ThS Trn Vit Bng 0949.505.767 19/26 ng 8, Linh Trung, Th c

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THI MINH HA K THI THPTQG NM 2015

Li gii chi tit: ThS Trn Vit Bng

Cu 7: Hn hp X gm Mg (0,10 mol), Al (0,04 mol) v Zn (0,15 mol). Cho X tc dng vi dung dch HNO3 long (d), sau phn ng khi lng dung dch tng 13,23 gam. S mol HNO3 tham gia phn ng l

A. 0,6200 mol. B. 1,2400 mol. C. 0,6975 mol. D. 0,7750 mol. mX = 0,1.24 + 0,04.27 + 0,15.65 = 13,23 phn ng khng to kh ne cho = 0,62 nNH4NO3 = 0,62 : 8 = 0,0775 nHNO3 = 0,775 mol. Cu 11. in phn vi in cc tr dung dch cha 0,2 mol Cu(NO3)2, cng dng in 2,68A, trong thi gian t (gi), thu c dung dch X. Cho 14,4 gam bt Fe vo X, thu c kh NO (sn phm kh duy nht ca N+5) v 13,5 gam cht rn. Bit cc phn ng xy ra hon ton v hiu sut ca qu trnh in phn l 100%. Gi tr ca t l

A. 0,60. B. 1,00. C. 0,25. D. 1,20. 2

14,4gFe3

Cu : xX NO : 0,2 13,5g NO

H : y

+

+

+

Gi s Cu2+ d m < 64.0,2 = 12,8 < 13,5 loi. Vy Fe d. Catot: Cu2+ + 2e Cu 0,2-x Anot: 4OH- 4e O2 + 2H2O y Bte: y = 2(0,2-x) (1)

Khi a Fe vo X: 4H+ + NO3- + 3e NO + 2H2O Cu2+ + 2e Cu y 0,2 3y/4 x 2x Gi s mol Fe phn ng l z: Fe 2e Fe2+ z 2z

Bte: 3y/4 + 2x = 2z (2) Btklg: 14,4 56z + 64x = 13,5 (3)

T (1) (2) (3) y = 0,1 t = (0,1.96500) : 2,68 = 1 gi. Cu 15. Dung dch X gm Al2(SO4)3 0,75M v H2SO4 0,75M. Cho V1 ml dung dch KOH 1M vo 100 ml dung dch X, thu c 3,9 gam kt ta. Mt khc, khi cho V2 ml dung dch KOH 1M vo 100 ml dung dch X cng thu c 3,9 gam kt ta. Bit cc phn ng xy ra hon ton. T l V2:V1 l

A. 4 : 3. B. 25 : 9 C. 13 : 9 D. 7 : 3 100 ml dung dch X gm: 0,15 mol Al3+ v 0,15 mol H+. C n = 0,05

TH1: nOH- = 0,15 + 3.0,05 = 0,3 TH2: nOH = 4.0,15 0,05 + 0,15 = 0,7 V2 : V1 = 7 : 3

Cu 16. Cho 115,3 gam hn hp hai mui MgCO3 v RCO3 vo dung dch H2SO4 long, thu c 4,48 lt kh CO2 (ktc), cht rn X v dung dch Y cha 12 gam mui. Nung X n khi lng khng i, thu c cht rn Z v 11,2 lt kh CO2 (ktc). Khi lng ca Z l

A. 92,1 gam. B. 80,9 C. 84,5 D. 88,5 (MgCO3 + RCO3) + H2SO4 Mui + rn X + CO2 + H2O vi nCO2 = nH2O = nH2SO4 = 0,2 Btklg: mX = 115,3 + 0,2.98 12 0,2(44+18) = 110,5 X Z + CO2 vi nCO2 = 0,5. Btklg: mZ = mX mCO2 = 110,5 0,5.44 = 88,5

ThS Trn Vit Bng 0949.505.767 19/26 ng 8, Linh Trung, Th c

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Cu 19. Cho m gam bt Fe vo 200 ml dung dch cha hai mui AgNO3 0,15M v Cu(NO3)2 0,1M, sau mt thi gian thu c 3,84 gam hn hp kim loi v dung dch X. Cho 3,25 gam bt Zn vo dung dch X, sau khi phn ng xy ra hon ton, thu c 3,895 gam hn hp kim loi v dung dch Y. Gi tr ca m l

