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MC LC LI 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 ..................................................... 97

Phn 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 ................................................................................................... 277 s 25 ................................................................................................... 284

Phn th ba: P N 25 THI TH TUYN SINH I HC, CAO NG 291

p n 01 ............................................................................................ 291p n 02 ............................................................................................ 291

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p 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 ............................................................................................ 297

LI 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 ton mt 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.

Cu 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 nhanh trong vng t 1-2 pht chng ta phi bit phn loi v nm chc cc phng php suy lun. Vic gii bi tp trc nghim khng nht thit phi theo ng qui trnh cc bc gii, khng nht thit phi s dng ht cc d kin u bi v i khi khng cn vit 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.

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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 2008

Phn th nht 10 PHNG PHP GII NHANH BI TP TRC NGHIM

HA HC

Phng php 1 P DNG NH LUT BO TON KHI LNG

Nguyn 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 ng m gam hn hp X nung

nng. Sau khi kt thc th nghim thu 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 gii

Cc phn ng kh st oxit c th c: 3Fe2O3 + CO

ot 2Fe3O4 + CO2 (1) Fe3O4 + CO

ot 3FeO + CO2 (2) FeO + CO

ot Fe + CO2 (3) Nh vy cht rn A c th gm 3 cht Fe, FeO, Fe3O4 hoc 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:

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44x + 28(0,5 x) = 0,5 20,4 2 = 20,4 nhn c x = 0,4 mol v cng chnh l s mol CO tham gia phn ng.

Theo LBTKL ta c: mX + mCO = mA +

2COm

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 gii Ta bit rng c 3 loi ru tch nc iu kin H2SO4 c, 140oC th to thnh 6 loi ete v tch ra 6

phn t H2O. Theo LBTKL ta c

2H O etem m m 132,8 11,2 21,6 ru gam

2H O

21,6n 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 1,2 0,26

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 gii Fe + 6HNO3 Fe(NO3)3 + 3NO2 + 3H2O Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O

2NOn 0,5 mol

3 2HNO NOn 2n 1 mol.

p dng nh lut bo ton khi lng ta c:

2 23

NOd HNOm m m m

1 63 10012 46 0,5 89 gam.63

2 2d mui h k.loi

t nFe = x mol, nCu = y mol ta c:

56x 64y 123x 2y 0,5

x 0,1y 0,1

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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 gii

M2CO3 + 2HCl 2MCl + CO2 + H2O R2CO3 + 2HCl 2MCl2 + CO2 + H2O

2CO4,88n 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 + 0,436,5 = mmui + 0,244 + 0,218 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 K2CO3 0,5M (va ) thu c kt ta C v dung dch D. Lng KCl trong dung dch D nhiu gp 22/3 ln lng KCl c trong A. % khi lng KClO3 c trong A l

A. 47,83%. B. 56,72%. C. 54,67%. D. 58,55%. Hng dn gii

o

o

o

2

t3 2

t3 2 2 2

t2 2 2 2

2 2

(A) (A)

h B

3KClO KCl O (1)2

Ca(ClO ) CaCl 3O (2)

83,68 gam A Ca(ClO ) CaCl 2O (3)CaCl CaClKCl KCl

2O

n 0,78 mol. p dng nh lut bo ton khi lng ta c: mA = mB + 2Om

mB = 83,68 320,78 = 58,72 gam. Cho cht rn B tc dng vi 0,18 mol K2CO3

Hn hp B 2 2 3 3

(B) (B)

CaCl K CO CaCO 2KCl (4)0,18 0,18 0,36 molKCl KCl

hn hp D

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( B) 2KCl B CaCl (B)m m m58,72 0,18 111 38,74 gam

( D )KCl KCl (B) KCl (pt 4)

m m m

38,74 0,36 74,5 65,56 gam

( A ) ( D )KCl KCl

3 3m m 65,56 8,94 gam22 22

(B) (A)KCl pt (1) KCl KClm = m m 38,74 8,94 29,8 gam.

Theo phn ng (1):

3KClO29,8m 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 gii

1,88 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: 444a + 183a = 46 a = 0,02 mol. Trong cht A c: nC = 4a = 0,08 mol

nH = 3a2 = 0,12 mol nO = 4a2 + 3a 0,0852 = 0,05 mol nC : nH : no = 0,08 : 0,12 : 0,05 = 8 : 12 : 5 Vy 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 ton vi NaOH thu c 6,4 gam ru v mt lng mi c khi lng nhiu hn lng este l 13,56% (so vi lng este). Xc nh cng thc cu to ca este.

A. CH3COO CH3. B. CH3OCOCOOCH3. C. CH3COOCOOCH3. D. CH3COOCH2COOCH3.

Hng dn gii R(COOR)2 + 2NaOH R(COONa)2 + 2ROH 0,1 0,2 0,1 0,2 mol

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R OH6,4M 320,2

Ru CH3OH. p dng nh lut bo ton khi lng ta c: meste + mNaOH = mmui + mru mmui meste = 0,240 64 = 1,6 gam.

m mmui meste = 13,56100 meste

meste = 1,6 100 11,8 gam13,56 Meste = 118 vC

R + (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 v C2H5COOCH3, B. C2H5COOCH3 v CH3COOC2H5. C. HCOOC3H7 v C2H5COOCH3. D. C B, C u ng.

Hng dn gii t cng thc trung bnh tng qut ca hai este n chc ng phn l RCOOR . RCOOR + NaOH RCOONa + ROH 11,44 11,08 5,56 gam p dng nh lut bo ton khi lng ta c: MNaOH = 11,08 + 5,56 11,44 = 5,2 gam

NaOH 5,2n 0,13 mol40

RCOONa 11,08M 85,230,13 R 18,23

R OH 5,56M 42,770,13 R 25,77

RCOOR 11,44M 880,13

CTPT ca este l C4H8O2 Vy cng thc cu to 2 este ng phn l: HCOOC3H7 v C2H5COOCH3

hoc C2H5COOCH3 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.

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- 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 l

A. 1,434 lt. B. 1,443 lt. C. 1,344 lt. D. 0,672 lt. Hng dn gii

Phn 1: V anehit no n chc nn 2 2CO H O

n n = 0,06 mol.

