Giai Chi Tiet Hoa Lan 1-Dhv2014-Www.dethithudaihoc.com

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TRNG I HC VINH TRNG THPT CHUYN

LI GII CHI TIEEYS THI KSCL LP 12 LN I, NM 2014 Mn: HA HC KHI A, B

(Thi gian lm bi: 90 pht; 50 cu trc nghim) H v tn .................................................................. S bo danh .......................... M thi 359

ThS. Quch Vn Long GV Trng THPT Chuyn H Vinh Cho khi lng nguyn t ca cc nguyn t (theo vC): C = 12; H = 1; O = 16; S = 32; Cu = 64; Fe = 56; Cr = 52; Al = 27; N = 14; Ca = 40; Ba = 137; Na = 23; Mg = 24; Cl = 35,5; Zn = 65; K = 39; Ag = 108; Br = 80; P = 31. A. PHN CHUNG CHO TT C TH SINH (40 cu: T cu 1 n cu 40) Cu 1: Hp cht thm X c cng thc phn t C7H8O2; 1 mol X phn ng va vi 2 lt dung dch NaOH 1M. S ng phn cu to tha mn iu kin trn ca X l

A. 8. B. 7. C. 5. D. 6. Gii

C7H8O2 = 7 2 2 8 4

2.

.

nNaOH = 2nX X c 1 nhm OH lin kt trc tip vi nguyn t cacbon ca vng benzen. Cng thc cu to ph hp ca X l

OHOH

CH3

OHOH

CH3

OH

CH3

OH

OH

OHCH3

OH

OH

CH3

OHCH3

OH

p n D Cu 2: Cho hai h cn bng sau trong hai bnh kn:

C (r) + CO2 (k) 2CO(k) ; H = 172 kJ CO (k) + H2O (k) CO2 (k) + H2 (k) ; H = - 41 kJ C bao nhiu iu kin trong cc iu kin sau y lm cc cn bng trn chuyn dch ngc chiu nhau (gi

nguyn cc iu kin khc)? (1) Tng nhit . (2) Thm kh CO2 vo. (3) Thm kh H2 vo. (4) Tng p sut. (5) Dng cht xc tc. (6) Thm kh CO vo. A. 5. B. 2. C. 4. D. 3.

Gii C (r) + CO2 (k) 2CO (k) ; H = 172 kJ (I) CO (k) + H2O (k) CO2 (k) + H2 (k) ; H = - 41 kJ (II)

(1) Tng nhit th cn bng (I) chuyn dch theo chiu thun cn (II) chuyn dch theo chiu nghch. (2) Thm kh CO2 th cn bng (I) chuyn dch theo chiu thun cn (II) chuyn dch theo chiu nghch. (3) thm kh H2 th c cn bng (I) khng chuyn dch, cn (II) chuyn dch theo chiu nghch. (4) Tng p sut th cn bng (I) chuyn dch theo chiu nghch cn cn bng (II) khng chuyn dch. (5) Dng cht xc tc th c hai cn bng khng chuyn dch. (6) Thm kh CO th cn bng (I) chuyn dch theo chiu nghch cn (II) chuyn dch theo chiu thun. Bao gm cc iu kin (1), (2) v (6). p n D

Cu 3: Cho s phn ng:

C2H2 0t xt, X 2 0

3

HPd PbCO t/ ,

Y 0HBr (1 : 1)

80 C Z

Trong X, Y, Z u l cc sn phm chnh. Cng thc cu to thu gn ca Z l A. CH2=CHCHBrCH3. B. CH2=CHCH2CH2Br. C. CH3CH=CHCH2Br. D. CH3CBr=CHCH3.

Gii 2CHCH

0t xt, CH2=CHCCH

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(X) CH2=CHCCH + H2

03Pd PbCO t/ , CH2=CHCH=CH2

(Y) CH2=CHCH=CH2 + HBr CH2=CHCHBrCH3 (Z) p n A Cu 4: Thy phn hon ton 7,55 gam Gly-Ala-Val-Gly trong dung dch cha 0,02 mol NaOH un nng, thu c dung dch X. Cho X tc dng vi 100 ml dung dch HCl 1M. Sau khi kt thc cc phn ng thu c dung dch Y. C cn cn thn dung dch Y thu c m gam cht rn khan. Gi tr ca m l

A. 11,21. B. 12,72. C. 11,57. D. 12,99. Gii

Bi ny c nhiu cch gii nhanh. Chng hn:

Cch 1: Gly Ala Val Gly7 55n302,

= 0,025 mol; nNaOH = 0,02 mol; nHCl = 0,1 mol.

