dng ht vi dung dch HNO3 thu c hn hp kh gm 0,01 mol NO v 0,04 mol NO2. Tnh khi lng mui to ra trong dung dch.A. 10,08 gam. B. 6,59 gam. C. 5,69 gam. D. 5,96 gam.Hng dn giiCch 1: t x, y, z ln lt l s mol Cu, Mg, Al.Nhng e:Cu = 2Cu+ + 2e Mg = 2Mg+ + 2e Al = 3Al+ + 3e x x 2xy y 2y zz 3zThu e: 5N+ + 3e =2N+ (NO)5N+ + 1e =4N+ (NO2)0,03 0,010,04 0,04Ta c:2x + 2y + 3z = 0,03 + 0,04 = 0,07v 0,07 cng chnh l s mol NO3 Khi lng mui nitrat l: 1,35 + 62 0,07 = 5,69 gam. (p n C)28Cch 2: Nhn nh mi: Khi cho kim loi hoc hn hp kim loi tc dng vi dung dch axit HNO3 to hn hp 2 kh NO v NO2 th3 2HNO NO NOn 2n 4n +3HNOn 2 0,04 4 0,01 0,12 + mol2H On 0,06 molp dng nh lut bo ton khi lng:3 2 2KL HNO mui NO NO H Om m m m m m + + + +1,35 + 0,12 63 = mmui + 0,01 30 + 0,04 46 + 0,06 18 mmui=5,69 gam.V d 8: (Cu 19 - M 182 - Khi A - TSH - 2007)Hatanhonton12gamhnhpFe, Cu(tlmol 1:1)bngaxit HNO3, thu c V lt ( ktc) hn hp kh X (gm NO v NO2) v dung dch Y (ch cha hai mui v axit d). T khi ca X i vi H2 bng 19. Gi tr ca V lA. 2,24 lt.B. 4,48 lt.C. 5,60 lt.D. 3,36 lt.Hng dn giit nFe = nCu = a mol 56a + 64a = 12 a = 0,1 mol.Cho e: Fe Fe3++3e CuCu2+ +2e0,10,3 0,10,2Nhn e: N+5+3e N+2N+5+1e N+43x xy yTng ne cho bng tng ne nhn. 3x + y=0,5Mt khc: 30x + 46y=19 2(x + y). x = 0,125 ;y = 0,125.Vhh kh (ktc)=0,125 2 22,4=5,6 lt. (p n C)29V d 9: Nung m gam bt st trong oxi, thu c 3 gam hn hp cht rn X. Ha tan ht hn hp X trong dung dch HNO3 (d), thot ra 0,56 lt ( ktc) NO (l sn phm kh duy nht). Gi tr ca m lA. 2,52 gam.B. 2,22 gam.C. 2,62 gam.D. 2,32 gam.Hng dn giim gam Fe + O2 3 gam hn hp cht rn X 3HNO d 0,56 lt NO.Thc cht cc qu trnh oxi ha - kh trn l:Cho e: Fe Fe3+ + 3e m56 3m56mol eNhn e: O2+4e 2O2N+5 +3eN+23 m32 4(3 m)32mol e 0,075 mol 0,025 mol3m56 = 4(3 m)32 + 0,075 m = 2,52 gam. (p n A)V d 10: Hn hp X gm hai kim loi A v B ng trc H trong dy in ha v c ha tr khng i trong cc hp cht. Chia m gam X thnh hai phn bng nhau:-Phn1: Hatanhontontrongdungdchchaaxit HClvH2SO4 long to ra 3,36 lt kh H2.- Phn 2: Tc dng hon ton vi dung dch HNO3 thu c V ltkh NO (sn phm kh duy nht).Bit cc th tch kh o iu kin tiu chun. Gi tr ca V lA. 2,24 lt. B. 3,36 lt. C. 4,48 lt. D. 6,72 lt.Hng dn giit hai kim loi A, B l M.- Phn 1: M+nH+ Mn++2nH2(1)- Phn 2: 3M+4nH++nNO3 3Mn+ +nNO+2nH2O (2)30Theo (1): S mol e ca M cho bng s mol e ca 2H+ nhn;Theo (2): S mol e ca M cho bng s mol e ca N+5 nhn.Vy s mol e nhn ca 2H+ bng s mol e nhn ca N+5.2H++2e H2 v N+5+ 3e N+20,30,15 mol0,30,1 mol VNO = 0,1 22,4 = 2,24 lt. (p n A)V d 11: Cho m gam bt Fe vo dung dch HNO3 ly d, ta c hn hp gm hai kh NO2v NO c VX= 8,96 lt (ktc) v t khi i vi O2bng 1,3125. Xc nh %NO v %NO2 theo th tch trong hn hp X v khi lng m ca Fe dng?A. 25% v 75%; 1,12 gam. B. 25% v 75%; 11,2 gam.C. 35% v 65%; 11,2 gam. D. 45% v 55%; 1,12 gam.Hng dn giiTa c: nX = 0,4 mol;MX = 42.S ng cho:22NO NONO NOn : n 12 : 4 3n n 0, 4 mol '+ 2NONOn 0,1 moln 0,3 mol' 2NONO%V 25%%V 75%'vFe 3e Fe3+ N+5+ 3e N+2N+5+1e N+43x x 0,3 0,1 0,3 0,3Theo nh lut bo ton electron:3x = 0,6 mol x = 0,2 molmFe = 0,2 56 = 11,2 gam. (p p B).312NO : 46 42 30 1242NO : 30 46 42 4 V d 12: Cho 3 kim loi Al, Fe, Cu vo 2 lt dung dch HNO3 phn ng va thu c 1,792 lt kh X (ktc) gm N2 v NO2 c t khi hi so vi He bng 9,25. Nng mol/lt HNO3 trong dung dch u lA. 0,28M. B. 1,4M. C. 1,7M. D. 1,2M.Hng dn giiTa c:( )2 2N NOXM MM 9, 25 4 372+ l trung bnh cng khi lng phn t ca hai kh N2 v NO2 nn:2 2XN NOnn n 0,04 mol2 vNO3+10e N2 NO3 +1e NO20,08 0,4 0,04 mol0,040,04 0,04 molM Mn++n.e 0,04 mol3HNO (bkh)n 0,12mol. Nhn nh mi: Kim loi nhng bao nhiu electron th cng nhn by nhiu gc NO3 to mui.3HNO ( ) ( ) ( )n n.e n.e 0,04 0,4 0,44 mol.to mui nh ng nhn + Do :3HNO ( )n 0, 44 0,12 0,56 molphn ng + [ ]30,56HNO 0, 28M.2 (p n A)V d 13: Khi cho 9,6 gam Mg tc dng ht vi dung dch H2SO4 m c, thy c 49 gam H2SO4tham gia phn ng, to mui MgSO4, H2O v sn phm khX. X lA. SO2B. S C. H2SD. SO2, H2S Hng dn giiDung dch H2SO4 m c va l cht oxi ha va l mi trng.Gi a l s oxi ha ca S trong X. 32Mg Mg2++ 2e S+6 + (6-a)e S a0,4 mol0,8 mol0,1 mol0,1(6-a) molTng s mol H2SO4 dng l : 490,598 (mol)S mol H2SO4 dng to mui bng s mol Mg = 9,6 : 24 = 0,4 mol.S mol H2SO4 dng oxi ha Mg l:0,5 0,4 = 0,1 mol.Ta c:0,1 (6 a) = 0,8 x = 2. Vy X l H2S. (p n C)V d 14: a gam bt st ngoi khng kh, sau mt thi gian s chuyn thnh hn hp A c khi lng l 75,2 gam gm Fe, FeO, Fe2O3v Fe3O4. Cho hn hp A phn ng ht vi dung dch H2SO4 m c, nng thu c 6,72 lt kh SO2 (ktc). Khi lng a gam l: A. 56 gam. B. 11,2 gam. C. 22,4 gam. D. 25,3 gam.Hng dn giiS mol Fe ban u trong a gam: Fean56 mol.S mol O2 tham gia phn ng: 2O75,2 an32 mol.Qu trnh oxi ha:3Fe Fe 3ea 3amol mol56 56+ +(1)S mol e nhng: e3an mol56Qu trnh kh: O2+4e2O2(2)SO42 + 4H+ + 2e SO2 + 2H2O (3)T (2), (3) cho 2 2e O SOn 4n 2n +

75, 2 a 3a4 2 0,332 56 + a = 56 gam. (p n A)33V d 15: Cho 1,35 gam hn hp A gm Cu, Mg, Al tc dng vi HNO3 d c 1,12 lt NO v NO2 (ktc) c khi lng mol trung bnh l 42,8. Tng khi lng mui nitrat sinh ra l:A. 9,65 gam B. 7,28 gam C. 4,24 gam D. 5,69 gam Hng dn giiDa vo s ng cho tnh c s mol NO v NO2 ln lt l 0,01 v 0,04 mol. Ta c cc bn phn ng:NO3 + 4H+ + 3e NO + 2H2ONO3 + 2H+ + 1e NO2 + H2ONh vy, tng electron nhn l 0,07 mol.Gi x, y, z ln lt l s mol Cu, Mg, Al c trong 1,35 gam hn hp kim loi. Ta c cc bn phn ng: Cu Cu2++ 2e Mg Mg2++ 2eAl Al3++ 3e 2x + 2y + 3z = 0,07.Khi lng mui nitrat sinh ra l:m= 3 2Cu( NO )m+ 3 2Mg( NO )m+ 3 3Al( NO )m = 1,35 + 62(2x + 2y + 3z) = 1,35 + 62 0,07 = 5,69 gam.MT S BI TP VN DNG GIAI THEO PHNG PHP BO TOM MOL ELECTRON01. Ho tan hon ton m gam Al vo dung dch HNO3 rt long th thu c hn hpgm0,015mol kh N2Ov 0,01mol kh NO(phnngkhngto NH4NO3). Gi tr ca m lA. 13,5 gam. B. 1,35 gam. C.0,81 gam. D. 8,1 gam.02.Chomt lung CO iqua ng s ng 0,04 mol hn hp A gm FeO v Fe2O3t nng. Sau khi kt thc th nghim thu c cht rn B gm 4 cht nng 4,784 gam. Kh i ra khi ng s hp th vo dung dch Ca(OH)2 d, th thu c 4,6 gam kt ta. Phn trm khi lng FeO trong hn hp A l A. 68,03%. B. 13,03%. C. 31,03%. D. 68,97%.3403. Mt hn hp gm hai bt kim loi Mg v Al c chia thnh hai phn bng nhau:- Phn 1: cho tc dng vi HCl d thu c 3,36 lt H2. - Phn 2: ho tan ht trong HNO3 long d thu c V lt mt kh khng mu, ho nu trong khng kh (cc th tch kh u o ktc). Gi tr ca V lA. 2,24 lt. B. 3,36 lt. C. 4,48 lt. D. 5,6 lt.04.Dung dch X gm AgNO3v Cu(NO3)2 c cng nng . Ly mt lng hn hp gm 0,03 mol Al; 0,05 mol Fe cho vo 100 ml dung dch X cho ti kh phn ng kt thc thu c cht rn Y cha 3 kim loi.Cho Y vo HCl d gii phng 0,07 gam kh. Nng ca hai mui l A. 0,3M. B. 0,4M. C. 0,42M. D. 0,45M.05. Cho 1,35 gam hn hp Cu, Mg, Al tc dng vi HNO3 d c 896 ml hn hp gm NO v NO2 c M 42 . Tnh tng khi lng mui nitrat sinh ra (kh ktc).A. 9,41 gam. B. 10,08 gam.C. 5,07 gam. D. 8,15 gam.06.Ha tan ht 4,43 gam hn hp Al v Mg trong HNO3long thu c dung dch A v 1,568 lt (ktc) hn hp hai kh (u khng mu) c khi lng 2,59 gam trong c mt kh b ha thnh mu nu trong khng kh. Tnh s mol HNO3 phn ng.A. 0,51 mol. B. A. 0,45 mol. C. 0,55 mol. D. 0,49 mol.07. Ha tan hon ton m gam hn hp gm ba kim loi bng dung dch HNO3 thu c 1,12 lt hn hp kh D (ktc) gm NO2 v NO. T khi hi ca D so vi hiro bng 18,2. Tnh th tch ti thiu dung dch HNO3 37,8% (d = 1,242g/ml) cn dng.A. 20,18 ml. B. 11,12 ml. C. 21,47 ml. D. 36,7 ml.08. Ha tan 6,25 gam hn hp Zn