A. 0,560. B. 2,240. C. 2,800. D. 1,435. nAg+ = 0,03 v nCu2+ = 0,02 nZn phn ng = (0,03 + 0,02.2) : 2 = 0,035 mZn d = 0,975 m(Cu, Ag) trong 3,895g = 2,92 m + 0,03.108 + 0,02.64 = 3,84 + 2,92 m = 2,24. Cu 21. Cho 46,8 gam hn hp CuO v Fe3O4 (t l mol 1:1) tan ht trong dung dch H2SO4 long, va , thu c dung dch A. Cho m gam Mg vo A, sau khi phn ng kt thc thu c dung dch B. Thm dung dch KOH d vo B c kt ta D. Nung D trong khng kh n khi lng khng i, thu c 45,0 gam cht rn E. Gi tr gn nht ca m l

A. 6,6. B. 11 C. 13,2 D. 8,8 nCuO = nFe3O4 = 0,15 to nCu2+= nFe2+ = 0,15 v nFe3+ = 0,3. Gi s khi cho Mg vo A cha to mD > 46,8 m mD = 45 loi. Gi s Mg d nMg2+ = (0,3 + 0,45.2 + 0,15.2) : 2 = 0,75 mMgO = 30 45 loi. TH1: Mg tc dng ht vi A to x mol Cu nCu2+ = 0,15 x

Bte: 0,3 + 2x = 2nMg2+ nMg2+ = x + 0,15 D gm: (x + 0,15) mol MgO; (0,15 x) mol CuO v 0,225 mol Fe2O3. x = 0,225 > 0,15 loi.

TH2: Mg tc dng ht vi A to 0,15 mol Cu; y mol Fe nFe2+ = 0,45 y Bte: 0,3 + 2.0,15 + 2y = 2nMg2+ nMg2+ = y + 0,3 D gm: (y + 0,3) mol MgO v (0,225 y/2) mol Fe2O3. y = 0,075 nMg = 0,375 m = 9.

Cu 22. Ho tan hon ton 12,2 gam hn hp gm FeCl2 v NaCl (c t l s mol tng ng l 1:2) vo mt lng nc d, thu c dung dch X. Cho dung dch AgNO3 d vo X, sau khi phn ng xy ra hon ton, thu c m gam cht rn. Gi tr ca m l

A. 34,1. B. 28,7 C. 10,8 D. 57,4 nFeCl2 = 0,05 v nNaCl = 0,1. Cht rn gm 0,05 mol Ag v 0,2 mol AgCl m = 34,1. Cu 25. Cho 66,2 gam hn hp X gm Fe3O4, Fe(NO3)2, Al tan hon ton trong dung dch cha 3,1 mol KHSO4 long. Sau khi cc phn ng xy ra hon ton thu c dung dch Y ch cha 466,6 gam mui sunfat trung ha v 10,08 lt (ktc) kh Z gm 2 kh trong c mt kh ha nu ngoi khng kh. Bit t khi ca Z so vi He l 23/18. Phn trm khi lng ca Al trong hn hp X gn nht vi gi tr no sau y?

A. 15. B. 20. C. 25. D. 30. D tnh c Z gm nNO = 0,05 v nH2 = 0,4

4KHSO :3,13 4

23 2

Al : xNO : 0,05

66,2g Fe O : yH : 0,4

Fe(NO ) : z

27x + 232y + 180z = 66,2 (1). To H2 nn NO3- ht nNH4+ = 2z 0,05 Y ch cha mui sunfat nn H+ ht 4H+ + NO3- + 3e NO + 2H2O 2H+ + 2e H2 0,2 0,05 0,15 0,05 0,8 0,8 0,4 10H+ + NO3- + 8e NH4+ + 3H2O 2H+ + O H2O 2z 0,05 8y 4y nH+ = 1 + 10(2z 0,05) + 8y = 3,1 8y + 20z = 2,6 (2). Khi lng mui sunfat = 27x + 168y + 56z + 3,1(39 + 96) + 18(2z 0,05) = 466,6