Gly-Ala-Val-Gly + 3H2O 0xt t, 2Gly + Ala + Val (1)

0,025 0,075 0,05 0,025 0,025 Coi X gm: Gly, Ala, Val v NaOH 0,02 mol NaOH + HCl NaCl + H2O (2) 0,02 0,02 0,02 0,02 Amino axit (Gly, Ala, Val) + HCl mui (3) V namino axit > nHCl cn = 0,08 mol nn HCl ht, amino axit cn. Theo nh lut bo ton khi lng: mcht rn + 2H O (2)m = mNaOH + mHCl + mGly-Ala-Val-Gly ban u + 2H O (1)m

m = mNaOH + mHCl + mGly-Ala-Val-Gly ban u + 2H O (1)m - 2H O(2)m = 12,99 gam Cch 2: Coi X gm Gly-Ala-Val-Gly v NaOH. NaOH + HCl NaCl + H2O (1) 0,02 0,02 0,02 Gly-Ala-Val-Gly + 3H2O + 4HCl Mui (2) 0,02 0,06 0,08 Gly-Ala-Val-Gly + 3H2O 3Amino axit (3) 0,005 0,015 mcht rn + 2H O (1)m = mNaOH + mHCl + mGly-Ala-Val-Gly ban u + 2H O(2)(3)m

m = mNaOH + mHCl + mGly-Ala-Val-Gly ban u + 2H O(2)(3)m - 2H O (1)m = 40.0,02 + 36,5.0,1 + 7,55 + 18(0,06 + 0,015 - 0,02) = 12,99 gam p n D Cu 5: Tin hnh cc th nghim sau:

(1) Cho dung dch Na2CO3 vo dung dch AlCl3. (2) Sc kh H2S vo dung dch FeSO4. (3) Sc kh CO2 ti d vo dung dch NaAlO2 (hoc Na[Al(OH)4]. (4) Sc kh NH3 ti d vo dung dch Al(NO3)3. (5) Sc kh H2S vo dung dch CuSO4. (6) Cho dung dch Na2S2O3 vo dung dch H2SO4 long. Sau khi kt thc cc phn ng, c bao nhiu th nghim thu c kt ta? A. 6. B. 5. C. 4. D. 3.

Gii (1) 2Al3+ + 3CO 23

+ 3H2O 2Al(OH)3 + 3CO2 (3) CO2 + NaAlO2 + H2O Al(OH)3 + NaHCO3 (4) Al3+ + 3NH3 + 3H2O Al(OH)3 + 3NH 4

(5) H2S + Br2 S + 2HBr (6) Na2S2O3 + H2SO4 Na2SO4 + S + SO2 + H2O p n B


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Cu 6: Hn hp kh X gm CO, CO2 v N2, t khi ca X so vi H2 l 19. Cho m gam X phn ng hon ton vi 100 ml dung dch Y cha NaOH 2M v Na2CO3 1,5M, thu c dung dch Z. Cho Z tc dng vi lng d dung dch CaCl2, sau khi kt thc phn ng thu c 10 gam kt ta. Gi tr ca m l

A. 15,2. B. 9,5. C. 13,3. D. 30,4. Gii

Ca2+ + CO 23 CaCO3

0,1 0,1 2

3COn cn = 0,1 mol

CO2 + 2OH CO 23 + H2O (1)

0,1 0,2 0,1 2

3COn sau (1) = 0,1 + 0,15 = 0,25 mol 2

3COn phn ng = 0,25 - 0,1 = 0,15 mol

CO2 + CO 23 + H2O 2HCO 3

0,15 0,15

2COn = 0,1 + 0,15 = 0,25 mol

CO2 + 2OH CO 23 + H2O

V N2 v CO cng khi lng mol phn t l 28 gam/mol v u khng phn ng vi dung dch NaOH hoc Na2CO3 nn ta coi X ch gm CO v CO2.