2CO Cn n 0,06(phn 2) (phn 2) mol.

Theo bo ton nguyn t v bo ton khi lng ta c: C C (A)n n 0,06(phn 2) mol.

2CO (A)n = 0,06 mol

2CO

V = 22,40,06 = 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 nng 4,784 gam. Kh i ra khi ng s cho hp th vo dung dch Ba(OH)2 d th thu c 9,062 gam kt ta. Phn trm khi lng Fe2O3 trong hn hp A l

A. 86,96%. B. 16,04%. C. 13,04%. D.6,01%. Hng dn gii

0,04 mol hn hp A (FeO v Fe2O3) + CO 4,784 gam hn hp B + CO2. CO2 + Ba(OH)2 d BaCO3 + H2O

2 3CO BaCOn n 0,046 mol

v 2CO( ) CO

n n 0,046 molp. p dng nh lut bo ton khi lng ta c: mA + mCO = mB +

2COm

mA = 4,784 + 0,04644 0,04628 = 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 72 101 13,04%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 LNG

01. 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 l

A. 31,45 gam. B. 33,99 gam. C. 19,025 gam. D. 56,3 gam.

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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 l A. 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 l

A. 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 l A. 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 l A. 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 l A. 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 l

A. 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 l

A. 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 l A. 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. D 6. B 7. D 8. a-D, b-B 9. B 10. A

Phng php 2 BO TON MOL NGUYN T

C 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 gp nhiu phng trnh phn ng li lm mt, qui gn vic tnh ton v nhm nhanh p s. Rt ph hp vi vic gii cc dng bi ton ha hc trc nghim.

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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 H2SO4 c thu c th tch kh SO2 (sn phm kh duy nht) iu kin tiu chun l

A. 448 ml. B. 224 ml. C. 336 ml. D. 112 ml. Hng dn gii

Thc cht phn ng kh cc oxit trn l H2 + O H2O 0,05 0,05 mol

t 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)

Fe 3,04 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 + 4H2O x x/2 2Fe3O4 + 10H2SO4 3Fe2(SO4)3 + SO2 + 10H2O y y/2 tng: SO2 x y 0,2n 0,01 mol2 2

Vy:

2SOV 224 ml. (p n B)

V 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 gii Thc cht phn ng kh cc oxit trn l CO + 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.

O 0,32n 0,02 mol16

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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 H2 qua 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 l


2hh (CO H )

2,24n 0,1 mol22,4

Thc cht phn ng kh cc oxit l: CO + O CO2 H2 + O H2O.

Vy: 2O CO H

n 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 l


CnH2n+1CH2OH + CuO ot CnH2n+1CHO + Cu + H2O

Khi lng cht rn trong bnh gim chnh l s gam nguyn t O trong CuO phn ng. Do nhn c:

mO = 0,32 gam O 0,32n 0,02 mol16

Hn hp hi gm: n 2n 12

C H CHO : 0,02 molH O : 0,02 mol.

Vy hn hp hi c tng s mol l 0,04 mol. C M = 31

mhh hi = 31 0,04 = 1,24 gam. mancol + 0,32 = mhh hi

mancol = 1,24 0,32 = 0,92 gam. (p n A) Ch : Vi ru bc (I) hoc ru bc (II) u tha mn u bi.

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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.

A. 0,5 lt. B. 0,7 lt. C. 0,12 lt. D. 1 lt. Hng dn gii

mO = moxit mkl = 5,96 4,04 = 1,92 gam. O

1,92n 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 mol HCl 0,24V 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 l


Axit 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 0,12 + nO (p.) = 0,32 + 0,21 nO (p.) = 0,6 mol

2On 0,3 mol

2O

V 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. Cng thc ca oxit st v phn trm th tch ca kh CO2 trong hn hp kh sau phn ng l

A. FeO; 75%. B. Fe2O3; 75%. C. Fe2O3; 65%. D. Fe3O4; 65%.

Hng dn gii FexOy + yCO xFe + yCO2

Kh thu c c M 40 gm 2 kh CO2 v CO d

2COCO

n 3n 1

2CO

%V 75% .

Mt khc: 2CO ( ) CO

75n n 0,2 0,15100p.

mol nCO d = 0,05 mol.

2CO

CO

n 44 1240

n 28 4

13

Thc cht phn ng kh oxit st l do CO + O (trong oxit st) CO2

nCO = nO = 0,15 mol mO = 0,1516 = 2,4 gam mFe = 8 2,4 = 5,6 gam nFe = 0,1 mol.

Theo phng trnh phn ng ta c:

2

Fe

CO

n 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 bng oxi d thu 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 l


Hng dn gii Gi M l kim loi i din cho ba kim loi trn vi ho tr l n. M + n

2O2 M2On (1)

M2On + 2nHCl 2MCln + nH2O (2) Theo phng trnh (1) (2)

2HCl On 4.n .

p dng nh lut bo ton khi lng 2O

m 44,6 28,6 16 gam

2On 0,5 mol nHCl = 40,5 = 2 mol

Cl

n 2 mol mmui = mhhkl + Clm = 28,6 + 235,5 = 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 gii

Hn hp A 2 3

FeO : 0,01 molFe O : 0,03 mol + CO 4,784 gam B (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 2H

n 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)

VNMATHS.TK

14

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 + 0,032 = 0,07 mol nFe (B) = a + 2b + c + 3d

a + 2b + c + 3d = 0,07 (4) T (1, 2, 3, 4) b = 0,006 mol c = 0,012 mol d = 0,006 mol. (p n A)

V 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 l


mO (trong oxit) = moxit mkloi = 24 17,6 = 6,4 gam. 2O H Om 6,4 gam ; 2H O

6,4n 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?


Fe3O4 (FeO, Fe) 3Fe2+ n mol

24 4Fe trong FeSO SOn n 0,3 mol p dng nh lut bo ton nguyn t Fe: 43 4 Fe FeSOFe Fe On n 3n = 0,3 n = 0,1

3 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 gii t cng thc tng qut ca mt trong ba ete l CxHyO, ta c: C

0,72m 12 0,4844

gam ; H 0,72m 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.