v Al vo 275 ml dung dch HNO3 thu c dung dch A, cht rn B gm cc kim loi cha tan ht cn nng 2,516 gam v 1,12 lt hn hp kh D ( ktc) gm NO v NO2. T khi ca hn hp D so vi H2 l 16,75. Tnh nng mol/l ca HNO3v tnh khi lng mui khan thu c khi c cn dung dch sau phn ng.A. 0,65M v 11,794 gam. B. 0,65M v 12,35 gam.C. 0,75M v 11,794 gam. D. 0,55M v 12.35 gam.09.t chy 5,6 gam bt Fe trong bnh ng O2thu c 7,36 gam hn hp A gm Fe2O3, Fe3O4 v Fe. Ha tan hon ton lng hn hp A bng dung dch HNO3 thu c V lt hn hp kh B gm NO v NO2. T khi ca B so vi H2 bng 19. Th tch V ktc l35A. 672 ml. B. 336 ml. C. 448 ml. D. 896 ml.10. Cho a gam hn hp A gm oxit FeO, CuO, Fe2O3 c s mol bng nhau tc dng hon ton vi lng va l 250 ml dung dch HNO3khi un nng nh, thu c dung dch B v 3,136 lt (ktc) hn hp kh C gm NO2 v NO c t khi so vi hiro l 20,143. Tnh a.A. 74,88 gam. B. 52,35 gam. C. 61,79 gam. D. 72,35 gam.p n cc bi tp vn dng1. B 2. B 3. A 4. B 5. C6. D 7. C 8. A 9. D 10. APhng php 4S DNG PHNG TRNH ION - ELETRON lm tt cc bi ton bng phng php ion iu u tin cc bn phi nm chc phng trnh phn ng di dng cc phn t t suy ra cc phng trnh ion, i khi c mt s bi tp khng th gii theo cc phng trnh phn t c m phi gii da theo phng trnh ion. Vic gii bi ton ha hc bng phng php ion gip chng ta hiu k hn v bn cht ca cc phng trnh ha hc. T mt phng trnh ion c th ng vi rt nhiu phng trnh phn t. V d phn ng gia hn hp dung dch axit vi dung dch baz u c chung mt phng trnh ion lH++OH H2OhocphnngcaCukimloi vi hnhpdungdchNaNO3vdungdch H2SO4 l3Cu+8H++2NO3 3Cu2++2NO+4H2O...Sau y l mt s v d:V d 1: Hn hp X gm (Fe, Fe2O3, Fe3O4, FeO) vi s mol mi cht l 0,1 mol, ha tan ht vo dung dch Y gm (HCl v H2SO4long) d thu c dung dch Z. Nh t t dung dch Cu(NO3)2 1M vo dung dch Z cho ti khi ngng thot kh NO. Th tch dung dch Cu(NO3)2cn dng v th tch kh thot ra ktc thuc phng n no?A. 25 ml; 1,12 lt. B. 0,5 lt; 22,4 lt.36C. 50 ml; 2,24 lt. D. 50 ml; 1,12 lt.Hng dn giiQuy hn hp 0,1 mol Fe2O3 v 0,1 mol FeO thnh 0,1 mol Fe3O4. Hn hp X gm: (Fe3O4 0,2 mol; Fe 0,1 mol) tc dng vi dung dch YFe3O4 + 8H+ Fe2+ +2Fe3++4H2O 0,2 0,2 0,4 molFe+2H+ Fe2++H2 0,1 0,1 molDung dch Z: (Fe2+: 0,3 mol; Fe3+: 0,4 mol) + Cu(NO3)2:3Fe2++NO3+4H+ 3Fe3++NO+2H2O 0,30,10,1 mol VNO =0,1 22,4 = 2,24 lt.3 23Cu( NO )NO1n n 0,052 mol3 2dd Cu( NO )0,05V 0,051 lt(hay 50 ml).(p n C)V d 2: Ha tan 0,1 mol Cu kim loi trong 120 ml dung dch X gm HNO3 1M v H2SO40,5M. Sau khi phn ng kt thc thu c V lt kh NO duy nht (ktc).Gi tr ca V lA. 1,344 lt. B. 1,49 lt. C. 0,672 lt. D. 1,12 lt.Hng dn gii3HNOn 0,12 mol ;2 4H SOn 0,06 molTng:Hn 0, 24+mol v 3NOn 0,12 mol.Phng trnh ion:3Cu + 8H+ +2NO3 3Cu2++2NO+4H2OBan u:0,1 0,24 0,12 molPhn ng: 0,090,24 0,06 0,06 mol37Sau phn ng: 0,01 (d)(ht)0,06 (d) VNO=0,06 22,4=1,344 lt. (p n A)V d 3: Dung dch X cha dung dch NaOH 0,2M v dung dch Ca(OH)2 0,1M. Sc 7,84 lt kh CO2(ktc) vo 1 lt dung dch X th lng kt ta thu c lA. 15 gam. B. 5 gam. C. 10 gam. D. 0 gam.Hng dn gii2COn= 0,35 mol ;nNaOH = 0,2 mol;2Ca(OH)n= 0,1 mol.Tng:OHn = 0,2 + 0,1 2 = 0,4 mol v2Can+ = 0,1 mol.Phng trnh ion rt gn:CO2+2OH CO32+H2O0,35 0,40,20,4 0,2 mol2CO ( )nd = 0,35 0,2 = 0,15 moltip tc xy ra phn ng:CO32+CO2+H2O 2HCO3Ban u: 0,2 0,15 molPhn ng: 0,15 0,15 mol23COn cn li bng 0,15 mol3CaCOn= 0,05 mol3CaCOm= 0,05 100 = 5 gam. (p n B)V d 4:Ha tan ht hn hp gm mt kim loi kim v mt kim loi kim th trong nc c dung dch A v c 1,12 lt H2 bay ra ( ktc). Cho dung dch cha 0,03 mol AlCl3 vo dung dch A. khi lng kt ta thu c lA. 0,78 gam. B. 1,56 gam. C. 0,81 gam. D. 2,34 gam.Hng dn giiPhn ng ca kim loi kim v kim loi kim th vi H2O:38M+nH2O M(OH)n+2nH2T phng trnh ta c:2HOHn 2n= 0,1mol.Dung dch A tc dng vi 0,03 mol dung dch AlCl3:Al3+ + 3OH Al(OH)3Ban u: 0,03 0,1 molPhn ng: 0,030,090,03 molOH ( )nd= 0,01moltip tc ha tan kt ta theo phng trnh:Al(OH)3+OH AlO2+2H2O 0,01 0,01 molVy:3Al(OH)m= 78 0,02 = 1,56 gam. (p n B)V d 5: Dung dch A cha 0,01 mol Fe(NO3)3 v 0,15 mol HCl c kh nng ha tan ti a bao nhiu gam Cu kim loi? (Bit NO l sn phm kh duy nht)A. 2,88 gam. B. 3,92 gam. C. 3,2 gam. D. 5,12 gam.Hng dn giiPhng trnh ion:Cu + 2Fe3+ 2Fe2+ + Cu2+ 0,005 0,01 mol3Cu + 8H+ +2NO3 3Cu2++2NO+4H2OBan u:0,150,03 mol H+ dPhn ng:0,045 0,12 0,03 mol mCu ti a=(0,045 + 0,005) 64=3,2 gam. (p n C)V d 6: Cho hn hp gm NaCl v NaBr tc dng vi dung dch AgNO3 d thu c kt ta c khi lng ng bng khi lng AgNO3 phn ng. Tnh phn trm khi lng NaCl trong hn hp u.A. 23,3% B. 27,84%. C. 43,23%. D. 31,3%.39Hng dn giiPhng trnh ion: Ag++Cl AgClAg++Br AgBrt: nNaCl = x mol ;nNaBr = y molmAgCl + mAgBr=3( )AgNOmp.3Cl Br NOm m m + 35,5x + 80y=62(x + y) x : y=36 : 53Chn x = 36, y = 53NaCl58,5 36 100%m58,5 36 103 53 + =27,84%. (p n B)V d 7: Trn 100 ml dung dch A (gm KHCO3 1M v K2CO3 1M) vo 100 ml dung dch B (gm NaHCO3 1M v Na2CO3 1M) thu c dung dch C.Nh t t 100 ml dung dch D (gm H2SO41M v HCl 1M) vo dung dchCthucVlt CO2(ktc) vdungdchE. Chodungdch Ba(OH)2 ti d vo dung dch E th thu c m gam kt ta. Gi tr ca m v V ln lt lA. 82,4 gam v 2,24 lt. B. 4,3 gam v 1,12 lt.C. 43 gam v 2,24 lt. D. 3,4 gam v 5,6 lt.Hng dn giiDung dch C cha: HCO3 : 0,2 mol ; CO32 : 0,2 mol.Dung dch D c tng:Hn+ = 0,3 mol.Nh t t dung dch C v dung dch D:CO32+H+ HCO3 0,2 0,2 0,2 molHCO3+H+ H2O+CO2Ban u: 0,40,1 molPhn ng: 0,1 0,10,1 mol40D: 0,3 molTip tc cho dung dch Ba(OH)2 d vo dung dch E:Ba2++ HCO3 + OH BaCO3 + H2O 0,3 0,3 molBa2+ + SO42 BaSO40,1 0,1 mol2COV= 0,1 22,4 = 2,24 lt.Tng khi lng kt ta:m = 0,3 197 + 0,1 233 = 82,4 gam.(p n A)V d 8: Ha tan hon ton 7,74 gam mt hn hp gm Mg, Al bng 500 ml dung dch gm H2SO4 0,28M v HCl 1M thu c 8,736 lt H2 (ktc) v dung dch X.Thm V lt dung dch cha ng thi NaOH 1M v Ba(OH)2 0,5M vo dung dch X thu c lng kt ta ln nht.a) S gam mui thu c trong dung dch X lA. 38,93 gam. B. 38,95 gam.C. 38,97 gam. D. 38,91 gam.b) Th tch V lA. 0,39 lt. B. 0,4 lt.C. 0,41 lt. D. 0,42 lt.c) Lng kt ta lA. 54,02 gam. B. 53,98 gam.C. 53,62 gam. D. 53,94 gam.Hng dn giia) Xc nh khi lng mui thu c trong dung dch X:2 4H SOn= 0,28 0,5 = 0,14 mol24SOn= 0,14 molv Hn+ = 0,28 mol.41 nHCl = 0,5 molHn+ = 0,5 molvCln = 0,5 mol.Vy tngHn+ = 0,28 + 0,5 = 0,78 mol.M2Hn= 0,39 mol. Theo phng trnh ion rt gn:Mg0+2H+ Mg2++H2(1)Al+3H+ Al3++32H2(2)Ta thy2HH (p-)n 2n+H+ ht. mhh mui=mhh k.loi + 24SO Clm m + =7,74 + 0,14 96 + 0,5 35,5 = 38,93gam. (p n A)b) Xc nh th tch V:2NaOHBa(OH)n 1V moln 0,5V mol '