2

CO

CO

n 3n 5

CO (M = 28)

CO2 (M = 44)

38

6

10

nCO = 0,15 mol nX = 0,15 + 0,25 = 0,4 mol m = 38.0,4 = 15,2 gam p n A Cu 7: Hn hp M gm anehit X v xeton Y. t chy hon ton m gam M cn dng va 0,4 mol O2, thu c 0,35 mol CO2 v 0,35 mol H2O. S mol ca Y trong m gam M c th l

A. 0,08 mol. B. 0,10 mol. C. 0,05 mol. D. 0,06 mol. Gii

2 2CO H On n X, Y u l hp cht no, n chc, mch h. t cng thc chung n 2nC H O .

n 2nC H O + 23n 1O

2 2nCO + 2nH O

nM = x + y = 2 2CO O2(1 5n n ), = 0,25 (1)

2COM

n 0 35n 1 4n 0 25

, ,,

X l HCHO

t Y l CnH2nO (n 3) 2COn = x + ny = 0,35 (2)

(1)(2) (n - 1)y = 0,1 y = 0 1 0 1 0 05n 1 3 1

, , ,

Kt hp p n y = 0,05 mol p n C Cu 8: Cho cc pht biu sau:

(1) Trong hp cht vi oxi, nit c cng ha tr cao nht bng 5. (2) CaOCl2 l mui kp. (3) dinh dng ca phn ln c nh gi bng hm lng phn trm photpho trong phn ln. (4) Amophot l hn hp cc mui (NH4)2HPO4 v (NH4)3PO4. (5) Phn hn hp cha nit, photpho, kali c gi chung l phn NPK. (6) Bn nhiu phn m amoni s lm tng chua ca t. S pht biu ng l


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A. 1. B. 4. C. 2. D. 3. Gii

cc pht biu (5), (6) p n C Cu 9: Dy gm cc ion (khng k s in li ca H2O) cng tn ti trong mt dung dch l

A. Fe2+, K+, OH , Cl . B. Ba2+, HSO 4 , K+, NO 3

.

C. Al3+, Na+, S 2 , NO 3 . D. Cu2+, NO 3

, H+, Cl . Gii

A. Fe2+ + 2OH Fe(OH)2 B. Ba2+ + HSO 4

BaSO4 + H+

C. 2Al3+ + 3S 2 + 6H2O 2Al(OH)3 + 3H2S p n D Cu 10: Pht biu no sau y l sai ?

A. Trong cc hp cht, nguyn t kim loi kim ch c s oxi ha + 1. B. Theo chiu tng dn in tch ht nhn, cc kim loi kim (t liti n xesi) c nhit nng chy gim

dn. C. Kim loi kim dng iu ch mt s kim loi khc bng phng php thy luyn. D. Cc kim loi kim u tc dng vi nc nhit thng. p n C v kim loi kim phn ng vi nc ngay nhit thng nn khng dng phng php thy

luyn iu ch kim loi khc. Cu 11: Dung dch X gm NaOH x mol/l v Ba(OH)2 y mol/l v dung dch Y gm NaOH y mol/l v Ba(OH)2 x mol/l. Hp th ht 0,04 mol CO2 vo 200 ml dung dch X, thu c dung dch M v 1,97 gam kt ta. Nu hp th ht 0,0325 mol CO2 vo 200 ml dung dch Y th thu c dung dch N v 1,4775 gam kt ta. Bit hai dung dch M v N phn ng vi dung dch KHSO4 u sinh ra kt ta trng, cc phn ng u xy ra hon ton. Gi tr ca x v y ln lt l

A. 0,1 v 0,075. B. 0,05 v 0,1. C. 0,075 v 0,1. D. 0,1 v 0,05. Gii

M v N phn ng vi dung dch KHSO4 u sinh ra kt ta trng Trong M v N u c cha Ba2+ hay CO 23

ht, Ba2+ cn. Th nghim 1:

OHn (X) = 0,2x + 0,4y

Ba2+ + CO 23 BaCO3

0,01 0,01 CO2 + 2OH CO 23

+ H2O 0,01 0,02 0,01 CO2 + OH HCO 3

0,03 0,03 0,2x + 0,4y = 0,05 (1) Th nghim 2:

OHn (Y) = 0,2y + 0,4x

Ba2+ + CO 23 BaCO3

0,0075 0,0075 CO2 + 2OH CO 23

+ H2O 0,0075 0,015 0,0075 CO2 + OH HCO 3

0,025 0,025 0,4x + 0,2y = 0,04 (2) Gii h (1)(2) x = 0,05 mol/l v y = 0,1 mol/l p n B Cu 12: Hp cht mch h X, c cng thc phn t C4H8O3. Khi cho X tc dng vi dung dch NaOH un nng, thu c mui Y v ancol Z. Ancol Z ha tan c Cu(OH)2. S ng phn cu to tha mn iu kin trn ca X l