15

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 T

01. Ha tan hon ton hn hp X gm 0,4 mol FeO v 0,1mol Fe2O3 vo dung dch HNO3 long, 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 l


02. Cho kh CO i qua ng s cha 16 gam Fe2O3 un nng, sau phn ng thu c hn hp rn X gm Fe, FeO, Fe3O4, Fe2O3. Ha tan hon ton X bng H2SO4 c, nng thu c dung dch Y. C cn dung dch Y, lng mui khan thu c l

A. 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 l


04. t chy hn hp hirocacbon X thu c 2,24 lt CO2 (ktc) v 2,7 gam H2O. Th tch O2 tham gia phn ng chy (ktc) l

A. 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 dch 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 l


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 l

A. 8,4%. B. 16,8%. C. 19,2%. D. 22,4%.

08. (Cu 2 - M 231 - TSC - Khi A 2007) t chy hon ton mt th tch kh thin nhin gm metan, 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 l

A. 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 lt

VNMATHS.TK

16

10. 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 l


p n cc bi tp vn dng: 1. D 2. C 3. C 4. D 5. C 6. C 7. B 8. A 9. C 10. C

Phng php 3 BO TON MOL ELECTRON

Trc 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. lt

Hng dn gii 1. Cc phn ng c th c: 2Fe + O2

ot 2FeO (1) 2Fe + 1,5O2

ot Fe2O3 (2) 3Fe + 2O2

ot Fe3O4 (3) Cc phn ng ha tan c th c: 3FeO + 10HNO3 3Fe(NO3)3 + NO + 5H2O (4) Fe2O3 + 6HNO3 2Fe(NO3)3 + 3H2O (5)

17

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 0,039

56 mol.

trong , n l s mol NO thot ra. Ta d dng rt ra n = 0,001 mol;

VNO = 0,00122,4 = 0,0224 lt = 22,4 ml. (p n B) 2. Cc phn ng c th c: 2Al + 3FeO

ot 3Fe + Al2O3 (7) 2Al + Fe2O3

ot 2Fe + Al2O3 (8) 8Al + 3Fe3O4

ot 9Fe + 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 Fe0 cui cng thnh Fe+2, Al0 thnh Al+3, O20 thnh

2O2 v 2H+ thnh H2 nn ta c phng trnh bo ton electron nh sau: 5,4 30,013 2 0,009 4 n 2

27

Fe0 Fe+2 Al0 Al+3 O20 2O2 2H+ H2 n = 0,295 mol

2HV 0,295 22,4 6,608 lt. (p n A)

Nhn 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 HNO3 un nng thu c V lt kh NO (sn phm kh duy nht) ktc. Gi tr ca V l


Tm tt theo s : o2 3 t

NO

Fe O0,81 gam Al V ?

CuO 3ha tan hon tondung dch HNO

hn hp A

Thc cht trong bi ton ny ch c qu trnh cho v nhn electron ca nguyn t Al v N. Al Al+3 + 3e 0,81

27 0,09 mol

v N+5 + 3e N+2

VNMATHS.TK

18

0,09 mol 0,03 mol VNO = 0,0322,4 = 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 A 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 l

A. 2M v 1M. B. 1M v 2M. C. 0,2M v 0,1M. D. kt qu khc.

Tm tt s :

Al Fe

8,3 gam hn hp X

(n = n )

AlFe + 100 ml dung dch Y

3

3 2

AgNO : x molCu(NO ) :y mol

Cht rn A(3 kim loi)

2

HCl d1,12 lt H

2,8 gam cht rn khng tan B

Hng dn gii Ta c: nAl = nFe =

8,30,1 mol.

83

t 3AgNO

n 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+ + 2e 0,1 0,3 0,1 0,2

Tng s mol e nhng bng 0,5 mol. Qu trnh kh: Ag+ + 1e Ag Cu2+ + 2e Cu 2H+ + 2e H2 x x x y 2y y 0,1 0,05

Tng 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. 108x + 64y = 28 (2)

19

Gii h (1), (2) ta c: x = 0,2 mol ; y = 0,1 mol.

3M AgNO

0,2C0,1

= 2M; 3 2M Cu( 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 l

A. 63% v 37%. B. 36% v 64%. C. 50% v 50%. D. 46% v 54%.

Hng dn gii t nMg = x mol ; nAl = y mol. Ta c: 24x + 27y = 15. (1)

Qu trnh oxi ha:

Mg Mg2+ + 2e Al Al3+ + 3e x 2x y 3y

Tng s mol e nhng bng (2x + 3y). Qu trnh kh: N+5 + 3e N+2 2N+5 + 24e 2N+1 0,3 0,1 0,8 0,2

N+5 + 1e N+4 S+6 + 2e S+4 0,1 0,1 0,2 0,1

Tng 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

%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 l

A. 11,2 lt. B. 21 lt. C. 33 lt. D. 49 lt. Hng dn gii

V Fe 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

VNMATHS.TK

20

60 mol56

60256

mol S S+4 + 4e 30 mol

32 304

32 mol

Thu e: Gi s mol O2 l x mol. O2 + 4e 2O-2 x mol 4x

Ta 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.


Trong 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 2

N

(NO). S mol e do R1 v R2 nhng ra l

5N

+ 3e 2N 0,15 05,0

4,2212,1

TN2: R1 v R2 trc tip nhng e cho 5

N

to ra N2. Gi x l s mol N2, th s mol e thu vo l 2

5N

+ 10e 02N 10x x mol Ta c: 10x = 0,15 x = 0,015

2NV = 22,4.0,015 = 0,336 lt. (p n B)

V d 7: Cho 1,35 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.


Cch 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 2x y y 2y z z 3z

21

Thu e: 5

N

+ 3e = 2

N

(NO) 5

N

+ 1e = 4

N

(NO2)

0,03 0,01 0,04 0,04 Ta c: 2x + 2y + 3z = 0,03 + 0,04 = 0,07

v 0,07 cng chnh l s mol NO3 Khi lng mui nitrat l: 1,35 + 620,07 = 5,69 gam. (p n C) Cch 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 th

3 2HNO NO NO

n 2n 4n

3HNOn 2 0,04 4 0,01 0,12 mol

2H O

n 0,06 mol p dng nh lut bo ton khi lng:

3 2 2KL HNO mui NO NO H Om m m m m m

1,35 + 0,1263 = mmui + 0,0130 + 0,0446 + 0,0618 mmui = 5,69 gam.