Tng OHn = 2V mol v2Ban+ = 0,5V mol.Phng trnh to kt ta:Ba2+ +SO42 BaSO4(3)0,5V mol0,14 molMg2++2OH Mg(OH)2(4)Al3++3OH Al(OH)3(5) kt ta t ln nht th s mol OH kt ta ht cc ion Mg2+v Al3+. Theo cc phng trnh phn ng (1), (2), (4), (5) ta c:Hn+ =OHn = 0,78 mol2V = 0,78 V = 0,39 lt. (p n A)c) Xc nh lng kt ta:2Ban+ = 0,5V = 0,5 0,39 = 0,195 mol > 0,14 mol Ba2+ d.4BaSOm= 0,14 233 = 32,62 gam.42Vymkt ta=4BaSOm+ m 2 k.loi + OHm =32,62 + 7,74 + 0,78 17=53,62 gam. (p n C)V d 9: (Cu 40 - M 182 - TS i Hc - Khi A 2007)Cho m gam hn hp Mg, Al vo 250 ml dung dch X cha hn hp axit HCl 1M v axit H2SO4 0,5M, thu c 5,32 lt H2 ( ktc) v dung dch Y (coi th tch dung dch khng i). Dung dch Y c pH lA. 1.B. 6.C. 7.D. 2.Hng dn gii nHCl = 0,25 mol ;2 4H SOn= 0,125.Tng:Hn+ = 0,5 mol ;2H ( )ntothnh= 0,2375 mol.Bit rng: c 2 mol ion H+ 1 mol H2vy 0,475 mol H+ 0,2375 mol H2H ( )nd+= 0,5 0,475 = 0,025 mol0,025H0, 25+ ] ] = 0,1 = 101M pH = 1. (p n A)V d 10: (Cu 40 - M 285 - Khi B - TSH 2007)Thc hin hai th nghim:1) Cho 3,84 gam Cu phn ng vi 80 ml dung dch HNO3 1M thot ra V1 lt NO.2) Cho 3,84 gam Cu phn ng vi 80 ml dung dch cha HNO3 1M v H2SO4 0,5 M thot ra V2 lt NO.Bit NO l snphm kh duy nht, cc th tch kh o cng iu kin. Quan h gia V1 v V2 lA. V2 = V1. B. V2 = 2V1. C. V2 = 2,5V1. D. V2 = 1,5V1.Hng dn gii43TN1: 3CuHNO3,84n 0,06 mol64n 0,08 mol ' 3HNOn 0,08 moln 0,08 mol+'3Cu+8H++2NO3 3Cu2++2NO+4H2OBan u: 0,060,080,08 mol H+ phn ng htPhn ng: 0,03 0,08 0,02 0,02 mol V1 tng ng vi 0,02 mol NO.TN2: nCu = 0,06 mol ;3HNOn= 0,08 mol ;2 4H SOn= 0,04 mol.Tng: Hn+ = 0,16 mol ;3NOn= 0,08 mol.3Cu+ 8H++2NO3 3Cu2++2NO+4H2OBan u: 0,060,160,08 mol Cu v H+ phn ng htPhn ng: 0,06 0,16 0,040,04 mol V2 tng ng vi 0,04 mol NO.Nh vy V2 = 2V1. (p n B)V d 11: (Cu 33 - M 285 - Khi B - TSH 2007)Trn 100 ml dung dch (gm Ba(OH)20,1M v NaOH 0,1M) vi 400 ml dung dch (gm H2SO40,0375M v HCl 0,0125M), thu c dung dch X. Gi tr pH ca dung dch X lA. 7.B. 2.C. 1.D. 6.Hng dn gii2Ba(OH)NaOHn 0,01 moln 0,01 mol'Tng OHn = 0,03 mol.2 4H SOHCln 0,015 moln 0,005 mol'Tng Hn+ = 0,035 mol.Khi trn hn hp dung dch baz vi hn hp dung dch axit ta c phng trnh ion rt gn:H+ +OHH2O44Bt u0,035 0,03 molPhn ng:0,03 0,03Sau phn ng:H ( )nd+ = 0,035 0,03 = 0,005 mol.Tng:Vdd (sau trn) = 500 ml(0,5 lt).0,005H0,5+ ] ] = 0,01 = 102 pH = 2. (p n B)V d 12: (Cu 18 - M 231 - TS Cao ng - Khi A 2007)Cho mt mu hp kim Na-Ba tc dng vi nc (d), thu c dung dch X v 3,36 lt H2( ktc). Th tch dung dch axit H2SO42M cn dng trung ho dung dch X lA. 150 ml.B. 75 ml.C. 60 ml.D. 30 ml.Hng dn giiNa+H2O NaOH+12H2Ba+2H2O Ba(OH)2+H22Hn= 0,15 mol, theo phng trnh tng s 22HOH (d X)n 2n= 0,3 mol.Phng trnh ion rt gn ca dung dch axit vi dung dch baz lH++OH H2OHn+ = OHn = 0,3 mol 2 4H SOn= 0,15 mol2 4H SO0,15V2 = 0,075 lt(75 ml). (p n B)V d 13:Ha tan hn hp X gm hai kim loi A v B trong dung dch HNO3 long. Kt thc phn ng thu c hn hp kh Y (gm 0,1 mol NO, 0,15molNO2v0,05 molN2O).Bitrngkhngcphn ng to mui NH4NO3. S mol HNO3 phn ng l: A. 0,75 mol. B. 0,9 mol. C. 1,05 mol.D. 1,2 mol.Hng dn giiTa c bn phn ng:NO3+2H++1e NO2+H2O (1)45 2 0,15 0,15NO3+4H++ 3eNO+2H2O (2) 4 0,1 0,12NO3+10H++8e N2O+5H2O (3) 10 0,05 0,05T (1), (2), (3)nhn c:3HNOHn np+ = 2 0,15 4 0,1 10 0, 05 + + = 1,2 mol. (p n D)V d 14:Cho 12,9 gam hn hp Al v Mg phn ng vi dung dch hn hp hai axit HNO3v H2SO4(c nng) thu c 0,1 mol mi kh SO2, NO, NO2. C cn dung dch sau phn ng khi lng mui khan thu c l:A. 31,5 gam. B. 37,7 gam. C. 47,3 gam. D. 34,9 gam.Hng dn giiTa c bn phn ng:2NO3+2H++1e NO2+H2O+NO3(1) 0,1 0,14NO3+ 4H++3e NO+2H2O+3NO3(2) 0,13 0,12SO42+4H++2e SO2+H2O+SO42(3)0,1 0,1T (1), (2), (3) s mol NO3 to mui bng 0,1 + 3 0,1 = 0,4 mol;s mol SO42 to mui bng 0,1 mol. mmui = mk.loi + 3NOm + 24SOm = 12,9 + 62 0,4 + 96 0,1 = 47,3. (p n C)V d 15: Ha tan 10,71 gam hn hp gm Al, Zn, Fe trong 4 lt dung dch HNO3 aM va thu c dung dch A v 1,792 lt hn hp kh gm N2v N2O c t l mol 1:1. C cn dung dch A thu c m (gam.) mui khan. gi tr ca m, a l:A. 55,35 gam. v 2,2M B. 55,35 gam. v 0,22MC. 53,55 gam. v 2,2M D. 53,55 gam. v 0,22M46Hng dn gii2 2N O N1,792n n 0,042 22, 4 mol.Ta c bn phn ng:2NO3+12H++10e N2+6H2O0,080,480,042NO3+10H++8e N2O+5H2O0,080,4 0,043HNOHn n 0,88+ mol.0,88a 0, 224 M.S mol NO3 to mui bng 0,88 (0,08 + 0,08) = 0,72 mol.Khi lng mui bng 10,71 + 0,72 62 = 55,35 gam. (p n B)V d 16:Ha tan 5,95 gam hn hp Zn, Al c t l mol l 1:2 bng dung dch HNO3 long d thu c 0,896 lt mt sn shm kh X duy nht cha nit. X l:A. N2O B. N2C. NO D. NH4+Hng dn giiTa c: nZn = 0,05 mol; nAl = 0,1 mol.Gi a l s mol ca NxOy, ta c:Zn Zn2++2e Al Al3++3e0,05 0,1 0,10,3xNO3 + (6x 2y)H++(5x 2y)e NxOy + (3x 2y)H2O 0,04(5x 2y)0,04 0,04(5x 2y) = 0,4 5x 2y = 10VyX l N2. (p n B)V d 17: Cho hn hp gm 0,15 mol CuFeS2 v 0,09 mol Cu2FeS2 tc dng vi dung dch HNO3 d thu c dung dch X v hn hp kh Y gm NO v NO2. Thm BaCl2 d vo dung dch X thu c m gam kt ta. Mt 47khc, nu thm Ba(OH)2d vo dung dch X, ly kt ta nung trong khng kh n khi lng khng i thu c a gam cht rn. Gi tr ca m v a l:A. 111,84g v 157,44g B. 111,84g v 167,44gC. 112,84g v 157,44g A. 112,84g v 167,44gHng dn giiTa c bn phn ng:CuFeS2 + 8H2O 17e Cu2+ + Fe3+ + 2SO42 + 16+ 0,150,150,15 0,3Cu2FeS2 + 8H2O 19e 2Cu2+ + Fe3+ + 2SO42 + 16+ 0,09 0,180,09 0,1824SOn 0, 48 mol; Ba2++SO42 BaSO40,48 0,48 m = 0,48 233 = 111,84 gam.nCu = 0,33 mol; nFe = 0,24 mol.Cu CuO2Fe Fe2O30,330,330,240,12 a = 0,33 80 + 0,12 160 + 111,84 = 157,44 gam. (p n A).V d 18:Ha tan 4,76 gam hn hp Zn, Al c t l mol 1:2 trong 400ml dung dch HNO3 1M va , dc dung dch X cha m gam mui khan v thy c kh thot ra. Gi tr ca m l:A. 25.8 gam. B. 26,9 gam. C. 27,8 gam. D. 28,8 gam.Hng dn giinZn = 0,04 mol; nAl = 0,08 mol.- Do phn ng khng to kh nn trong dung dch to NH4NO3. Trong dung dch c:0,04 mol Zn(NO3)2 v 0,08 mol Al(NO3)3Vy s mol NO3 cn li to NH4NO3 l: 480,4 0,04 2 0,08 3 = 0,08 mol- Do trong dung dch to 0,04mol NH4NO3 m = 0,04 189 + 0,08 213 + 0,04 80 = 27,8 gam. (p n C)Phng php 5S DNG CC GI TR TRUNG BNHy l mt trong mt s phng php hin i nht cho php gii nhanh chng v n gin nhiu bi ton ha hc v hn hp cc cht rn, lng cng nh kh.Nguyntc caphngphpnhsau: Khi lngphnttrungbnh (KLPTTB) (k hiu M) cng nh khi lng nguyn t trung bnh (KLNTTB) chnh l khi lng ca mt mol hn hp, nn n c tnh theo cng thc:M tng khi l ng hn h p (tnh theo gam)tng s mol cc cht trong hn h p.i i1 1 2 2 3 31 2 3 iM nM n M n M n ...Mn n n ... n+ + + + + +(1)trong M1, M2,... l KLPT (hoc KLNT) ca cc cht trong hn hp; n1, n2,... l s mol tng ng ca cc cht.Cng thc (1) c th vit thnh:1 2 31 2 3i i in n nM M . M . M . ...n n n + + + 491 1 2 2 3 3M M x M x M x ... + + + (2)trong x1, x2,... l % s mol tng ng (cng chnh l % khi lng) ca cc cht. c bit i vi cht kh th x1, x2, ... cng chnh l % th tch nn cng thc (2) c th vit thnh:i i1 1 2 2 3 31 2 3 iM VM V M V M V ...MV V V ... V+ + + + + +(3)trong V1, V2,... l th tch ca cc cht kh. Nu hn hp ch c 2 cht th cc cng thc (1), (2), (3) tng ng tr thnh (1), (2), (3) nh sau:1 1 2 1M n M (n n )Mn+ (1)trong n l tng s s mol ca cc cht trong hn hp,1 1 2 1M M x M (1 x ) + (2)trong con s 1 ng vi 100% v1 1 2 1M V M (V V )MV+ (3)trong V1 l th tch kh th nht v V l tng th tch hn hp.T cng thc tnh KLPTTB ta suy ra cc cng thc tnh KLNTTB.Vi cc cng thc:x y z 1x y z 2C H O ; n molC H O ; n mol ta c:- Nguyn t cacbon trung bnh:1 1 2 21 2x n x n ...xn n ...+ ++ +- Nguyn t hiro trung bnh:1 1 2 21 2y n y n ...yn n ...