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A. 3. B. 1. C. 2. D. 4. Gii

C4H8O3 ( = 1). X phn ng vi NaOH to ra mui v ancol X cha chc este. Z ha tan Cu(OH)2 Z c hai nhm OH hai cacbon cnh nhau tr ln. Cng thc cu to ph hp ca X l CH3COOCH2CH2OH; HCOOCH2CH(OH)CH3; HCOOCH(CH3)CH2OH. p n A Cu 13: S ipeptit mch h ti a c th to ra t mt dung dch gm: H2NCH2CH2COOH, CH3CH(NH2)COOH, H2NCH2COOH l

A. 3. B. 2. C. 9. D. 4. Gii

C 2 - amino axit nn c ti a 22 = 4 ipeptit mch h. Ala-Gly, Gly-Gly, Ala-Ala, Gly-Ala p n D Cu 14: Cho 7,6 gam hn hp X gm Mg v Ca phn ng va vi 4,48 lt (ktc) hn hp kh Y gm Cl2 v O2 thu c 19,85 gam cht rn Z ch gm cc mui clorua v cc oxit kim loi. Khi lng ca Mg trong 7,6 gam X l

A. 2,4 gam. B. 1,8 gam. C. 4,6 gam. D. 3,6 gam. Gii

Gi x, y ln lt l s mol ca Cl2 v O2 c trong 0,2 mol hn hp kh Y. Ta c: x + y = 0,2 (1) Theo nh lut bo ton khi lng: mY = mZ - mX = 19,85 - 7,6 = 12,25 gam 71x + 32y = 12,25 (2) Gii h (1)(2) ta c:

x 0 15moly 0 05mol

,,

Gi a, b ln lt l s mol ca Mg v Ca c trong 7,6 gam X. Ta c: 24a + 40b = 7,6 (3) Mg Mg2+ + 2e a 2a Ca Ca2+ + 2e b 2b

Cl2 + 2e 2Cl 0,15 0,3 O2 + 4e 2O 2 0,05 0,2

2a + 2b = 0,5 (4) Gii h (3)(4) ta c:

a 0 15molb 0 1mol

,,

mMg = 24.0,15 = 3,6 gam p n D Cu 15: Hn hp kh X gm O2 v O3, t khi ca X so vi H2 l 17,6. Hn hp kh Y gm C2H4 v CH4, t khi ca Y so vi H2 l 11. Th tch hn hp kh X (ktc) ti thiu cn dng t chy hon ton 0,044 mol hn hp kh Y l

A. 3,36 lt. B. 2,24 lt. C. 1,12 lt. D. 4,48 lt. Gii

XM = 35,2 gam/mol; YM = 13,8.2 = 27,6 gam/mol.

2

3

O

O

n 12 8 4n 3 2

,,

O2 (M = 32)

O3 (M = 48)

35,2

(35,2 - 32) = 3,2

(48 - 35,2) = 12,8

Y28 16M 22

22

2 4 4C H CHn n = 0,022 mol


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Gi x l s mol X ti thiu cn dng t chy ht 0,044 mol Y 2On = 0,8x v 3On = 0,2x

Quy X v O O (X)n = 2.0,8x + 3.0,2x = 2,2x mol C2H4 + 6O 2CO2 + 2H2O 0,022 0,132 CH4 + 4O CO2 + 2H2O 0,022 0,088 nO = 2,2x = 0,22 x = 0,1 mol VX = 2,24 lt p n B Cu 16: Cho mt s tnh cht sau: (1) cu trc mch khng phn nhnh; (2) tan trong nc; (3) phn ng vi Cu(OH)2; (4) b thy phn trong mi trng kim long, nng; (5) tham gia phn ng trng bc; (6) tan trong dung dch [Cu(NH3)4](OH)2; (7) phn ng vi HNO3 c (xc tc H2SO4 c). Cc tnh cht ca xenluloz l

A. (3), (6), (7). B. (1), (4), (6), (7). C. (2), (3), (5), (6). D. (1), (6), (7). Theo SGK p n D

Cu 17: Cho 200 ml dung dch H3PO4 1M vo 500 ml dung dch KOH 1M. Mui to thnh v khi lng tng ng khi kt thc cc phn ng (b qua s thy phn ca cc mui) l

A. K2HPO4 17,4 gam; K3PO4 21,2 gam. B. KH2PO4 13,6 gam; K2HPO4 17,4 gam. C. KH2PO4 20,4 gam; K2HPO4 8,7 gam. D. KH2PO4 26,1 gam; K3PO4 10,6 gam.