V d 8: (Cu 19 - M 182 - Khi A - TSH - 2007) Ha tan hon ton 12 gam hn hp Fe, Cu (t l mol 1:1) bng axit 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 l


t nFe = nCu = a mol 56a + 64a = 12 a = 0,1 mol. Cho e: Fe Fe3+ + 3e Cu Cu2+ + 2e

0,1 0,3 0,1 0,2 Nhn e: N+5 + 3e N+2 N+5 + 1e N+4

3x x y y Tng ne cho bng tng ne nhn.

3x + y = 0,5 Mt khc: 30x + 46y = 192(x + y).

x = 0,125 ; y = 0,125. Vhh kh (ktc) = 0,125222,4 = 5,6 lt. (p n C)

V 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 l


m gam Fe + O2 3 gam hn hp cht rn X 3HNO d 0,56 lt NO.

VNMATHS.TK

22

Thc cht cc qu trnh oxi ha - kh trn l: Cho e: Fe Fe3+ + 3e m

56 3m

56mol e

Nhn e: O2 + 4e 2O2 N+5 + 3e N+2 3 m

32 4(3 m)

32 mol e 0,075 mol 0,025 mol

3m56

= 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: - Phn 1: Ha tan hon ton trong dung dch cha axit HCl v H2SO4 long to ra 3,36 lt kh H2. - Phn 2: Tc dng hon ton vi dung dch HNO3 thu c V lt kh NO (sn phm kh duy

nht). Bit cc th tch kh o iu kin tiu chun. Gi tr ca V l


t hai kim loi A, B l M. - Phn 1: M + nH+ Mn+ + 2n H2 (1) - Phn 2: 3M + 4nH+ + nNO3 3Mn+ + nNO + 2nH2O (2) Theo (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+2 0,3 0,15 mol 0,3 0,1 mol VNO = 0,122,4 = 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 NO2 v NO c VX = 8,96 lt (ktc) v t khi i vi O2 bng 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 gii Ta c: nX = 0,4 mol; MX = 42. S ng cho:

2NO : 46 42 30 1242

NO : 30 46 42 4

23

22

NO NO

NO NO

n : n 12 : 4 3

n n 0,4 mol

2

NO

NO

n 0,1 moln 0,3 mol

2

NO

NO

%V 25%%V 75%

v Fe 3e Fe3+ N+5 + 3e N+2 N+5 + 1e N+4 3x x 0,3 0,1 0,3 0,3 Theo nh lut bo ton electron: 3x = 0,6 mol x = 0,2 mol mFe = 0,256 = 11,2 gam. (p p B).

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 l

A. 0,28M. B. 1,4M. C. 1,7M. D. 1,2M. Hng dn gii

Ta c: 2 2N NO

X

M MM 9,25 4 37

2

l trung bnh cng khi lng phn t ca hai kh N2 v NO2 nn:

2 2

XN NO

nn n 0,04 mol2

v NO3 + 10e N2 NO3 + 1e NO2

0,08 0,4 0,04 mol 0,04 0,04 0,04 mol M Mn+ + n.e 0,04 mol

3HNO (b kh)

n 0,12 mol. 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 nhng nhn

Do : 3HNO ( )

n 0,44 0,12 0,56 molphn ng

3 0,56HNO 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 H2SO4 tham gia

phn ng, to mui MgSO4, H2O v sn phm kh X. X l A. SO2 B. S C. H2S D. SO2, H2S

Hng dn gii Dung dch H2SO4 m c va l cht oxi ha va l mi trng. Gi a l s oxi ha ca S trong X. Mg Mg2+ + 2e S+6 + (6-a)e S a

VNMATHS.TK

24

0,4 mol 0,8 mol 0,1 mol 0,1(6-a) mol

Tng s mol H2SO4 dng l : 49 0,598 (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, Fe2O3 v 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 gii

S mol Fe ban u trong a gam: Fe an 56 mol.

S mol O2 tham gia phn ng: 2O75,2 an

32 mol.

Qu trnh oxi ha: 3Fe Fe 3ea 3amol mol56 56

(1)

S mol e nhng: e 3an mol56 Qu trnh kh: O2 + 4e 2O2 (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) V 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 gii Da 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 + 2H2O NO3 + 2H+ + 1e NO2 + H2O Nh 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+ + 2e Al Al3+ + 3e 2x + 2y + 3z = 0,07.

25

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 ELECTRON

01. Ho tan hon ton m gam Al vo dung dch HNO3 rt long th thu c hn hp gm 0,015 mol kh N2O v 0,01mol kh NO (phn ng khng to NH4NO3). Gi tr ca m l

A. 13,5 gam. B. 1,35 gam. C. 0,81 gam. D. 8,1 gam. 02. 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 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%. 03. 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 l

A. 2,24 lt. B. 3,36 lt. C. 4,48 lt. D. 5,6 lt. 04. Dung dch X gm AgNO3 v 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 HNO3 long 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 HNO3 v 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.

VNMATHS.TK

26

09. t chy 5,6 gam bt Fe trong bnh ng O2 thu 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 l

A. 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 HNO3 khi 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.


p n cc bi tp vn dng 1. B 2. B 3. A 4. B 5. C 6. D 7. C 8. A 9. D 10. A

Phng php 4 S 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 l

H+ + OH H2O hoc phn ng ca Cu kim loi vi hn hp dung dch NaNO3 v dung dch H2SO4 l

3Cu + 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 H2SO4 long) 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)2 cn 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 gii Quy 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 Y Fe3O4 + 8H+ Fe2+ + 2Fe3+ + 4H2O 0,2 0,2 0,4 mol Fe + 2H+ Fe2+ + H2 0,1 0,1 mol Dung 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

27

VNO = 0,122,4 = 2,24 lt.