+ ++ +v i khi tnh c c s lin kt , s nhm chc trung bnh theo cng thc trn.50V d 1: Ha tan hon ton 2,84 gam hn hp hai mui cacbonat ca hai kim loi phn nhm IIA v thuc hai chu k lin tip trong bng tun hon bng dung dch HCl ta thu c dung dch X v 672 ml CO2 ( ktc).1. Hy xc nh tn cc kim loi.A. Be, Mg. B. Mg, Ca.C. Ca, Ba. D. Ca, Sr.2. C cn dung dch X th thu c bao nhiu gam mui khan?A. 2 gam. B. 2,54 gam. C. 3,17 gam.D. 2,95 gam.Hng dn gii1. Gi A, B l cc kim loi cn tm. Cc phng trnh phn ng lACO3 + 2HCl ACl2 + H2O + CO2(1)BCO3 + 2HCl BCl2 + H2O + CO2(2)(C th gi M l kim loi i din cho 2 kim loi A,B lc ch cn vit mt phng trnh phn ng).Theo cc phn ng (1), (2) tng s mol cc mui cacbonat bng:2CO0,672n 0,0322, 4 mol.Vy KLPTTB ca cc mui cacbonat l2,84M 94,670,03 vA,B M 94,67 60 34,67 V thuc 2 chu k lin tip nn hai kim loi l Mg (M = 24) v Ca (M = 40). (p n B)2. KLPTTB ca cc mui clorua:M 34,67 71 105,67 + mui clorua.Khi lng mui clorua khan l 105,67 0,03 = 3,17 gam. (p n C)V d 2: Trong t nhin, ng (Cu) tn ti di hai dng ng v 6329Cu v 6529Cu. KLNT (xp x khi lng trung bnh) ca Cu l 63,55. Tnh % v khi lng ca mi loi ng v.A. 65Cu: 27,5% ;63Cu: 72,5%. B. 65Cu: 70% ;63Cu: 30%.51C. 65Cu: 72,5% ;63Cu: 27,5%.D. 65Cu: 30% ;63Cu: 70%.Hng dn giiGi x l % ca ng v 6529Cu ta c phng trnh:M = 63,55 = 65.x + 63(1 x) x = 0,275Vy: ng v 65Cu chim 27,5% v ng v 63Cu chim 72,5%. (p n C)V d 3:Hn hp kh SO2v O2c t khi so vi CH4bng 3. Cn thm bao nhiu lt O2 vo 20 lt hn hp kh cho t khi so vi CH4 gim i 1/6, tc bng 2,5. Cc hn hp kh cng iu kin nhit v p sut.A. 10 lt. B. 20 lt.C. 30 lt. D. 40 lt.Hng dn giiCch 1: Gi x l % th tch ca SO2 trong hn hp ban u, ta c:M = 16 3 = 48 = 64.x + 32(1 x) x = 0,5Vy: mi kh chim 50%. Nh vy trong 20 lt, mi kh chim 10 lt. Gi V l s lt O2 cn thm vo, ta c:64 10 32(10 V)M 2,5 16 4020 V + + +.Gii ra c V = 20 lt. (p n B)Cch 2:Ghi ch:C th coi hn hp kh nh mt kh c KLPT chnh bng KLPT trung bnh ca hn hp, v d, c th xem khng kh nh mt kh vi KLPT l 29.Hn hp kh ban u coi nh kh th nht (20 lt c M = 16 3 = 48), cn O2 thm vo coi nh kh th hai, ta c phng trnh:48 20 32VM 2,5 16 4020 V + +,Rt ra V = 20 lt. (p n B)52V d 4: C 100 gam dung dch 23% ca mt axit n chc (dung dch A). Thm 30 gam mt axit ng ng lin tip vo dung dch ta c dung dch B. Trung ha 1/10 dung dch B bng 500 ml dung dch NaOH 0,2M (va ) ta c dung dch C.1. Hy xc nh CTPT ca cc axit.A. HCOOHvCH3COOH.B. CH3COOHvC2H5COOH.C. C2H5COOHvC3H7COOH.D. C3H7COOHvC4H9COOH.2. C cn dung dch C th thu c bao nhiu gam mui khan?A. 5,7 gam. B. 7,5 gam.C. 5,75 gam. D. 7,55 gam.Hng dn gii1. Theo phng php KLPTTB:RCOOH1 23m 2,310 10 gam,2RCH COOH1 30m 310 10 gam.2,3 3M 530,1+ .Axit duy nht c KLPT < 53 l HCOOH (M = 46) v axit ng ng lin tip phi l CH3COOH (M = 60). (p n A)2. Theo phng php KLPTTB:V Maxit= 53 nnM= 53+23 1 75 mui. V s mol mui bng s mol axit bng 0,1 nn tng khi lng mui bng 75 0,1 = 7,5 gam. (p n B)V d 5: C V lt kh A gm H2 v hai olefin l ng ng lin tip, trong H2 chim 60% v th tch. Dn hn hp A qua bt Ni nung nng c hn hp kh B. t chy hon ton kh B c 19,8 gam CO2 v 13,5 gam H2O. Cng thc ca hai olefin lA. C2H4 v C3H6. B. C3H6 v C4H8.C. C4H8 v C5H10. D. C5H10 v C6H12.53Hng dn giit CTTB ca hai olefin l n 2nC H . cng iu kin nhit v p sut th th tch t l vi s mol kh.Hn hp kh A c:n 2n2C HHn0,4 2n 0,6 3 .p dng nh lut bo ton khi lng v nh lut bo ton nguyn t t chy hn hp kh B cng chnh l t chy hn hp kh A. Ta c:n 2nC H+ 23nO2n CO2+ n H2O (1)2H2+O2 2H2O (2)Theo phng trnh (1) ta c:2 2CO H On n = 0,45 mol.n 2nC H0, 45nn mol.Tng:2H O13,5n18= 0,75 mol2H O ( pt 2)n= 0,75 0,45 = 0,3 mol2Hn= 0,3 mol.Ta c:n 2n2C HHn0,45 2n 0,3 n 3 n= 2,25Hai olefin ng ng lin tip l C2H4 v C3H6. (p n B)V d 6: t chy hon ton a gam hn hp hai ru no, n chc lin tip trong dy ng ng thu c 3,584 lt CO2 ktc v 3,96 gam H2O. Tnh a v xc nh CTPT ca cc ru.A. 3,32 gam ; CH3OH v C2H5OH.B. 4,32 gam ; C2H5OH v C3H7OH.C. 2,32 gam ; C3H7OH v C4H9OH.54D. 3,32 gam ; C2H5OH v C3H7OH.Hng dn giiGinl s nguyn t C trung bnh v x l tng s mol ca hai ru.CnH2n+1OH + 23nO2 2nCO+2(n 1) H O + x mol n x mol (n 1) +x mol2CO3,584n n.x 0,1622, 4 mol (1)2H O3,96n (n 1)x 0,2218 + mol (2)T (1) v (2) gii ra x = 0,06 vn= 2,67.Ta c: a = (14 n+ 18).x = (14 2,67) + 18 0,06 = 3,32 gam.n= 2,672 53 7C H OHC H OH(p n D)V d 7: Hn hp 3 ru n chc A, B, C c tng s mol l 0,08 v khi lng l 3,38gam. Xc nh CTPT ca ru B, bit rng B v C c cng s nguyn t cacbon v s mol ru A bng5 3 tng s mol ca ru B v C, MB> MC.A. CH3OH. B. C2H5OH. C. C3H7OH. D. C4H9OH.Hng dn giiGi M l nguyn t khi trung bnh ca ba ru A, B, C. Ta c:3,38M 42, 20,08 Nh vy phi c t nht mt ru c M < 42,25. Ch c CH3OH c (M = 32)Ta c: A0,08 5n 0,055 3 +;mA = 32 0,05 = 1,6 gam.mB + C = 3,38 1,6 = 1,78 gam;55B C0,08 3n 0,035 3+ + mol ;B C1,78M 59,330.03+ .Gi y l s nguyn t H trung bnh trong phn t hai ru B v C. Ta c: x yC H OH 59,33 hay12x + y + 17 = 59,33 12x + y = 42,33Bin lun:x 1 2 3 4y30,33 18,33 6,33 < 0Ch c nghim khi x = 3. B, C phi c mt ru c s nguyn t H < 6,33 v mt ru c s nguyn t H > 6,33.Vy ru B l C3H7OH.C 2 cp nghim: C3H5OH (CH2=CHCH2OH)vC3H7OH C3H3OH (CH CCH2OH) vC3H7OH (p n C)V d 8: Cho 2,84 gam hn hp 2 ru n chc l ng ng lin tip nhau tc dng vi mt lng Na va to ra 4,6 gam cht rn v V lt kh H2 ktc. Tnh V.A. 0,896 lt. B. 0,672 lt. C. 0,448 lt. D. 0,336 lt.Hng dn giitR l gc hirocacbon trung bnh v x l tng s mol ca 2 ru.ROH+Na RONa+ 21H2 x molxx2.Ta c: ( )( )R 17 x 2,84R 39 x 4,6+ '+ Gii ra c x = 0,08. 56Vy : 2H0,08V 22,4 0,8962 lt.(p n A)V d 9: (Cu 1 - M 182 - Khi A - TSH nm 2007)Cho 4,48 lt hn hp X ( ktc) gm 2 hirocacbon mch h li t t qua bnh cha 1,4 lt dung dch Br2 0,5M. Sau khi phn ng hon ton, s mol Br2 gim i mt na v khi lng bnh tng thm 6,7 gam. Cng thc phn t ca 2 hirocacbon lA. C2H2 v C4H6. B. C2H2 v C4H8.C. C3H4 v C4H8.D. C2H2 v C3H8.Hng dn giihh X4, 48n 0, 222, 4 moln 1, 4 0,5 0,72Br ban u mol0,7n22Br p.ng = 0,35 mol.Khi lng bnh Br2tng 6,7 gam l s gam ca hirocabon khng no. t CTTBcahai hirocacbonmchhln 2n 2 2aC H+ ( a lslinkt trung bnh).Phng trnh phn ng:n 2n 2 2aC H+ +2aBrn 2n 2 2a 2aC H Br+ 0,2 mol 0,35 mol0,35a0, 2 = 1,756,714n 2 2a0, 2+ n= 2,5.Dohai hirocacbonmchhphnnghontonvi dungdchBr2nn chngu l hirocacbonkhng no. Vy hai hirocacbon l C2H2v C4H8. (p n B)57V d 10: Tch nc hon ton t hn hp X gm 2 ancol A v B ta c hn hp Y gm cc olefin. Nu t chy hon ton X th thu c 1,76 gam CO2. Khi t chy hon ton Y th tng khi lng H2O v CO2 to ra lA. 2,94 gam. B. 2,48 gam. C. 1,76 gam. D. 2,76 gam.Hng dn giiHn hp X gm hai ancol A v B tch nc c olefin (Y) hai ancol l ru no, n chc.t CTTB ca hai ancol A, B l n 2n 1C H OH+ta c cc phng trnh phn ng sau:n 2n 1C H OH+ + 23nO2 2nCO+ 2(n 1)H O +n 2n 1C H OH+ 2oH SO170 C4n 2nC H+H2O(Y) n 2nC H+ 23nO2 2nCO+ 2n H ONhn xt:- Khi t chy X v t chy Y cng cho s mol CO2 nh nhau.- t chy Y cho 2 2CO H On n .Vy t chy Y cho tng( )2 2CO H Om m 0,04 (44 18) 2,48 + + gam.(p n B)MT S BI TP VN DNG GII THEP PHNG PHP TRUNG BNH01. t chy hon ton 0,1 mol hn hp hai axit cacboxylic l ng ng k tip thu c 3,36 lt CO2 (ktc) v 2,7 gam H2O. S mol ca mi axit ln lt lA. 0,05 mol v 0,05 mol. B. 0,045 mol v 0,055 mol.C. 0,04 mol v 0,06 mol. D. 0,06 mol v 0,04 mol.02. C 3 ancol bn khng phi l ng phn ca nhau. t chy mi cht u c s mol CO2 bng 0,75 ln s mol H2O. 3 ancol lA. C2H6O; C3H8O; C4H10O. B. C3H8O; C3H6O2; C4H10O.58C. C3H8O; C3H8O2; C3H8O3. D. C3H8O; C3H6O; C3H8O2.03. Cho axit oxalic HOOCCOOH tc dng vi hn hp hai ancol no, n chc, ng ng lin tip thu c 5,28 gam hn hp 3 este trung tnh. Thy phn lng este trn bng dung dch NaOH thu c 5,36 gam mui. Hai ru c cng thcA. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.C. C3H7OH v C4H9OH. D. C4H9OH v C5H11OH.04. Nitro ha benzen c 14,1 gam hn hp hai cht nitro c khi lng phn t hn km nhau 45 vC. t chy hon ton hn hp hai cht nitro ny c 0,07 mol N2. Hai cht nitro lA. C6 H5NO2 v C6H4(NO2)2.B. C6 H4(NO2)2 v C6H3(NO2)3.C. C6 H3(NO2)3 v C6H2(NO2)4.D. C6 H2(NO2)4 v C6H(NO2)5.05.Mt hn hp X gm 2 ancol thuc cng dy ng ng c khi lng 30,4 gam. Chia X thnh hai phn bng nhau.- Phn 1: cho tc dng vi Na d, kt thc phn ng thu c 3,36 lt H2 (ktc).