Gii H3PO4 + 2KOH K2HPO4 + H2O 0,2 0,4 0,2 nKOH cn = 0,5 - 0,4 = 0,1 mol KOH + K2HPO4 K3PO4 + H2O 0,1 0,1 0,1

2 4K HPOn cn = 0,2 - 0,1 = 0,1 mol

2 4K HPOm = 174.0,1 = 17,4 gam; 3 4K POm = 212.0,1 = 21,2 gam.

p n A Cu 18: Khi ni v kim loi kim th, pht biu no sau y l sai ?

A. Cc kim loi canxi v stronti c cng kiu mng tinh th lp phng tm din. B. T beri n bari kh nng phn ng vi H2O gim dn. C. Phng php c bn iu ch kim loi kim th l in phn mui clorua nng chy ca chng. D. Khi t nng, cc kim loi kim th u bc chy trong khng kh. Theo SGK p n B

Cu 19: Mt mu nc cng c cha cc ion: Ca2+, Mg2+, HCO 3 , Cl , SO 24

. Cht lm mm mu nc cng trn l

A. HCl. B. NaHCO3. C. Na3PO4. D. BaCl2. Gii

3Ca2+ + 2PO 34 Ca3(PO4)2

3Mg2+ + 2PO 34 Mg3(PO4)2

p n C Cu 20: Cho dy cc cht: phenyl clorua, benzyl clorua, etylmetyl ete, mantoz, tinh bt, nilon - 6, poli(vinyl axetat), t visco, protein, metylamoni clorua. S cht trong dy tc dng c vi dung dch NaOH long, un nng l

A. 8. B. 7. C. 6. D. 5. Gii

Bao gm: benzyl clorua, nilon - 6, poli(vinyl axetat), protein, metylamoni clorua. p n D

Cu 21: Hn hp X gm C3H6, C4H10, C2H2 v H2. Cho m gam X vo bnh kn c cha mt t bt Ni lm xc tc. Nung nng bnh thu c hn hp Y. t chy hon ton Y cn dng va V lt O2 (ktc). Sn phm chy cho hp th ht vo bnh ng nc vi trong d, thu c mt dung dch c khi lng gim 21,45 gam. Nu cho Y i qua bnh ng lng d dung dch brom trong CCl4 th c 24 gam brom phn ng. Mt khc, cho 11,2 lt (ktc) hn hp X i qua bnh ng dung dch brom d trong CCl4, thy c 64 gam brom phn ng. Bit cc phn ng u xy ra hon ton. Gi tr ca V l


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A. 21,00. B. 14,28. C. 10,50. D. 28,56. Gii

Gi x, y, z, t ln lt l s mol ca C3H6, C4H10, C2H2 v H2 c trong m gam X. V Y c phn ng vi dung dch brom nn H2 ht. Ta c: nlk (X) = 2 2H Brn n (Y) x + 2z = t + 0,15

t = x + 2z - 0,15 nX = x + y + z + t = 2x + y + 3z - 0,15

0,5 mol X + Br2: 2Br64n 0 4mol

160,

C3H6 + Br2 C3H6Br2 C2H2 + 2Br2 C2H2Br4

x 2z 0 4 42x y 3z 0 15 0 5 5

, , ,

3x + 4y + 2z = 0,6 = 2 2CO COn (X) = n (Y) = 3CaCOn

3 22 2

CaCO COH O H O

m m mn (X) = n (Y) = 0 675mol

18,

dung dch gim

mX = mY = mC + mH = 12.0,6 + 2.0,675 = 8,55 gam Theo nh lut bo ton nguyn t oxi, ta c:

2 2 2O CO H O

1n = n n2

= 0,9375 mol 2OV = 0,9375.22,4 = 21 lt

p n A Nhn xt: Bi tp ny kh hay! Cu 22: Cho nguyn t ca cc nguyn t: X (Z = 11); Y (Z = 12); L (Z = 17); E (Z = 16); G (Z = 8); Q (Z = 9); T (Z = 18); M (Z = 19). Trng hp no sau y ch gm cc nguyn t v ion c cng cu hnh electron?