ThS Trn Vit Bng 0949.505.767 19/26 ng 8, Linh Trung, Th c

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27x + 168y + 92z = 49 (3) T (1), (2) v (3) x = 0,4 %mAl = 16,3%. Cu 32. Hn hp T gm ba cht hu c X, Y, Z (50 < MX < MY < MZ v u to nn t cc nguyn t C, H, O). t chy hon ton m gam T thu c H2O v 2,688 lt kh CO2 (ktc). Cho m gam T phn ng vi dung dch NaHCO3 d, thu c 1,568 lt kh CO2 (ktc). Mt khc, cho m gam T phn ng hon ton vi lng d dung dch AgNO3 trong NH3, thu c 10,8 gam Ag. Gi tr ca m l

A. 4,6. B. 4,8. C. 5,2. D. 4,4. T khng th cha HCHO v 30 < 50 nC/T = nCO2 = 0,12; nnhm -COOH = 0,07 v nnhm -CHO = 0,05 nO/T = 0,19. Trong T cc cht u c t nht 1 lin kt i nn nH 2nC = 0,24 m 0,12.12 + 0,19.16 + 0,24 = 4,72 loi B v C. Mt khc: mC + mO = 0,12.12 + 0,19.16 = 4,48 loi D chn A. Cu 38. Ln men m gam tinh bt thnh ancol etylic vi hiu sut ca c qu trnh l 75%. Lng CO2 sinh ra c hp th hon ton vo dung dch Ca(OH)2, thu c 50 gam kt ta v dung dch X. Thm dung dch NaOH 1M vo X, thu c kt ta. lng kt ta thu c l ln nht th cn ti thiu 100 ml dung dch NaOH. Gi tr ca m l

A. 72,0. B. 90 C. 64,8 D. 75,6 nCaCO3 = 0,5. Gi nCa(HCO3)2 = x nCa2+ = x v nHCO3- = 2x. max th ch cn to ra x mol CO32 thu x mol CaCO3, tc cn dng x mol OH- x = 0,1 nCO2 = 0,7 ntinh bt = 0,35 m = (0,35.162) : 0,75 = 75,6. Cu 39. X phng ho hon ton m gam mt este no, n chc, mch h E bng 26 gam dung dch MOH 28% (M l kim loi kim). C cn hn hp sau phn ng thu c 24,72 gam cht lng X v 10,08 gam cht rn khan Y. t chy hon ton Y, thu c sn phm gm CO2, H2O v 8,97 gam mui cacbonat khan. Mt khc, cho X tc dng vi Na d, thu c 12,768 lt kh H2 (ktc). Phn trm khi lng mui trong Y c gi tr gn nht vi

A. 67,5. B. 85,0. C. 80,0. D. 97,5. 2MOH M2CO3 7,28 8,97 d suy c M l K v nKOH = 0,13 X gm 1,04 mol H2O v ancol nancol = 2nH2 nH2O = 0,57.2 1,04 = 0,1 Y c cha 0,03 mol KOH d %mKOH/Y = 16,67% %m mui = 83,34% chn B. Cu 43. un nng 0,16 mol hn hp E gm hai peptit X (CxHyOzN6) v Y (CnHmO6Nt) cn dng 600 ml dung dch NaOH 1,5M ch thu c dung dch cha a mol mui ca glyxin v b mol mui ca alanin. Mt khc t chy 30,73 gam E trong O2 va thu c hn hp CO2, H2O v N2, trong tng khi lng ca CO2 v nc l 69,31 gam. Gi tr a : b gn nht vi

A. 0,730. B. 0,810. C. 0,756. D. 0,962. Gi X c x mol v Y c y mol. X (6 mt xch) + 6NaOH v Y (5 mt xch) + 5NaOH nn x + y = 0,16 v 6x + 5y = 0,9 x = 0,1 v y = 0,06 Gi s: E ch gm X v X ch gm Alanin M < 6.89 5.18 = 444

E ch gm Y v Y ch gm Glyxin M > 5.75 4.18 = 303 303.0,16 < mE < 444.0,16 48,48 < mE < 71,04 D on: E trong TN1 gp 2 ln trong TN2 tc mE = 30,73.2 = 61,46. Gi X c n mt xch Ala v (6-n) mt xch Gly Gi Y c m mt xch Ala v (5-m) mt xch Gly 0,1[89n + 75(6-n) 5.18] + 0,06[89m + 75(5-m) 4.18] = 61,46 0,1(14n + 360) + 0,06(14m + 303) = 61,46 n = 4 v m = 2 X l Ala4Gly2 0,1 mol v Y l Al2Gly3 0,06 mol a = 0,52 v b = 0,38 b/a = 0,73