-Phn2:tchnchonton 180oC,xc tcH2SO4cthu cmt anken cho hp th vo bnh ng dung dch Brom d thy c 32 gam Br2 b mt mu. CTPT hai ancol trn lA. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.C. CH3OH v C3H7OH. D. C2H5OH v C4H9OH.06. Chia hn hp gm 2 anehit no n chc lm hai phn bng nhau:- Phn 1: em t chy hon ton thu c 1,08 gam nc.- Phn 2: tc dng vi H2 d (Ni, to) th thu c hn hp A. em A t chy hon ton th th tch kh CO2 (ktc) thu c lA. 1,434 lt. B. 1,443 lt. C. 1,344 lt. D. 1,444 lt.07. Tch nc hon ton t hn hp Y gm hai ru A, B ta c hn hp X gm cc olefin. Nu t chy hon ton Y th thu c 0,66 gam CO2. Vy khi t chy hon ton X th tng khi lng H2O v CO2 to ra lA. 0,903 gam. B. 0,39 gam. C. 0,94 gam. D. 0,93 gam.5908. Cho 9,85 gam hn hp 2 amin n chc no bc 1 tc dng va vi dung dch HCl th thu c 18,975 gam mui. Vy khi lng HCl phi dng lA. 9,521 gam. B. 9,125 gam. C. 9,215 gam. D. 0,704 gam.09. Cho 4,2 gam hn hp gm ru etylic, phenol, axit fomic tc dng va vi Na thy thot ra 0,672 lt kh (ktc) v mt dung dch. C cn dung dch thu c hn hp X. Khi lng ca X lA. 2,55 gam. B. 5,52 gam. C. 5,25 gam. D. 5,05 gam.10. Hn hp X gm 2 este A, B ng phn vi nhau v u c to thnh t axit n chc v ru n chc. Cho 2,2 gam hn hp X bay hi 136,5oC v 1 atm th thu c 840 ml hi este. Mt khc em thu phn hon ton 26,4 gam hn hp X bng 100 ml dung dch NaOH 20% (d = 1,2 g/ml) ri em c cn th thu c 33,8 gam cht rn khan. Vy cng thc phn tca este lA. C2H4O2. B. C3H6O2. C. C4H8O2. D. C5H10O2.p n cc bi tp trc nghim vn dng:1. A 2. C 3. A 4. A 5. C6. C 7. D 8. B 9. B 10. CPhng php 6TNG GIM KHI LNGNguyntccaphngphplxemkhi chuyntcht Athnhcht B (khng nht thit trc tip, c th b qua nhiu giai on trung gian) khi lng tng hay gim bao nhiu gam thng tnh theo 1 mol) v da vo khi lng thay i ta d dng tnh c s mol cht tham gia phn ng hoc ngc li. V d trong phn ng:MCO3 + 2HCl MCl2 + H2O + CO260Ta thy rng khi chuyn 1 mol MCO3 thnh MCl2 th khi lng tng(M + 2 35,5) (M + 60) = 11 gamv c 1 mol CO2 bay ra. Nh vy khi bit lng mui tng, ta c th tnh lng CO2 bay ra.Trong phn ng este ha:CH3COOH + ROH CH3COOR+ H2Oth t 1 mol ROH chuyn thnh 1 mol este khi lng tng(R+ 59) (R+ 17) = 42 gam.Nh vy nu bit khi lng ca ru v khi lng ca este ta d dng tnh c s mol ru hoc ngc li.Vi bi tp cho kim loi A y kim loi B ra khi dung dch mui di dng t do:- Khi lng kim loi tng bngmB (bm) mA (tan).- Khi lng kim loi gim bngmA (tan) mB (bm).Sau y l cc v d in hnh:V d 1: C 1 lt dung dch hn hp Na2CO3 0,1 mol/l v (NH4)2CO3 0,25 mol/l. Cho 43 gam hn hp BaCl2 v CaCl2 vo dung dch . Sau khi cc phn ng kt thc ta thu c 39,7 gam kt ta A v dung dch B.Tnh % khi lng cc cht trong A.A. 3BaCO%m= 50%,3CaCO%m= 50%.B. 3BaCO%m= 50,38%,3CaCO%m= 49,62%.C. 3BaCO%m= 49,62%,3CaCO%m= 50,38%. D.Khng xc nh c.Hng dn giiTrong dung dch:Na2CO3 2Na++CO3261(NH4)2CO3 2NH4++CO32BaCl2 Ba2++2ClCaCl2 Ca2++2ClCc phn ng:Ba2++CO32 BaCO3(1)Ca2++CO32 CaCO3(2)Theo (1) v (2) c 1 mol BaCl2, hoc CaCl2 bin thnh BaCO3 hoc CaCO3 th khi lng mui gim (71 60) = 11 gam. Do tng s mol hai mui BaCO3 v CaCO3 bng:43 39,711 = 0,3 molm tng s mol CO32 = 0,1 + 0,25 = 0,35, iu chng t d CO32.Gi x, y l s mol BaCO3 v CaCO3 trong A ta c:x y 0,3197x 100y 39,7+ '+ x = 0,1 mol ;y = 0,2 mol.Thnh phn ca A:3BaCO0,1 197%m 10039,7 = 49,62%;3CaCO%m= 100 49,6 = 50,38%. (p n C)V d 2: Ho tan hon ton 23,8 gam hn hp mt mui cacbonat ca kim loi ho tr (I) v mt mui cacbonat ca kim loi ho tr (II) bng dung dch HCl thy thot ra 4,48 lt kh CO2 (ktc). C cn dung dch thu c sau phn ng th khi lng mui khan thu c l bao nhiu?A. 26,0 gam.B. 28,0 gam. C. 26,8 gam. D. 28,6 gam.Hng dn giiC 1 mol mui cacbonat to thnh 1 mol mui clorua cho nn khi lng mui khan tng (71 60) = 11 gam, m622COn= nmui cacbonat = 0,2 mol.Suy ra khi lng mui khan tng sau phn ng l 0,2 11 = 2,2 gam.Vy tng khi lng mui khan thu c l 23,8 + 2,2 = 26 gam. (p n A)V d 3:Cho 3,0 gam mt axit no, n chc A tc dng va vi dung dch NaOH. C cn dung dch sau phn ng thu c 4,1 gam mui khan. CTPT ca A lA. HCOOH B. C3H7COOHC. CH3COOH D. C2H5COOH.Hng dn giiC 1 mol axit n chc to thnh 1 mol mui th khi lng tng (23 1) = 22 gam, m theo u bi khi lng mui tng (4,1 3) = 1,1 gam nn s mol axit lnaxit = 1,122 = 0,05 mol. Maxit = 30,05 = 60 gam.t CTTQ ca axit no, n chc A l CnH2n+1COOH nn ta c:14n + 46 = 60 n = 1.Vy CTPT ca A l CH3COOH. (p n C)V d 4:Cho dung dch AgNO3d tc dng vi dung dch hn hp c ha tan 6,25 gam hai mui KCl v KBr thu c 10,39 gam hn hp AgCl v AgBr. Hy xc nh s mol hn hp u.A. 0,08 mol. B. 0,06 mol. C. 0,03 mol. D. 0,055 mol.Hng dn giiC 1 mol mui halogen to thnh 1 mol kt takhi lng tng: 108 39 = 69 gam; 0,06 mol khi lng tng: 10,39 6,25 = 4,14 gam.Vy tng s mol hn hp u l 0,06 mol. (p n B)V d 5:Nhng mtthanhgraphit c ph mt lp kim loi ha tr (II) vo dung dch CuSO4d. Sau phn ng khi lng ca thanh graphit gim 63i 0,24gam. Cngthanhgraphit nynucnhngvodungdch AgNO3thkhi phnng xong thy khi lng thanh graphit tng ln 0,52 gam. Kim loi ha tr (II) l kim loi no sau y?A. Pb. B. Cd. C. Al. D. Sn.Hng dn giit kim loi ha tr (II) l M vi s gam l x (gam).M+CuSO4 d MSO4+CuC M gam kim loi tan ra th s c 64 gam Cu bm vo. Vy khi lng kim loi gim (M 64) gam;Vy:x (gam) = 0,24.MM 64 khi lng kim loi gim 0,24 gam.Mt khc: M+2AgNO3 M(NO3)2+2AgC M gam kim loi tan ra th s c 216 gam Ag bm vo. Vy khi lng kim loi tng (216 M) gam;Vy:x (gam) = 0,52. M216 M khi lng kim loi tng 0,52 gam.Ta c:0,24.MM 64 = 0,52. M216 M M = 112 (kim loi Cd). (p n B)V d 6:Ho tan hon ton 104,25 gam hn hp X gm NaCl v NaI vo nc c dung dch A. Sc kh Cl2 d vo dung dch A. Kt thc th nghim, c cn dung dch thu c 58,5 gam mui khan. Khi lng NaCl c trong hn hp X lA. 29,25 gam. B. 58,5 gam.C. 17,55 gam. D. 23,4 gam.Hng dn giiKh Cl2 d ch kh c mui NaI theo phng trnh2NaI+Cl2 2NaCl+I2C 1 mol NaI to thnh 1 mol NaClKhi lng mui gim 127 35,5 = 91,5 gam.Vy: 0,5 molKhi lng mui gim 104,25 58,5 = 45,75 gam.64 mNaI = 150 0,5 = 75 gam mNaCl = 104,25 75 = 29,25 gam. (p n A)V d 7:Ngm mt vt bng ng c khi lng 15 gam trong 340 gam dung dch AgNO36%. Sau mt thi gian ly vt ra thy khi lng AgNO3 trong dung dch gim 25%. Khi lng ca vt sau phn ng lA. 3,24 gam. B. 2,28 gam. C. 17,28 gam. D. 24,12 gam.Hng dn gii3AgNO ( )340 6n= 170 100ban u = 0,12 mol; 3AgNO ( )25n= 0,12100ph.ng= 0,03 mol. Cu + 2AgNO3 Cu(NO3)2+2Ag0,015 0,030,03 molmvt sau phn ng=mvt ban u + mAg (bm) mCu (tan)=15 + (108 0,03) (64 0,015) = 17,28 gam.(p n C)V d 8: Nhng mt thanh km v mt thanh st vo cng mt dung dch CuSO4. Sau mt thi gian ly hai thanh kim loi ra thy trong dung dch cn li c nng mol ZnSO4 bng 2,5 ln nng mol FeSO4. Mt khc, khi lng dung dch gim 2,2 gam.Khi lng ng bm ln thanh km v bm ln thanh st ln lt lA. 12,8 gam; 32 gam. B. 64 gam; 25,6 gam.C. 32 gam; 12,8 gam. D. 25,6 gam; 64 gam.Hng dn giiV trong cng dung dch cn li (cng th tch) nn:[ZnSO4] = 2,5 [FeSO4]4 4ZnSO FeSOn 2,5n Zn + CuSO4 ZnSO4+Cu(1)2,5x 2,5x 2,5x mol65Fe + CuSO4 FeSO4+Cu(2) x x x x molT (1), (2) nhn c gim khi lng ca dung dch lmCu (bm) mZn (tan) mFe (tan) 2,2 = 64 (2,5x + x) 65 2,5x 56x x = 0,4 mol.Vy: mCu (bm ln thanh km) = 64 2,5 0,4 = 64 gam;mCu (bm ln thanh st) = 64 0,4= 25,6 gam.