A. X+, Y2+, G 2 , L . B. L , E 2 , T, M+. C. X+, Y2+, G 2 , Q. D. Q , E 2 , T, M+. Gii

Cch 1: Cc nguyn t v ion mun cng cu hnh electron th phi c cng s electron. Nhn vo cc p n th ch

c p n B l tha mn. Cch 2:

X (Z = 11): 1s22s22p63s1 X+: 1s22s22p6 Y (Z = 12): 1s22s22p63s2 Y2+: 1s22s22p6 L (Z = 17): 1s22s22p63s23p5 L : 1s22s22p63s23p6 E (Z = 16): 1s22s22p63s23p4 E 2 : 1s22s22p63s23p6 G (Z = 8): 1s22s22p4 G 2 : 1s22s22p6 Q (Z = 9): 1s22s22p5 Q : 1s22s22p6 T (Z = 18): 1s22s22p63s23p6 M (Z = 19): 1s22s22p63s23p6 M+: 1s22s22p63s23p6

X+, Y2+, G2-, H hoc L , E2-, T, M+ c cng cu hnh electron. p n B

Cu 23: Cho dy gm 7 dung dch ring bit: H2N[CH2]4CH(NH2)COOH, HOOC[CH2]2CH(NH2)COOH, H2NCH2COONa, ClH3NCH2COOH, CH3CH(NH2)COOH, C6H5ONa (natri phenolat), C6H5NH3Cl (phenylamoni clorua). S dung dch trong dy c pH > 7 l

A. 2. B. 5. C. 3. D. 4. Gii

Bao gm: H2N[CH2]4CH(NH2)COOH, H2NCH2COONa, C6H5ONa p n C Cu 24: Cht no sau y cha c 3 loi lin kt (ion, cng ho tr, cho - nhn)?

A. K2CO3. B. NaHCO3. C. NaNO3. D. HNO3. Gii

Cng thc phn t Cng thc cu to Cc loi lin kt


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NaNO3 Na+ O N

O

O

1 lin kt ion, 3 lin kt cng ha tr v 1 lin kt cho - nhn.

K2CO3 K+ O

O

C OK+

2 lin kt ion, 4 lin kt cng ha tr.

NaHCO3 Na+ O

O

C O

H

1 lin kt ion v 5 lin kt cng ha tr.

HNO3 O N

O

O

H

4 lin kt cng ha tr v 1 lin kt cho - nhn.

p n C Cu 25: Cho cc pht biu sau:

(1) Anehit va c tnh oxi ha va c tnh kh. (2) Cc phn t phenol khng to lin kt hiro lin phn t. (3) Xiclopropan khng lm mt mu dung dch KMnO4. (4) Benzen khng lm mt mu dung dch brom. (5) Natri fomat tham gia phn ng trng bc. Cc pht biu ng l A. (2), (4), (5). B. (1), (5). C. (1), (3), (5). D. (1), (3), (4), (5).

Gii (2) Cc phn t phenol c to lin kt hiro lin phn t. p n D

Cu 26: Trng hp no sau y khng xy ra phn ng ha hc? A. Sc kh CO2 vo dung dch NaClO. B. Cho kim loi Be vo H2O. C. Sc kh Cl2 vo dung dch FeSO4. D. Cho kim loi Al vo dung dch HNO3 long, ngui.

Gii A. CO2 + NaClO + H2O NaHCO3 + HClO B. Be khng phn ng vi H2O mi nhit . C. 3Cl2 + 6FeSO4 2FeCl3 + 2Fe2(SO4)3 D. Al tan trong dung dch HNO3 long, ngui. Al + HNO3 Al(NO3)3 + (N2, N2O, NO, NH4NO3) + H2O p n B

Cu 27: Cho dy cht: C2H4, C2H5OH, CH3COOH, CH3COOC2H3, C2H2. S cht trong dy trc tip to ra t CH3CHO bng mt phn ng l

A. 2. B. 5. C. 4. D. 3. Gii

CH3CHO + H2 0Ni t, CH3CH2OH

CH3CHO + 12

O2 2 0Mn t, CH3COOH

p n A Cu 28: C cc dung dch ring bit: Cu(NO3)2, HCl, FeCl3, AgNO3, Mg(NO3)2, NiSO4. Nhng vo mi dung dch mt thanh st nguyn cht. S trng hp xut hin n mn ha hc l

A. 5. B. 3. C. 6. D. 4. Gii

Bao gm cc dung dch: Cu(NO3)2, NaHSO4, FeCl3, AgNO3. NiSO4. Fe + Cu(NO3)2 Fe(NO3)2 + Cu


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/18 - M thi 359

Fe + 2HCl FeCl2 + H2 Fe + 2FeCl3 3FeCl2 Fe + 2AgNO3 Fe(NO3)2 + 2Ag Fe + NiSO4 FeSO4 + Ni p n A Cu 29: Cho Zn ti d vo dung dch gm HCl; 0,05 mol NaNO3 v 0,1 mol KNO3. Sau khi kt thc cc phn ng thu c dung dch X cha m gam mui; 0,125 mol hn hp kh Y gm hai kh khng mu, trong c mt kh ha nu trong khng kh. T khi ca Y so vi H2 l 12,2. Gi tr ca m l