3 2 3Cu( NO ) NO

1n n 0,052

mol

3 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 H2SO4 0,5M. Sau khi phn ng kt thc thu c V lt kh NO duy nht (ktc). Gi tr ca V l


3HNO

n 0,12 mol ; 2 4H SO

n 0,06 mol Tng:

Hn 0,24 mol v

3NOn 0,12 mol.

Phng trnh ion: 3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O

Ban u: 0,1 0,24 0,12 mol Phn ng: 0,09 0,24 0,06 0,06 mol Sau phn ng: 0,01 (d) (ht) 0,06 (d)

VNO = 0,0622,4 = 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 l A. 15 gam. B. 5 gam. C. 10 gam. D. 0 gam.

Hng dn gii

2COn = 0,35 mol ; nNaOH = 0,2 mol; 2Ca(OH)n = 0,1 mol.

Tng: OH

n = 0,2 + 0,12 = 0,4 mol v 2Can = 0,1 mol. Phng trnh ion rt gn: CO2 + 2OH CO32 + H2O 0,35 0,4

0,2 0,4 0,2 mol

2CO ( )n d = 0,35 0,2 = 0,15 mol

tip tc xy ra phn ng: CO32 + CO2 + H2O 2HCO3

Ban u: 0,2 0,15 mol Phn ng: 0,15 0,15 mol

23CO

n cn li bng 0,15 mol

3CaCOn

= 0,05 mol

3CaCO

m = 0,05100 = 5 gam. (p n B)

VNMATHS.TK

28

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 l


Phn ng ca kim loi kim v kim loi kim th vi H2O: M + nH2O M(OH)n + 2n H2

T phng trnh ta c:

2HOHn 2n = 0,1mol.

Dung dch A tc dng vi 0,03 mol dung dch AlCl3: Al3+ + 3OH Al(OH)3

Ban u: 0,03 0,1 mol Phn ng: 0,03 0,09 0,03 mol

OH ( )

nd = 0,01mol

tip tc ha tan kt ta theo phng trnh: Al(OH)3 + OH AlO2 + 2H2O

0,01 0,01 mol Vy:

3Al(OH)m = 780,02 = 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)


Phng trnh ion: Cu + 2Fe3+ 2Fe2+ + Cu2+ 0,005 0,01 mol 3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O

Ban u: 0,15 0,03 mol H+ d Phn 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%.

Hng dn gii Phng trnh ion: Ag+ + Cl AgCl Ag+ + Br AgBr

29

t: nNaCl = x mol ; nNaBr = y mol mAgCl + mAgBr = 3( )AgNOm p.

3Cl Br NO

m m m 35,5x + 80y = 62(x + y) x : y = 36 : 53

Chn x = 36, y = 53 NaCl 58,5 36 100%m 58,5 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 H2SO4 1M v HCl 1M) vo dung dch C thu c V lt CO2 (ktc) v dung dch E. Cho dung dch Ba(OH)2 ti d vo dung dch E th thu c m gam kt ta. Gi tr ca m v V ln lt l

A. 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 gii Dung 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 mol HCO3 + H+ H2O + CO2

Ban u: 0,4 0,1 mol Phn ng: 0,1 0,1 0,1 mol

D: 0,3 mol

Tip tc cho dung dch Ba(OH)2 d vo dung dch E: Ba2+ + HCO3 + OH BaCO3 + H2O 0,3 0,3 mol Ba2+ + SO42 BaSO4 0,1 0,1 mol

2COV = 0,122,4 = 2,24 lt.

Tng khi lng kt ta: m = 0,3197 + 0,1233 = 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 l

VNMATHS.TK

30

A. 38,93 gam. B. 38,95 gam. C. 38,97 gam. D. 38,91 gam. b) Th tch V l A. 0,39 lt. B. 0,4 lt. C. 0,41 lt. D. 0,42 lt. c) Lng kt ta l A. 54,02 gam. B. 53,98 gam. C. 53,62 gam. D. 53,94 gam.

Hng dn gii a) Xc nh khi lng mui thu c trong dung dch X:

2 4H SOn = 0,280,5 = 0,14 mol

24SO

n = 0,14 mol v Hn = 0,28 mol.

nHCl = 0,5 mol

H

n = 0,5 mol v Cln = 0,5 mol.

Vy tng H

n = 0,28 + 0,5 = 0,78 mol.

M 2H

n = 0,39 mol. Theo phng trnh ion rt gn: Mg0 + 2H+ Mg2+ + H2 (1) Al + 3H+ Al3+ + 3

2H2 (2)

Ta thy 2HH (p-)

n 2n H+ ht. mhh mui = mhh k.loi + 2

4SO Clm m

= 7,74 + 0,1496 + 0,535,5 = 38,93gam. (p n A) b) Xc nh th tch V:

2

NaOH

Ba(OH)

n 1V mol

n 0,5V mol

Tng OH

n = 2V mol v 2Ban = 0,5V mol.

Phng trnh to kt ta: Ba2+ + SO42 BaSO4 (3) 0,5V mol 0,14 mol

Mg2+ + 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:

H

n = OHn = 0,78 mol

2V = 0,78 V = 0,39 lt. (p n A) c) Xc nh lng kt ta:

31

2Ban = 0,5V = 0,50,39 = 0,195 mol > 0,14 mol Ba2+ d.

4BaSOm = 0,14233 = 32,62 gam.

Vy mkt 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 l

A. 1. B. 6. C. 7. D. 2. Hng dn gii

nHCl = 0,25 mol ; 2 4H SOn = 0,125.

Tng: H

n = 0,5 mol ;

2H ( )

n to thnh = 0,2375 mol.

Bit rng: c 2 mol ion H+ 1 mol H2 vy 0,475 mol H+ 0,2375 mol H2

H ( )n

d = 0,5 0,475 = 0,025 mol

0,025H0,25

= 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 sn phm kh duy nht, cc th tch kh o cng iu kin. Quan h gia V1 v V2 l

A. V2 = V1. B. V2 = 2V1. C. V2 = 2,5V1. D. V2 = 1,5V1. Hng dn gii

TN1:

3

Cu

HNO

3,84n 0,06 mol64

n 0,08 mol

3

H

NO

n 0,08 moln 0,08 mol

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O Ban u: 0,06 0,08 0,08 mol H+ phn ng ht Phn 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: H

n = 0,16 mol ;

VNMATHS.TK

32

3NO

n = 0,08 mol.