(p n B)V d 9: (Cu 15 - M 231 - TSC - Khi A 2007)Cho 5,76 gam axit hu c X n chc, mch h tc dng ht vi CaCO3 thu c 7,28 gam mui ca axit hu c. Cng thc cu to thu gn ca X lA. CH2=CHCOOH.B. CH3COOH.C. HC CCOOH. D. CH3CH2COOH.Hng dn giit CTTQ ca axit hu c X n chc l RCOOH.2RCOOH+CaCO3 (RCOO)2Ca+CO2+H2OC 2 mol axit phn ng to mui th khi lng tng (40 2) = 38 gam. x mol axit (7,28 5,76) = 1,52 gam. x = 0,08 molRCOOH5,76M 720,08 R = 27Axit X: CH2=CHCOOH. (p n A)V d 10:Nhng thanh km vo dung dch cha 8,32 gam CdSO4. Sau khi kh hon ton ion Cd2+ khi lng thanh km tng 2,35% so vi ban u. Hi khi lng thanh km ban u.A. 60 gam. B. 70 gam. C. 80 gam. D. 90 gam.Hng dn gii66Gikhilngthanhkmbanulagamthkhilngtngthml 2,35a100 gam.Zn+CdSO4 ZnSO4 +Cd65 1 mol 112, tng (112 65) = 47 gam 8,32208 (=0,04 mol)2,35a100 gamTa c t l: 1 472,35a0,04100 a = 80 gam. (p n C)V d 11: Nhng thanh kim loi M ho tr 2 vo dung dch CuSO4, sau mt thi gianlythanhkimloi rathykhi lnggim0,05%. Mt khc nhng thanh kim loi trn vo dung dch Pb(NO3)2, sau mt thi gian thykhi lngtng 7,1%. Xc nh M, bit rng s mol CuSO4v Pb(NO3)2 tham gia 2 trng hp nh nhau.A. Al. B. Zn. C. Mg. D. Fe.Hng dn giiGi m l khi lng thanh kim loi, M l nguyn t khi ca kim loi, x l s mol mui phn ng.M+ CuSO4 MSO4 + CuM (gam) 1 mol64 gam, gim (M 64)gam. x molgim 0,05.m100 gam. x = 0,05. m100M 64 (1)M + Pb(NO3)2 M(NO3)2 + PbM (gam) 1 mol 207, tng (207 M) gamx mol tng 7,1. m100 gam67 x = 7,1.m100207 M (2)T (1) v (2) ta c:0,05. m100M 64 = 7,1.m100207 M (3)T (3) gii ra M = 65. Vy kim loi M l km. (p n B)V d 12:Cho 3,78 gam bt Al phn ng va vi dung dch mui XCl3to thnh dung dch Y. Khi lng cht tan trong dung dch Y gim 4,06 gam so vi dung dch XCl3. xc nh cng thc ca mui XCl3.A. FeCl3. B. AlCl3. C. CrCl3. D. Khng xc nh.Hng dn giiGi A lnguyn t khi ca kim loi X.Al +XCl3 AlCl3+X

3,7827 = (0,14 mol) 0,140,14 mol.Ta c : (A + 35,5 3) 0,14 (133,5 0,14) = 4,06Gii ra c: A = 56. Vy kim loi X l Fe v mui FeCl3. (p n A)V d 13: Nung 100 gam hn hp gm Na2CO3 v NaHCO3 cho n khi khi lng hn hp khng i c 69 gam cht rn. Xc nh phn trm khi lng ca mi cht tng ng trong hn hp ban u.A. 15,4% v 84,6%. B. 22,4% v 77,6%.C. 16% v 84%. D. 24% v 76%.Hng dn giiCh c NaHCO3 b phn hy. t x l s gam NaHCO3. 2NaHCO3 ot Na2CO3 + CO2 + H2OC nung168 gam khi lng gim: 44 + 18 = 62 gamx khi lng gim: 100 69 = 31 gamTa c: 168 62x 31x = 84 gam.68Vy NaHCO3 chim 84% v Na2CO3 chim 16%. (p n C)V d 14:Ha tan 3,28 gam hn hp mui CuCl2v Cu(NO3)2vo nc c dung dch A. Nhng Mg vo dung dch A cho n khi mt mu xanh ca dung dch. Ly thanh Mg ra cn li thy tng thm 0,8 gam. C cn dung dch sau phn ng thu c m gam mui khan. Tnh m?A. 1.28 gam. B. 2,48 gam. C. 3,1 gam. D. 0,48 gam.Hng dn giiTa c:mtng = mCu mMg phn ng =( ) 2 2 2Cu Mg Mgm m 3,28 m m 0,8gc axit+ + + + m = 3,28 0,8 = 2,48 gam. (p n B)V d 15: Ha tan 3,28 gam hn hp mui MgCl2v Cu(NO3)2vo nc c dung dch A. Nhng vo dung dch A mt thanh st. Sau mt khong thi gian ly thanh st ra cn li thy tng thm 0,8 gam. C cn dung dch sau phn ng thu c m gam mui khan. Gi tr m lA. 4,24 gam. B. 2,48 gam. C. 4,13 gam. D. 1,49 gam.Hng dn giip dng nh lut bo ton khi lng: Sau mt khong thi gian tng khi lng ca thanh Fe bng gim khi lng ca dung dch mui. Do :m = 3,28 0,8=2,48 gam. (p n B)MT S BI TP VN DNG GII THEO PHNG PHP TNG GIM KHI LNG01. Cho 115 gam hn hp gm ACO3, B2CO3, R2CO3 tc dng ht vi dung dch HCl thy thot ra 22,4 lt CO2(ktc). Khi lng mui clorua to ra trong dung dch lA. 142 gam. B. 126 gam. C. 141 gam. D. 132 gam.02. Ngm mt l st trong dung dch CuSO4. Nu bit khi lng ng bm trn l st l 9,6 gam th khi lng l st sau ngm tng thm bao nhiu gam so vi ban u?A. 5,6 gam. B. 2,8 gam. C. 2,4 gam. D. 1,2 gam.6903. Cho hai thanh st c khi lng bng nhau.- Thanh 1 nhng vo dung dch c cha a mol AgNO3.- Thanh 2 nhng vo dung dch c cha a mol Cu(NO3)2.Sau phn ng, ly thanh st ra, sy kh v cn li thy s cho kt qu no sau y?A. Khi lng hai thanh sau nhng vn bng nhau nhng khc ban u.B. Khi lng thanh 2 sau nhng nh hn khi lng thanh 1 sau nhng.C. Khi lng thanh 1 sau nhng nh hn khi lng thanh 2 sau nhng.D. Khi lng hai thanh khng i vn nh trc khi nhng.04. Cho V lt dung dch A cha ng thi FeCl3 1Mv Fe2(SO4)3 0,5M tc dng vi dung dch Na2CO3 c d, phn ng kt thc thy khi lng dung dch sau phn ng gim 69,2 gam so vi tng khi lng ca cc dung dch ban u. Gi tr ca V l:A. 0,2 lt.B. 0,24 lt. C. 0,237 lt. D.0,336 lt.05. Cho lung kh CO i qua 16 gam oxit st nguyn cht c nung nng trong mt ci ng. Khi phn ng thc hin hon ton v kt thc, thy khi lng ng gim 4,8 gam.Xc nh cng thc v tn oxit st em dng.06.DngCOkh40gam oxitFe2O3thuc33,92 gam chtrn Bgm Fe2O3, FeO v Fe. Cho 1B2tc dng vi H2SO4long d, thu c 2,24 lt kh H2 (ktc).Xc nh thnh phn theo s mol cht rn B, th tch kh CO (ktc) ti thiu c c kt qu ny.07. Nhng mt thanh st nng 12,2 gam vo 200 ml dung dch CuSO4 0,5M. Sau mt thi gian ly thanh kim loi ra, c cn dung dch c 15,52 gam cht rn khan.a)Vit phngtrnhphnngxyra, tmkhi lngtngcht ctrong 15,52 gam cht rn khan.b) Tnh khi lng thanh kim loi sau phn ng. Ha tan hon ton thanh kim loi ny trong dung dch HNO3 c nng, d thu c kh NO2 duy nht, th tch V lt (o 27,3 oC, 0,55 atm). Vit cc phng trnh phn ng xy ra. Tnh V.7008.Ngm mt thanh ng c khi lng 140,8 gam vo dung dch AgNO3sau mt thi gian ly thanh ng em cn li thy nng 171,2 gam. Tnh thnh phn khi lng ca thanh ng sau phn ng.09. Ngm mt l km nh trong mt dung dch c cha 2,24 gam ion kim loi c in tch 2+. Phn ng xong, khi lng l km tng thm 0,94 gam.Hy xc nh tn ca ion kim loi trong dung dch.10. C hai l kim loi cng cht, cng khi lng, c kh nng to ra hp cht c s oxi ha +2. Mt l c ngm trong dung dch Pb(NO3)2 cn l kia c ngm trong dung dch Cu(NO3)2.Sau mt thi gian ngi ta ly l kim loi ra khi dung dch, ra nh. Nhn thy khi lng l kim loi c ngm trong mui ch tng thm 19%, khi lnglkimloikiagim 9,6%. Bit rng, trong hai phn ng trn,khi lng cc kim loi b ha tan nh nhau.Hy xc nh tn ca hai l kim loi ang dng.p n cc bi tp vn dng:01. B 02. D. 03. B.04. A.05. Fe2O3. 06. VCO = 8,512 lt ;%nFe = 46,51% ;%nFeO = 37,21% ; 2 3Fe O%n 16, 28%. 07. a) 6,4 gam CuSO4 v 9,12 gam FeSO4.b) mKL = 12,68 gam ;2NOV 26,88 lt.08. Thanh Cu sau phn ng c mAg (bm) = 43,2 gam v mCu (cn li) = 128 gam.09. Cd2+10. CdPhng php 771QUI I HN HP NHIU CHT V S LNG CHT T HNMt s bi ton ha hc c th gii nhanh bng cc phng php bo ton electron, botonnguynt, botonkhi lngsongphngphpquyi cng tm ra p s rt nhanh v l phng php tng i u vit, c th vn dng vo cc bi tp trc nghim phn loi hc sinh.Cc ch khi p dng phng php quy i:1. Khi quy i hn hp nhiu cht (hn hp X) (t ba cht tr ln) thnh hn hp hai cht hay ch cn mt cht ta phi bo ton s mol nguyn t v bo ton khi lng hn hp.2. C th quy i hn hp X v bt k cp cht no, thm ch quy i v mt cht. Tuy nhin ta nn chn cp cht no n gin c t phn ng oxi ha kh nht n gin vic tnh ton.3. Trong qu trnh tnh ton theo phng php quy i i khi ta gp s m l do s b tr khi lng ca cc cht trong hn hp. Trong trng hp ny ta vn tnh ton bnh thng v kt qu cui cng vn tha mn.4. Khi quy i hn hp X v mt cht l FexOy th oxit FexOy tm c ch l oxit gi nh khng c thc.V d 1:Nung 8,4 gam Fe trong khng kh, sau phn ng thu c m gam cht rn X gm Fe, Fe2O3, Fe3O4, FeO. Ha tan m gam hn hp X vo dung dch HNO3d thu c 2,24 lt kh NO2(ktc) l sn phm kh duy nht. Gi tr ca m lA. 11,2 gam. B. 10,2 gam. C. 7,2 gam. D. 6,9 gam.Hng dn gii Quy hn hp X v hai cht Fe v Fe2O3:Ha tan hn hp X vo dung dch HNO3 d ta cFe + 6HNO3 Fe(NO3)3+ 3NO2+ 3H2O