A. 61,375. B. 64,05. C. 57,975. D. 49,775. Gii

BM 12 2 2 24 4 , . , Y cha H2. Kh khng mu ha nu trong khng kh l NO Y gm NO v H2. Gi x, y ln lt l s mol H2 v NO. Ta c:

x y 0 125 x 0 025mol2x 30y 0 125 24 4 y 0 1mol

, ,, . , ,

V c kh H2 thot ra v Zn d H+ v 3NO ht mui thu c l mui clorua

Do 3NO

n ban u = 0,15 mol > nNO = 0,1 mol X cha mui NH 4 . Theo bo ton nguyn t N

4NH

n = 0,15 - 0,1 = 0,05 mol.

Theo bo ton s mol electron:

24

Zn NO HNH2n 3n 8n 2n 0 75 phan ng , nZn phn ng = 2Zn

0 75n 0 375mol2

, ,

2 4Cl NaCl KCl

m m 136 0 375 53 5 0 05 58 5 0 05 74 5 0 1 muoi ZnCl NHm m + m . , , . , , . , , . ,

= 64,05 gam p n B Nhn xt: Bi ny kh hay! Cu 30: Hn hp M gm mt anken v hai amin no, n chc, mch h X v Y l ng ng k tip nhau (MX < MY). t chy hon ton mt lng M cn dng va 0,225 mol O2, thu c H2O, N2 v 0,12 mol CO2. Cng thc phn t ca Y l

A. C3H9N. B. C2H7N. C. C4H11N. D. CH5N. Gii

nY = 2 2O CO4 (n 1 5n )3

, = 0,06 mol

aminn = 2CO (amin)amin

n 0 12 2n 0 06

,,

nX = 1 (CH3NH2) nY = 2 (C2H7N)

p n B Cu 31: Cho m gam hn hp bt X gm Mg v Fe vo 200 ml dung dch cha CuCl2 0,5M v HCl 1M. Sau khi cc phn ng xy ra hon ton, thu c m gam hn hp Y gm hai kim loi. Khi lng ca Mg trong m gam hn hp X l


V Y gm hai kim loi Mg, Cu2+ v H+ ht, Fe cn hoc cha phn ng. Mg Mg2+ + 2e x 2x Fe Fe2+ + 2e Ban u: y Phn ng: y1 2y1 Cn: y - y1

Cu2+ + 2e Cu 0,1 0,2 0,1 2H+ + 2e H2 0,2 0,2

mcht rn = 56(y - y1) + 64.0,1 = 24x + 56y 24x + 56y1 = 6,4 (1) Theo nh lut bo ton s mol electron: 2x + 2y1 = 0,4 hay x + y1 = 0,2 (2)


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/18 - M thi 359

Gii h (1)(2) ta c:

1

x 0 15moly 0 05mol

,,

mMg = 0,15.24 = 3,6 gam p n C Cu 32: t chy hon ton mt amin n chc, mch h X bng mt lng khng kh (cha 20% th tch O2, cn li l N2) va , thu c 0,08 mol CO2; 0,1 mol H2O v 0,54 mol N2. Khng nh no sau y l ng ?

A. S nguyn t H trong phn t X l 7. B. Gia cc phn t X khng c lin kt hiro lin phn t. C. X khng phn ng vi HNO2. D. S ng phn cu to tha mn iu kin trn ca X l 1.

Gii Theo nh lut bo ton nguyn t O:

2 2 2O CO H O

1n n n2

= 0,13 mol 2Nn khng kh = 0,52 mol

2Nn (X) = 0,54 - 0,52 = 0,02 mol nX = 2.0,02 = 0,04 mol

t X l CxHyN ta c:

x = 2COX

n 0 08n 0 02

,,

= 2; y = 2H OX

2n 0 2 5n 0 04

,,

(C2H5N)

S nguyn t H trong phn t X l 5. Cng thc cu to ca X: CH2=CHNH2 V c lin kt NH nn X to c lin kt hiro lin phn t. X l amin bc I nn c phn ng vi HNO2. S ng phn cu to ca X l 1. p n D Cu 33: Hn hp M gm ancol X, axit cacboxylic Y (u no, n chc, mch h) v este Z to ra t X v Y. t chy hon ton m gam M cn dng va 0,18 mol O2, sinh ra 0,14 mol CO2. Cho m gam M trn vo 500 ml dung dch NaOH 0,1M un nng, sau khi kt thc cc phn ng thu c dung dch N. C cn dung dch N cn li 3,68 gam cht rn khan. Cng thc ca Y l