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O Ban u: 0,06 0,16 0,08 mol Cu v H+ phn ng ht Phn ng: 0,06 0,16 0,04 0,04 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)2 0,1M v NaOH 0,1M) vi 400 ml dung dch (gm H2SO4 0,0375M v HCl 0,0125M), thu c dung dch X. Gi tr pH ca dung dch X l

A. 7. B. 2. C. 1. D. 6. Hng dn gii

2Ba(OH)

NaOH

n 0,01 mol

n 0,01 mol

Tng

OHn = 0,03 mol.

2 4H SO

HCl

n 0,015 mol

n 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+ + OH H2O

Bt u 0,035 0,03 mol Phn ng: 0,03 0,03 Sau phn ng:

H ( )n

d = 0,035 0,03 = 0,005 mol. Tng: Vdd (sau trn) = 500 ml (0,5 lt). 0,005H

0,5 = 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 H2SO4 2M cn dng trung ho dung dch X l

A. 150 ml. B. 75 ml. C. 60 ml. D. 30 ml. Hng dn gii

Na + H2O NaOH + 12 H2 Ba + 2H2O Ba(OH)2 + H2

2Hn = 0,15 mol, theo phng trnh tng s 2 2HOH (d X)n 2n = 0,3 mol. Phng trnh ion rt gn ca dung dch axit vi dung dch baz l H+ + OH H2O

Hn = OHn = 0,3 mol 2 4H SOn = 0,15 mol

2 4H SO

0,15V2

= 0,075 lt (75 ml). (p n B)

33

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, 0,15 mol NO2 v 0,05 mol N2O). Bit rng khng c phn 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 gii

Ta c bn phn ng: NO3 + 2H+ + 1e NO2 + H2O (1) 2 0,15 0,15 NO3 + 4H+ + 3e NO + 2H2O (2) 4 0,1 0,1 2NO3 + 10H+ + 8e N2O + 5H2O (3) 10 0,05 0,05 T (1), (2), (3) nhn c:

3HNO Hn n

p = 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 HNO3 v 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:


Ta c bn phn ng: 2NO3 + 2H+ + 1e NO2 + H2O + NO3 (1) 0,1 0,1 4NO3 + 4H+ + 3e NO + 2H2O + 3NO3 (2) 0,1 3 0,1 2SO42 + 4H+ + 2e SO2 + H2O + SO42 (3) 0,1 0,1 T (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 + 2

4SOm

= 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 N2 v 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 0,22M C. 53,55 gam. v 2,2M D. 53,55 gam. v 0,22M

Hng dn gii

2 2N O N1,792n n 0,04

2 22,4 mol.

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34

Ta c bn phn ng: 2NO3 + 12H+ + 10e N2 + 6H2O 0,08 0,48 0,04

2NO3 + 10H+ + 8e N2O + 5H2O 0,08 0,4 0,04

3HNO H

n 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. N2 C. NO D. NH4+ Hng dn gii

Ta c: nZn = 0,05 mol; nAl = 0,1 mol. Gi a l s mol ca NxOy, ta c: Zn Zn2+ + 2e Al Al3+ + 3e 0,05 0,1 0,1 0,3

xNO3 + (6x 2y)H+ + (5x 2y)e NxOy + (3x 2y)H2O 0,04(5x 2y) 0,04 0,04(5x 2y) = 0,4 5x 2y = 10

Vy X 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 khc, nu thm Ba(OH)2 d 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 167,44g C. 112,84g v 157,44g A. 112,84g v 167,44g

Hng dn gii Ta c bn phn ng: CuFeS2 + 8H2O 17e Cu2+ + Fe3+ + 2SO42 + 16+ 0,15 0,15 0,15 0,3

Cu2FeS2 + 8H2O 19e 2Cu2+ + Fe3+ + 2SO42 + 16+ 0,09 0,18 0,09 0,18

24SO

n 0,48 mol; Ba2+ + SO42 BaSO4 0,48 0,48

m = 0,48 233 = 111,84 gam.

35

nCu = 0,33 mol; nFe = 0,24 mol.

Cu CuO 2Fe Fe2O3 0,33 0,33 0,24 0,12

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 gii nZn = 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)3

Vy s mol NO3 cn li to NH4NO3 l: 0,4 0,04 2 0,08 3 = 0,08 mol - Do trong dung dch to 0,04 mol NH4NO3 m = 0,04 189 + 0,08 213 + 0,04 80 = 27,8 gam. (p n C)

Phng php 5 S DNG CC GI TR TRUNG BNH

y 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.

Nguyn tc ca phng php nh sau: Khi lng phn t trung bnh (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 lng hn hp (tnh theo gam)tng s mol cc cht trong hn hp

.

i i1 1 2 2 3 31 2 3 i

M 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:

VNMATHS.TK

36

1 2 31 2 3i i i

n n nM M . M . M . ...n n n

1 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 i

M 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% v

1 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 1

x y z 2

C H O ; n molC H O ; n mol

ta c: - Nguyn t cacbon trung bnh: 1 1 2 2

1 2

x n x n ...xn n ...

- Nguyn t hiro trung bnh: 1 1 2 2

1 2

y n y n ...yn n ...

v i khi tnh c c s lin kt , s nhm chc trung bnh theo cng thc trn. V 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 gii

1. Gi A, B l cc kim loi cn tm. Cc phng trnh phn ng l

37

ACO3 + 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 l 2,84M 94,67

0,03 v A,BM 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 105,670,03 = 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%. C. 65Cu: 72,5% ; 63Cu: 27,5%. D. 65Cu: 30% ; 63Cu: 70%.