0,13 0,1 molS mol ca nguyn t Fe to oxit Fe2O3 l72Fe8, 4 0,1 0,35n56 3 3 2 3Fe O0,35n3 2Vy:2 3X Fe Fe Om m m +X0,1 0,35m 56 1603 3 + = 11,2 gam. Quy hn hp X v hai cht FeO v Fe2O3:FeO + 4HNO3 Fe(NO3)3+NO2+ 2H2O 0,1 0,1 molta c:22 2 32Fe O 2FeO0,1 0,1 mol0,15 mol4Fe 3O 2Fe O0,05 0,025 mol + '+ 2h Xm= 0,1 72 + 0,025 160 = 11,2 gam. (p n A)Ch :Vn c th quy hn hp X v hai cht (FeO v Fe3O4) hoc (Fe v FeO), hoc (Fe v Fe3O4) nhng vic gii tr nn phc tp hn (c th l ta phi t n s mol mi cht, lp h phng trnh, gii h phng trnh hai n s). Quy hn hp X v mt cht l FexOy:FexOy + (6x2y)HNO3 Fe(NO3)3+ (3x2y) NO2 + (3xy)H2O

0,13x 2y mol 0,1 mol. Fe8, 4 0,1.xn56 3x 2y x 6y 7 mol.Vy cng thc quy i l Fe6O7 (M = 448) v6 7Fe O0,1n3 6 2 7 = 0,025 mol. mX = 0,025 448 = 11,2 gam.Nhn xt: Quy i hn hp gm Fe, FeO, Fe2O3, Fe3O4 v hn hp hai cht l FeO, Fe2O3 l n gin nht.73V d 2: Ha tan ht m gam hn hp X gm FeO, Fe2O3, Fe3O4 bng HNO3 c nng thu c 4,48 lt kh NO2(ktc). C cn dung dch sau phn ng thu c 145,2 gam mui khan gi tr ca m lA. 35,7 gam. B. 46,4 gam. C. 15,8 gam. D. 77,7 gam.Hng dn giiQuy hn hp X v hn hp hai cht FeO v Fe2O3 ta cFeO + 4HNO3 Fe(NO3)3 + NO2 + 2H2O0,2 mol 0,2 mol 0,2 molFe2O3 + 6HNO32Fe(NO3)3 + 3H2O0,2 mol 0,4 mol3 3Fe( NO )145, 2n242= 0,6 mol. mX = 0,2 (72 + 160) = 46,4 gam. (p n B)V d 3: Ha tan hon ton 49,6 gam hn hp X gm Fe, FeO, Fe2O3, Fe3O4 bng H2SO4 c nng thu c dung dch Y v 8,96 lt kh SO2 (ktc).a) Tnh phn trm khi lng oxi trong hn hp X.A. 40,24%. B. 30,7%. C. 20,97%. D. 37,5%.b) Tnh khi lng mui trong dung dch Y.A. 160 gam. B.140 gam. C. 120 gam. D. 100 gam.Hng dn giiQuy hn hp X v hai cht FeO, Fe2O3, ta c:2 4 2 4 3 2 22 3 2 4 2 4 3 22FeO 4H SO Fe (SO ) SO 4H O0,8 0, 4 0, 4 mol49,6 gamFe O 3H SO Fe (SO ) 3H O0,05 0,05 mol + + + '+ + 2 3Fe Om= 49,6 0,8 72 = 8 gam (0,05 mol) nO (X) = 0,8 + 3 (0,05) = 0,65 mol.74Vy: a) O0,65 16 100%m49,9 = 20,97%. (p n C) b) 2 4 3Fe (SO )m= [0,4 + (-0,05)] 400 = 140 gam. (p n B)V d 4: kh hon ton 3,04 gam hn hp X gm FeO, Fe2O3, Fe3O4 th cn 0,05molH2.Mtkhc ha tan hon ton 3,04 gam hn hp X trong dung dch H2SO4 c nng th thu c th tch kh SO2 (sn phm kh duy nht ktc) l.A. 224 ml. B. 448 ml. C. 336 ml. D. 112 ml.Hng dn giiQuy hn hp X v hn hp hai cht FeO v Fe2O3 vi s mol l x, y, ta c:FeO+H2 ot Fe+H2O x yFe2O3+3H2 ot 2Fe+3H2O x3yx 3y 0,0572x 160y 3,04+ '+ x 0,02 moly 0,01 mol'2FeO+4H2SO4 Fe2(SO4)3+SO2+4H2O 0,02 0,01 molVy:2SOV= 0,01 22,4 = 0,224 lt(hay 224 ml). (p n A)V d 5: Nung m gam bt st trong oxi, thu c 3 gam hn hp cht rn X. Ha tan ht hn hp X trong dung dch HNO3(d) thot ra 0,56 lt NO ( ktc) (l sn phm kh duy nht). Gi tr ca m lA. 2,52 gam. B. 2,22 gam. C. 2,62 gam. D. 2,32 gam.Hng dn giiQuy hn hp cht rn X v hai cht Fe, Fe2O3:Fe + 4HNO3 Fe(NO3)3 + NO+ 2H2O0,025 0,0250,025 mol2 3Fe Om= 3 56 0,025 = 1,6 gam752 3Fe ( trong Fe O )1,6m 2160 = 0,02 mol mFe = 56 (0,025 + 0,02) = 2,52 gam. (p n A)V d 6: Hn hp X gm (Fe, Fe2O3, Fe3O4, FeO) vi s mol mi cht l 0,1 mol, ha tan ht vo dung dch Y gm (HCl v H2SO4long) d thu c dung dch Z. Nh t t dung dch Cu(NO3)2 1M vo dung dch Z cho ti khi ngng thot kh NO. Th tch dung dch Cu(NO3)2cn dng v th tch kh thot ra ktc thuc phng n no?A. 25 ml; 1,12 lt. B. 0,5 lt; 22,4 lt.C. 50 ml; 2,24 lt.D. 50 ml; 1,12 lt.Hng dn giiQuy hn hp 0,1 mol Fe2O3 v 0,1 mol FeO thnh 0,1 mol Fe3O4. Hn hp X gm:Fe3O4 0,2 mol; Fe 0,1 mol + dung dch YFe3O4 + 8H+ Fe2++2Fe3++4H2O0,2 0,2 0,4 molFe+2H+ Fe2++H2 0,1 0,1 molDung dch Z: (Fe2+: 0,3 mol; Fe3+: 0,4 mol) + Cu(NO3)2:3Fe2++NO3+4H+ 3Fe3++NO+2H2O 0,3 0,1 0,1 mol VNO =0,1 22,4 = 2,24 lt.3 23Cu( NO )NO1n n2 = 0,05 mol. 23 2d Cu( NO )0,05V1 = 0,05 lt(hay 50 ml). (p n C)V d 7: Nung 8,96 gam Fe trong khng kh c hn hp A gm FeO, Fe3O4, Fe2O3. A ha tan va vn trong dung dch cha 0,5 mol HNO3, bay ra kh NO l sn phm kh duy nht. S mol NO bay ra l.A. 0,01. B. 0,04. C. 0,03. D. 0,02.76Hng dn giiFe8,96n 0,1656 molQuy hn hp A gm (FeO, Fe3O4, Fe2O3) thnh hn hp (FeO, Fe2O3) ta c phng trnh:2Fe+O2 2FeOx x4Fe+3O2 2Fe2O3yy/23FeO+ 10HNO3 3Fe(NO3)3+NO +2H2Ox 10x/3 x/3Fe2O3+6HNO3 2Fe(NO3)3+3H2O y/2 3yH phng trnh:x y 0,1610x3y 0,53+ '+ x 0,06 moly 0,1 mol'NO0,06n 0,023 mol. (p n D)77Phng php 8S NG CHOBi ton trn ln cc cht vi nhau l mt dng bi tp hay gp trong chng trnh ha hc ph thng cng nh trong cc thi kim tra v thi tuyn sinh i hc, cao ng. Ta c th gii bi tp dng ny theo nhiu cch khc nhau, song vic gii loi dng bi tp ny theo phng php s ng cho theo tc gi l tt nht.Nguyn tc: Trn ln hai dung dch:Dung dch 1: c khi lng m1, th tch V1, nng C1 (nng phn trm hoc nng mol), khi lng ring d1.Dung dch 2: c khi lng m2, th tch V2, nng C2 (C2 > C1 ), khi lng ring d2.Dung dch thu c: c khi lng m = m1 + m2, th tch V = V1 + V2, nng C (C1 < C < C2) v khi lng ring d.S ng cho v cng thc tng ng vi mi trng hp l:a. i vi nng % v khi lng:2 12 1C C mm C C(1)b. i vi nng mol/lt:2 12 1C C VV C C(2)c. i vi khi lng ring:2 12 1C C VV C C(3)Khi s dng s ng cho cn ch :- Cht rn coi nh dung dch c C = 100%- Dung mi coi nh dung dch c C = 0%- Khi lng ring ca H2O l d = 1g/ml.78C1C2C| C2 - C || C1 - C |C| C2 - C || C1 - C |`CM1CM2d1d2| d2 - d || d1 - d |dSau y l mt s v d s dng phng php s ng cho trong tnh ton cc bi tp.V d 1: thu c dung dch HCl 25% cn ly m1gam dung dch HCl 45% pha vi m2 gam dung dch HCl 15%. T l m1/m2 lA. 1:2. B. 1:3. C. 2:1. D. 3:1.Hng dn giip dng cng thc (1):1245 25 m 20 2m 15 25 10 1 .(p n C)V d 2: pha c 500 ml dung dch nc mui sinh l (C = 0,9%) cn ly V ml dung dch NaCl 3% pha vi nc ct. Gi tr ca V lA. 150 ml. B. 214,3 ml. C. 285,7 ml. D. 350 ml.Hng dn giiTa c s : V1 = 0,95002,1 0,9+ = 150 ml.(p n A)V d 3: Ha tan 200 gam SO3 vo m2 gam dung dch H2SO4 49% ta c dung dch H2SO4 78,4%. Gi tr ca m2 lA. 133,3 gam. B. 146,9 gam. C. 272,2 gam. D. 300 gam.Hng dn giiPhng trnh phn ng:SO3+H2O H2SO4 100 gam SO3 98 10080 = 122,5 gam H2SO4.Nng dung dch H2SO4 tng ng 122,5%.Gi m1, m2 ln lt l khi lng ca SO3 v dung dch H2SO4 49% cn ly. Theo (1) ta c:79V1 (NaCl)V2 (H2O)0,930| 0,9 - 0 || 3 - 0,9 |1249 78, 4 m 29,4m 122,5 78,4 44,1 244,1m 20029, 4 = 300 gam.(p n D)V d 4: Nguyn t khi trung bnh ca brom l 79,319. Brom c hai ng v bn: 7935Brv 8135Br . Thnh phn % s nguyn t ca 8135Br lA. 84,05. B. 81,02. C. 18,98. D. 15,95.Hng dn giiTa c s ng cho:81357935% Br 0,319% Br 1,68181350,319% Br1,681 0,319+ 100% = 15,95%. (p n D)V d 5: Mt hn hp gm O2, O3 iu kin tiu chun c t khi hi vi hiro l 18. Thnh phn % v th tch ca O3 trong hn hp lA. 15%. B. 25%. C. 35%. D. 45%.Hng dn giip dng s ng cho:32OOV4 1V 12 3 3O1%V3 1+ 100% = 25%.(p n B)8081357935Br (M 81) 79,319 79 0,319A 79,319Br (M 79) 81 79,319 1,681 32OOV M 48 32 36M 18 2 36V M 32 48 36 V d 6: Cn trn hai th tch metan vi mt th tch ng ng X ca metan thu c hn hp kh c t khi hi so vi hiro bng 15. X lA. C3H8. B. C4H10. C. C5H12. D. C6H14.Hng dn giip dng s ng cho:42CH 2MV M 30 2V 14 1 M2 30= 28 M2 = 58 14n + 2 = 58 n = 4.Vy: X l C4H10.(p n B)V d 7: Thm 250 ml dung dch NaOH 2M vo 200 ml dung dch H3PO4 1,5M. Mui to thnh v khi lng tng ng lA. 14,2 gam Na2HPO4 ;32,8 gam Na3PO4.B. 28,4 gam Na2HPO4 ;16,4 gam Na3PO4.C. 12 gam NaH2PO4 ;28,4 gam Na2HPO4.D. 24 gam NaH2PO4;14,2 gam Na2HPO4.Hng dn giiC:3 4NaOHH POn 0, 25 2 51 2n 0, 2 1,5 3< 1 : 4.Hng dn giiTrn a mol AlCl3 vi b mol NaOH thu c kt ta th33 2 232 2Al3 3OH Al(OH)Al(OH) OH AlO 2H OAl 4OH AlO 2H Oa 4 mol+ + + +'+ ++ + kt ta tan hon ton th 3OHAlnn+ 4 ba 4.Vy c kt ta th ba < 4 a : b > 1 : 4. (p n D)87V d 4: (Cu 37 - M 182 - Khi A - TSH 2007)t chy hon ton a mol axit hu c Y c 2a mol CO2. Mt khc, trung ha a mol Y cn va 2a mol NaOH. Cng thc cu to thu gn ca Y lA. HOOCCH2CH2COOH. B. C2H5COOH.C. CH3COOH. D. HOOCCOOH.Hng dn gii- t a mol axit hu c Y c 2a mol CO2 axit hu c Y c hai nguyn t C trong phn t.