ThS Trn Vit Bng 0949.505.767 19/26 ng 8, Linh Trung, Th c

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Cu 44. Amino axit X c cng thc (H2N)2C3H5COOH. Cho 0,02 mol X tc dng vi 200 ml dung dch hn hp H2SO4 0,1M v HCl 0,3M, thu c dung dch Y. Cho Y phn ng va vi 400 ml dung dch NaOH 0,1M v KOH 0,2M, thu c dung dch cha m gam mui. Gi tr ca m l

A. 10,43. B. 6,38. C. 10,45. D. 8,09. nX + nH+ = nOH+ = 0,12 nn phn ng va to mui.

m = m(H2N)2C3H5COO- + mSO42 + mCl- + mNa+ + mK+ m = 0,02.117 + 0,02.96 + 0,06.35,5 + 0,04.23 + 0,08.39 = 10,43.

Cu 47. Ancol X (MX = 76) tc dng vi axit cacboxylic Y thu c hp cht Z mch h (X v Y u ch c mt loi nhm chc). t chy hon ton 17,2 gam Z cn va 14,56 lt kh O2 (ktc), thu c CO2 v H2O theo t l s mol tng ng l 7 : 4. Mt khc, 17,2 gam Z li phn ng va vi 8 gam NaOH trong dung dch. Bit Z c cng thc phn t trng vi cng thc n gin nht. S cng thc cu to ca Z tha mn l

A. 1. B. 3. C. 2. D. 4. T MX = 76 d suy c X l C3H8O2 c 2 cu to. nO2 = 0,65 17,2 + 0,65.32 = 44.7x + 18.4x x = 0,1 nCO2 = 0,7 v nH2O = 0,4 Bt O nO/Z = 0,5 nC : nH : nO = 0,7 : 0,8 : 0,5 Z l C7H8O5 v nZ = 0,1. Z c th l: HOOCCCCOOCH2CH2CH2OH HOOCCCCOOCH2CH(OH)CH3 HOOCCCCOOCH(CH3)CH2OH Cu 49. Hn hp X gm 2 ancol CH3OH, C2H5OH c cng s mol v 2 axit C2H5COOH v HOOC[CH2]4COOH. t chy hon ton 1,86 gam X cn dng va 10,08 lt khng kh (ktc, 20% O2 v 80% N2 theo th tch) thu c hn hp Y gm kh v hi. Dn Y qua nc vi trong d, sau khi cc phn ng xy ra hon ton thy khi lng dung dch gim m gam, m gn nht vi gi tr

A. 2,75. B. 4,25. C. 2,25 D. 3,75 V nCH4O = nC2H6O nn t cng thc 2 ancol ny l: C1,5H5O.

2

1,5 5

32O :

60,09

22

6 10 4

C H O : xC H O : yC H

CO : a1,86g

: bO

H: z

O+

Btklg: 44a + 18b = 4,74 (1) Bt O: x + 2y + 4z + 0,18 = 2a + b x + 2y + 4z = 2a + b 0,18 Bt C: 1,5x + 3y + 6z = 1,5(x + 2y + 4z) = 1,5(2a + b 0,18) = a 2a + 1,5b = 0,27 (2) (1) v (2) a = 0,075 v b = 0,08 m = 4,74 7,5 = 2,76 chn A.

Cu 50. Cho m gam hn hp gm hai ancol no, n chc, k tip nhau trong dy ng ng, tc dng vi CuO d, nung nng, thu c hn hp X gm kh v hi c t khi hi so vi H2 l 13,75. Cho X phn ng vi lng d dung dch AgNO3 trong NH3 un nng, thu c 64,8 gam Ag. Gi tr ca m l

A. 3,2. B. 7,8 C. 4,6 D. 11 MX = 27,5 Manehit = 27,5.2 18 = 37 X gm HCHO x mol, CH3CHO y mol v H2O (x + y) mol 4x + 2y = 0,6 v 30x + 44y + 18(x + y) = 27,5(2x + 2y) x = y = 0,1 m = 0,1(32 + 46) = 7,8