Hng dn gii Gi x l % ca ng v 6529Cu ta c phng trnh: M = 63,55 = 65.x + 63(1 x) x = 0,275 Vy: ng v 65Cu chim 27,5% v ng v 63Cu chim 72,5%. (p n C)

V d 3: Hn hp kh SO2 v O2 c t khi so vi CH4 bng 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 gii

Cch 1: Gi x l % th tch ca SO2 trong hn hp ban u, ta c: M = 163 = 48 = 64.x + 32(1 x) x = 0,5

Vy: 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 40

20 V .

Gii ra c V = 20 lt. (p n B)

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38

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 = 163 = 48), cn O2 thm vo coi nh kh th

hai, ta c phng trnh: 48 20 32VM 2,5 16 40

20 V ,

Rt ra V = 20 lt. (p n B) V 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. HCOOH v CH3COOH. B. CH3COOH v C2H5COOH. C. C2H5COOH v C3H7COOH. D. C3H7COOH v C4H9COOH.

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 gii 1. Theo phng php KLPTTB: RCOOH

1 23m 2,310 10

gam,

2RCH COOH

1 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 nn M = 53+23 1 75 mui . V s mol mui bng s mol axit bng 0,1 nn tng khi

lng mui bng 750,1 = 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 l

A. C2H4 v C3H6. B. C3H6 v C4H8. C. C4H8 v C5H10. D. C5H10 v C6H12.

Hng dn gii t 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:

39

n 2 n2

C H

H

n 0,4 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 + 23n O2

n CO2 + n H2O (1) 2H2 + O2 2H2O (2) Theo phng trnh (1) ta c:

2 2CO H On n = 0,45 mol.

n 2 nC H

0,45nn

mol.

Tng: 2H O

13,5n18

= 0,75 mol

2H O (pt 2)n = 0,75 0,45 = 0,3 mol

2H

n = 0,3 mol.

Ta c: n 2 n2

C H

H

n 0,45 2n 0,3 n 3

n = 2,25 Hai 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. D. 3,32 gam ; C2H5OH v C3H7OH.

Hng dn gii Gi n l s nguyn t C trung bnh v x l tng s mol ca hai ru. CnH2n+1OH + 2

3n O2

2n CO + 2(n 1) H O x mol n x mol (n 1) x mol

2CO3,584n n.x 0,1622,4

mol (1)

2H O

3,96n (n 1)x 0,2218

mol (2) T (1) v (2) gii ra x = 0,06 v n = 2,67. Ta c: a = (14 n + 18).x = (142,67) + 180,06 = 3,32 gam. n = 2,67 2 5

3 7

C H OHC H OH

(p n D)

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40

V d 7: Hn hp 3 ru n chc A, B, C c tng s mol l 0,08 v khi lng l 3,38 gam. Xc nh CTPT ca ru B, bit rng B v C c cng s nguyn t cacbon v s mol ru A bng 5 3 tng s mol ca ru B v C, MB > MC.

A. CH3OH. B. C2H5OH. C. C3H7OH. D. C4H9OH. Hng dn gii

Gi M l nguyn t khi trung bnh ca ba ru A, B, C. Ta c: 3,38M 42,2

0,08

Nh vy phi c t nht mt ru c M < 42,25. Ch c CH3OH c (M = 32) Ta c: A

0,08 5n 0,055 3

; mA = 320,05 = 1,6 gam. mB + C = 3,38 1,6 = 1,78 gam;

B C0,08 3n 0,03

5 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 hay 12x + y + 17 = 59,33 12x + y = 42,33 Bin lun:

x 1 2 3 4 y 30,33 18,33 6,33 < 0

Ch 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) v C3H7OH C3H3OH (CHCCH2OH) v C3H7OH (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.


t R l gc hirocacbon trung bnh v x l tng s mol ca 2 ru. ROH + Na RONa + 21 H2

x mol x x2

.

41

Ta c: R 17 x 2,84

R 39 x 4,6

Gii ra c x = 0,08.

Vy : 2H

0,08V 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 l

A. C2H2 v C4H6. B. C2H2 v C4H8. C. C3H4 v C4H8. D. C2H2 v C3H8.

Hng dn gii hh X

4,48n 0,222,4

mol

n 1,4 0,5 0,72Br ban u

mol

0,7n22Br p.ng

= 0,35 mol. Khi lng bnh Br2 tng 6,7 gam l s gam ca hirocabon khng no. t CTTB ca hai hirocacbon

mch h l n 2n 2 2aC H ( a l s lin kt trung bnh). Phng trnh phn ng: n 2n 2 2aC H + 2aBr n 2n 2 2a 2aC H Br 0,2 mol 0,35 mol 0,35a

0,2 = 1,75

6,714n 2 2a0,2

n = 2,5. Do hai hirocacbon mch h phn ng hon ton vi dung dch Br2 nn chng u l hirocacbon

khng no. Vy hai hirocacbon l C2H2 v C4H8. (p n B) V 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 l


Hn 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 + 2

3n O2

2nCO + 2(n 1)H O

n 2n 1C H OH 2 oH SO170 C

4 n 2nC H + H2O (Y)

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42

n 2nC H + 23n O2

2nCO + 2n H O Nhn 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 BNH 01. 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 l A. 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 l

A. C2H6O; C3H8O; C4H10O. B. C3H8O; C3H6O2; C4H10O. C. 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 thc

A. 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 l

A. 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). - Phn 2: tch nc hon ton 180oC, xc tc H2SO4 c thu c mt anken cho hp th vo bnh

ng dung dch Brom d thy c 32 gam Br2 b mt mu. CTPT hai ancol trn l A. 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 l A. 1,434 lt. B. 1,443 lt. C. 1,344 lt. D. 1,444 lt.

43

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 l


08. 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 l


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 l


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 t ca este l

A. 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. C 6. C 7. D 8. B 9. B 10. C

Phng php 6 TNG GIM KHI LNG

Nguyn tc ca phng php l xem khi chuyn t cht A thnh cht 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 + CO2 Ta thy rng khi chuyn 1 mol MCO3 thnh MCl2 th khi lng tng

(M + 235,5) (M + 60) = 11 gam v 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 + H2O

th 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.