- Trung ha a mol axit hu c Y cn dng 2a mol NaOH axit hu c Y c 2 nhm chc cacboxyl (COOH).Cng thc cu to thu gn ca Y l HOOCCOOH. (p n D)V d 5: (Cu 39 - M 182 - Khi A - TSH 2007)Dung dch HCl v dung dch CH3COOH c cng nng mol/l, pH ca hai dung dch tng ng l x v y. Quan h gia x v y l (gi thit, c 100 phn t CH3COOH th c 1 phn t in li)A. y = 100x. B. y = 2x. C. y = x 2. D. y = x + 2.Hng dn giipHHCl = x [H+]HCl = 10x3CH COOHpH y 3yCH COOH[H ] 10+ Ta c: HCl H++Cl 10x 10x (M)CH3COOH H++CH3COO100.10y 10y (M).Mt khc: [HCl] = [CH3COOH] 10x = 100.10y y = x + 2. (p n D)V d 6: (Cu 53 - M 182 - Khi A - TSH 2007) thu ly Ag tinh khit t hn hp X (gm a mol Al2O3, b mol CuO, c mol Ag2O), ngi ta ho tan X bi dung dch cha (6a + 2b + 2c) mol 88HNO3 c dung dch Y, sau thm (gi thit hiu sut cc phn ng u l 100%)A. c mol bt Al vo Y. B. c mol bt Cu vo Y.C. 2c mol bt Al vo Y. D. 2c mol bt Cu vo Y.Hng dn giiHa tan hn hp X trong dung dch HNO3Al2O3+6HNO3 2Al(NO3)3+3H2Oa 6a 2a molCuO+2HNO3 Cu(NO3)2+H2O b2bb molAg2O+2HNO3 2AgNO3+H2Oc 2c2c molDung dch HNO3va .DungdchYgm2a mol Al(NO3)3, bmol Cu(NO3)2,2c mol AgNO3. thu Ag tinh khit cn cho thm kim loi Cu vo phng trnhCu+2AgNO3 Cu(NO3)2+2Agc mol 2cVy cn c mol bt Cu vo dung dch Y. (p n B)V d 7: (Cu 32 - M 285 - Khi B - TSH 2007)in phn dung dch cha a mol CuSO4v b mol NaCl (vi in cc tr, c mng ngn xp). dung dch sau in phn lm phenolphtalein chuyn sang mu hng th iu kin ca a v b l (bit ion SO42 khng b in phn trong dung dch)A. b > 2a. B. b = 2a. C. b < 2a. D. 2b = a.Hng dn giiPhng trnh in phn dung dchCuSO4+2NaClpdd Cu+Cl2+Na2SO4(1) a 2a mol89Dung dch sau in phn lm phenolphtalein chuyn sang mu hng sau phn ng (1) th dung dch NaCl cn d v tip tc b in phn theo phng trnh2NaCl+2H2Opddmng ngn 2NaOH+H2+Cl2(2)Vy: b > 2a. (p n A)Ch : Tng t cng cu hi trn chng ta c th hi:+ dung dch sau in phn c mi trng axit th iu kin ca a v b l.A. b > 2a. B. b = 2a. C. b < 2a. D. a = 2b.+ dung dch sau in phn c kh nng ha tan kt ta Al(OH)3th iu kin ca a, b lA. b > 2a. B. b < 2a. C. b 2a. D. b 2a.V d 8: t chy hon ton a mol mt anehit X (mch h) to ra b mol CO2 v c mol H2O (bit b = a + c). Trong phn ng trng gng, mt phn t X ch cho 2 electron. X thuc dy ng ng anehitA. no, n chc.B. khng no c hai ni i, n chc.C. khng no c mt ni i, n chc.D. no, hai chc.Hng dn giiTrong phn ng trng gng mt anehit X ch cho 2e X l anehit n chc bi v:1RCHO+ 34RCOONH+trong : C+1 2e C+3.t cng thc phn t ca anehit n chc X l CxHyO ta c phng trnhCxHyO+2y 1x O4 2| `+ . ,xCO2+y2H2O a a.x a.y2 mol(b mol)(c mol) 90Ta c: b = a + c ax = a + a.y2 y = 2x 2.Cng thc tng qut ca anehit n chc Xl CxH2x2Oc dng Cx1H2(x1)1CHO l anehit khng no c mt lin kt i, n chc. (p n C)V d 9: Cng thc phn t ca mt ancol A l CnHmOx. cho A l ancol no th m phi c gi trA. m = 2n. B. m = 2n + 2.C. m = 2n 1. D. m = 2n + 1.Hng dn giiTheophngphpng nht h s: Cng thc tng qut ca ancol no l CnH2n+2-x(OH)x hay CnH2n+2Ox. Vy m = 2n+2. (p n B)V d 10:Hi t l th tch CO2 v hi nc (T) bin i trong khong no khi t chy hon ton cc ankin.A. 1 < T 2. B. 1 T < 1,5.C. 0,5 < T 1. D. 1 < T < 1,5.Hng dn giiCnH2n-2 nCO2+(n 1)H2Oiu kin: n 2 v n N.T = 22COH Onn = n 1.1n 11nVi mi n 2 T > 1; mt khc n tng T gim. n = 2 T = 2 l gi tr ln nht.Vy: 1 < T 2. (p n A)V d 11: t chy 1 mol aminoaxitNH2(CH2)nCOOH phi cn s mol O2 lA. 2n 3.2+B. 6n 3.2+C. 6n 3.4+D. 2n 3.4+Hng dn gii91Phng trnh t chy amino axit lH2N(CH2)nCOOH+6n 34+O2 (n + 1)CO2+2n 32+H2O (p n C)V d 12: Mt dung dch hn hp cha a mol NaAlO2 v a mol NaOH tc dng vi mt dung dch cha b mol HCl. iu kin thu c kt ta sau phn ng lA. a = b. B. a = 2b. C. b = 5a. D. a < b < 5a.Hng dn giiPhng trnh phn ng:NaOH+HCl NaCl+H2O (1)a mol a molNaAlO2+HCl+H2O Al(OH)3+NaCl (2)Al(OH)3+3HCl AlCl3+3H2O (3)NaAlO2+4HCl AlCl3+NaCl+2H2O (4)a mol 4a moliu kin khng c kt ta khinHCl2NaAlO4n+ nNaOH= 5a.Vy suy ra iu kin c kt ta:nNaOH < nHCl < 2NaAlO4n+ nNaOH a < b < 5a. (p n D)V d 13: Dung dch cha a mol NaOH tc dng vi dung dch cha b mol H3PO4 sinh ra hn hp Na2HPO4 + Na3PO4. T s ab lA. 1 < ab < 2. B. ab 3.C. 2 < ab < 3. D. ab 1.Hng dn giiCc phng trnh phn ng:92NaOH+H3PO4 NaH2PO4+H2O (1)2NaOH+H3PO4 Na2HPO4+2H2O (2)3NaOH+H3PO4 Na3PO4+3H2O (3)Ta c: nNaOH = a mol ;3 4H POn= b mol. thu c hn hp mui Na2HPO4 + Na3PO4 th phn ng xy ra c hai phng trnh (2 v 3), do :2 < 3 4NaOHH POnn < 3, tc l 2 < ab < 3. (p n C)V d 14: Hn hp X gm Na v Al.- Th nghim 1: Nu cho m gam X tc dng vi H2O d th thu c V1 lt H2.- Th nghim 2: nu cho m gam X tc dng vi dung dch NaOH d th thu c V2 lt H2.Cc kh o cng iu kin. Quan h gia V1 v V2 lA. V1 = V2. B. V1 > V2. C. V1 < V2. D. V1 V2. Hng dn giiCc phng trnh phn ng khi ha tan hn hp Na v Al vi H2O v vi dung dch NaOH d:Na+H2O NaOH+ 12H2(1)2Al+6H2O+2NaOH Na[Al(OH)4]+ 3H2(2)t s mol Na v Al ban u ln lt l x v y (mol).TN1: x y nNaOH va hoc d khi ha tan Al c hai th nghim cng to thnh x 3x2 2| `+ . , mol H2. V1 = V2.TN2: x < y trong TN1 (1) Al d, TN2 (2) Al tan ht 2 2H ( TN2) H (TN2)n n . > V2 > V1.Nh vy (x,y > 0) th V2 V1. (p n D)93V d 15: Mt bnh kn cha V lt NH3 v V lt O2 cng iu kin. Nung nng bnh c xc tc NH3 chuyn ht thnh NO, sau NO chuyn ht thnh NO2. NO2vlngO2cn li trong bnh hp th va vn ht trong nc thnh dung dch HNO3. T s VV lA. 1. B. 2. C. 3. D. 4.Hng dn giiCc phng trnh phn ng:4NH3+5O2 oxtt4NO+6H2O V 5V/4V2NO+O2

2NO2 VV/2 V4NO2 + O2+2H2O 4HNO3V 5V VV4 2| ` . , V = 45V VV4 2| ` . , VV = 2. (p n B)V d 16: Cht X c khi lng phn t l M. Mt dung dch cht X c nng a mol/l, khi lng ring d gam/ml. Nng C% ca dung dch X lA. a.M10d.B. d.M10a. C. 10aM.d. D. a.M1000d.Hng dn giiXt 1 lt dung dch cht X: nX = a mol mX = a.M mdd X = a.M.100C% = 1000d C% = a.M10d. (p n A)94V d 17: Hn hp X c mt s ankan. t chy 0,05 mol hn hp X thu c a mol CO2 v b mol H2O. Kt lun no sau y l ng?A. a = b. B. a = b 0,02.C. a = b 0,05. D. a = b 0,07.Hng dn giit cng thc tng qut ca 1 s ankan l x 2x 2C H+x 2x 2C H+ + 23x 1O2+ x CO2 + (x 1) +H2O0,5 0,05 x0,05(x 1) +mol0,05x a0,05(x 1) b'+ a = b 0,05.(p n C)V d 18: (Cu 40 - M 285 - Khi B - TSH 2007)Thc hin hai th nghim:1) Cho 3,84 gam Cu phn ng vi 80 ml dung dch HNO3 1M thot ra V1 lt NO.2) Cho 3,84 gam Cu phn ng vi 80 ml dung dch cha HNO3 1M v H2SO4 0,5 M thot ra V2 lt NO.Bit NO l snphm kh duy nht, cc th tch kh o cng iu kin. Quan h gia V1 v V2 lA. V2 = V1. B. V2 = 2V1. C. V2 = 2,5V1. D. V2 = 1,5V1.Hng dn giiTN1: 3CuHNO3,84n 0,06 mol64n 0,08 mol ' 3HNOn 0,08 moln 0,08 mol+'3Cu+8H++2NO3 3Cu2++2NO+4H2Ou bi: 0,060,08 0,08 H+ phn ng htPhn ng: 0,03 0,08 0,02 0,02 molV1 tng ng vi 0,02 mol NO.TN2: nCu = 0,06 mol ;3HNOn 0,08 mol ;2 4H SOn 0,04 mol. 95Tng Hn+ = 0,16 mol ;3NOn= 0,08 mol.3Cu+8H++2NO3 3Cu2++2NO+4H2Ou bi: 0,06 0,160,08 Cu v H+ phn ng htPhn ng: 0,06 0,16 0,04 0,04 molV2 tng ng vi 0,04 mol NO.Nh vy V2 = 2V1. (p n B) MT S BI TP VN DNG GII THEO PHNG PHP CC I LNG DNG TNG QUT01. Dung dch A c a mol NH4+, b mol Mg2+, c mol SO42 v d mol HCO3. Biu thc no biu th s lin quan gia a, b, c, d sau y l ng?A. a + 2b =c + d. B. a + 2b = 2c + d.C. a + b = 2c + d. D. a + b =c+ d.02. Cho a mol Fe vo dung dch cha b mol dung dch AgNO3.