VNMATHS.TK

44

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 bng mB (bm) mA (tan). - Khi lng kim loi gim bng

mA (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 gii

Trong dung dch: Na2CO3 2Na+ + CO32 (NH4)2CO3 2NH4+ + CO32 BaCl2 Ba2+ + 2Cl CaCl2 Ca2+ + 2Cl Cc 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 39,7

11 = 0,3 mol

m 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 100

39,7 = 49,62%;

3CaCO

%m = 100 49,6 = 50,38%. (p n C)

45

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?


C 1 mol mui cacbonat to thnh 1 mol mui clorua cho nn khi lng mui khan tng (71 60) = 11 gam, m

2CO

n = nmui cacbonat = 0,2 mol.

Suy ra khi lng mui khan tng sau phn ng l 0,211 = 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 l A. HCOOH B. C3H7COOH

C. CH3COOH D. C2H5COOH. Hng dn gii

C 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 l

naxit = 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 AgNO3 d 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 gii C 1 mol mui halogen to thnh 1 mol kt ta khi 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 mt thanh graphit c ph mt lp kim loi ha tr (II) vo dung dch CuSO4 d. Sau phn ng khi lng ca thanh graphit gim i 0,24 gam. Cng thanh graphit ny nu c nhng vo dung dch AgNO3 th khi phn ng 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 gii

t kim loi ha tr (II) l M vi s gam l x (gam). M + CuSO4 d MSO4 + Cu

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46

C 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.M

M 64 khi lng kim loi gim 0,24 gam. Mt khc: M + 2AgNO3 M(NO3)2 + 2Ag

C 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 l


Hng dn gii Kh Cl2 d ch kh c mui NaI theo phng trnh 2NaI + Cl2 2NaCl + I2 C 1 mol NaI to thnh 1 mol NaCl Khi lng mui gim 127 35,5 = 91,5 gam. Vy: 0,5 mol Khi lng mui gim 104,25 58,5 = 45,75 gam. mNaI = 1500,5 = 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 AgNO3 6%. Sau mt thi gian ly vt ra thy khi lng AgNO3 trong dung dch gim 25%. Khi lng ca vt sau phn ng l


3AgNO ( )

340 6n = 170 100ban u

= 0,12 mol;

3AgNO ( )

25n = 0,12100ph.ng

= 0,03 mol. Cu + 2AgNO3 Cu(NO3)2 + 2Ag 0,015 0,03 0,03 mol mvt sau phn ng = mvt ban u + mAg (bm) mCu (tan) = 15 + (1080,03) (640,015) = 17,28 gam.

(p n C)

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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 l

A. 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 gii V trong cng dung dch cn li (cng th tch) nn: [ZnSO4] = 2,5 [FeSO4]

4 4ZnSO FeSO

n 2,5n Zn + CuSO4 ZnSO4 + Cu (1) 2,5x 2,5x 2,5x mol Fe + CuSO4 FeSO4 + Cu (2) x x x x mol T (1), (2) nhn c gim khi lng ca dung dch l mCu (bm) mZn (tan) mFe (tan) 2,2 = 64(2,5x + x) 652,5x 56x x = 0,4 mol.

Vy: mCu (bm ln thanh km) = 642,50,4 = 64 gam; mCu (bm ln thanh st) = 640,4 = 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 l

A. CH2=CHCOOH. B. CH3COOH. C. HCCCOOH. D. CH3CH2COOH.

Hng dn gii t CTTQ ca axit hu c X n chc l RCOOH. 2RCOOH + CaCO3 (RCOO)2Ca + CO2 + H2O C 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 mol RCOOH 5,76M 720,08 R = 27

Axit 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 gii

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Gi khi lng thanh km ban u l a gam th khi lng tng thm l 2,35a100

gam.

Zn + CdSO4 ZnSO4 + Cd 65 1 mol 112, tng (112 65) = 47 gam 8,32

208 (=0,04 mol) 2,35a

100 gam

Ta 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 gian ly thanh kim loi ra thy khi lng gim 0,05%. Mt khc nhng thanh kim loi trn vo dung dch Pb(NO3)2, sau mt thi gian thy khi lng tng 7,1%. Xc nh M, bit rng s mol CuSO4 v Pb(NO3)2 tham gia 2 trng hp nh nhau.

A. Al. B. Zn. C. Mg. D. Fe. Hng dn gii

Gi 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 + Cu M (gam) 1 mol 64 gam, gim (M 64)gam. x mol gim 0,05.m

100 gam.

x = 0,05.m

100M 64 (1)

M + Pb(NO3)2 M(NO3)2 + Pb M (gam) 1 mol 207, tng (207 M) gam x mol tng 7,1.m

100 gam

x = 7,1.m100

207 M (2)

T (1) v (2) ta c: 0,05.m

100M 64 =

7,1.m100

207 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 XCl3 to 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 gii

Gi A l nguyn t khi ca kim loi X. Al + XCl3 AlCl3 + X

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3,7827

= (0,14 mol) 0,14 0,14 mol. Ta c : (A + 35,53)0,14 (133,50,14) = 4,06 Gii 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 gii Ch c NaHCO3 b phn hy. t x l s gam NaHCO3. 2NaHCO3

ot Na2CO3 + CO2 + H2O C nung 168 gam khi lng gim: 44 + 18 = 62 gam x khi lng gim: 100 69 = 31 gam Ta c: 168 62

x 31 x = 84 gam.

Vy NaHCO3 chim 84% v Na2CO3 chim 16%. (p n C) V d 14: Ha tan 3,28 gam hn hp mui CuCl2 v Cu(NO3)2 vo 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 gii

Ta 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 MgCl2 v Cu(NO3)2 vo 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 l


p 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 LNG

01. 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 l

A. 142 gam. B. 126 gam. C. 141 gam. D. 132 gam.

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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?


03. 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 1M v 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. Dng CO kh 40 gam oxit Fe2O3 thu c 33,92 gam cht rn B gm Fe2O3, FeO v Fe. Cho 1 B2 tc

dng vi H2SO4 long 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 phng trnh phn ng xy ra, tm khi lng tng cht c trong 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.

08. Ngm mt thanh ng c khi lng 140,8 gam vo dung dch AgNO3 sau mt thi gian ly thanh ng e

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