A. C2H5COOH. B. HCOOH. C. C3H7COOH. D. CH3COOH. Gii

t (X): R'OH; (Y): RCOOH (Z): RCOOR' Gi x, y, z ln lt l s mol ca X, Y v Z cha trong m gam M. Khi t ancol X th

2 2O (X) CO (X)n 1 5n, nn

y + z = 2 2CO O1 5n n, = 0,03

Khi cho M phn ng vi dung dch NaOH th ancol X khng phn ng. RCOOH + NaOH RCOONa + H2O y y y RCOOR' + NaOH RCOONa + R'OH z z z nNaOH cn = 0,05 - 0,03 = 0,02 mol mcht rn = mRCOONa + mNaOH cn (R + 67).0,03 + 40.0,02 = 3,68 R = 29 (C2H5-) Cng thc ca Y l C2H5COOH p n A Cu 34: ng dng no sau y khng phi ca kh SO2?

A. Sn xut axit sunfuric. B. Ty trng giy, bt giy. C. Kh trng nc sinh hot. D. Chng nm mc cho lng thc, thc phm. Theo SGK p n C

Cu 35: Hn hp X gm hai ancol X1 v X2 ( 1 2X XM M ). un nng X vi H2SO4 c thu c 0,03 mol H2O v hn hp Y gm hai anken ng ng lin tip, ba ete v hai ancol d. t chy hon ton Y thu c 0,13 mol CO2 v 0,15 mol H2O. Cng thc phn t ca X1 l

A. C2H5OH. B. C3H7OH. C. CH3OH. D. C3H5OH.


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/18 - M thi 359

Gii Tch nc X thu c hn hp c cha 2 anken ng ng lin tip nn X gm hai ancol no, n chc, mch h (C 2). Ta c:

2 2 2 2CO (X) CO (Y) H O(X) H O(Y)n n 0 13mol; n n 0 03 0 18mol, , ,

22 2

CO

H O CO

nn

n n

= 0 13

0 18 0 13,

, , = 2,6 n1 = 2 (C2H5OH) < n = 2,6 < n2 = 3 (C3H7OH)

p n A Cu 36: Dy gm cc polime c cu trc mch phn nhnh l

A. amilopectin, glicogen, poli(metyl metacrylat). B. amilopectin, glicogen. C. t visco, amilopectin, poli isopren. D. nha novolac, t nitron, poli(vinyl clorua). Theo SGK p n B

Cu 37: Cho 200 ml dung dch gm KOH 1M v NaOH 0,75M vo 100 ml dung dch AlCl3 1M. Sau khi kt thc cc phn ng thu c m gam kt ta. Gi tr ca m l

A. 3,90. B. 11,70. C. 7,80. D. 5,85. Gii

KOH NaOHOHn n n 0 2 1 0 2 0 75 0 35mol;, . , . , , 3 3AlClAln n = 0,1 mol.

Al3+ + 3OH Al(OH)3 0,1 0,3 0,1 Al(OH)3 + OH Al(OH) 4

0,05 0,05 mkt ta = 3Al(OH)m = 78.0,05 = 3,9 gam p n A

Cu 38: Hn hp M gm hai este n chc. Cho m gam M tc dng vi lng d dung dch NaOH un nng, thu c 17 gam mt mui v 12,4 gam hn hp N gm hai anehit thuc cng dy ng ng. T khi hi ca N so vi H2 l 24,8. Cho m gam M phn ng vi lng d dung dch AgNO3 trong NH3 un nng, thu c ti a a gam Ag. Bit cc phn ng u xy ra hon ton. Gi tr ca m v a ln lt l

A. 25,15 v 108. B. 25,15 v 54. C. 19,40 v 108. D. 19,40 v 54. Gii

Manehit = 2.24,8 = 49,6 gam/mol M1 = 44 (CH3CHO) < Manehit = 49,6 < M2 (RCH2CHO)

nmui = nRCOONa = nanehit = nNaOH = 0,25 mol R + 67 = 17

0 25, = 68 R = 1

Cng thc ca hai este: HCOOCH=CH2; HCOOCH=CHR m = 95.0,15 + 109.0,1 = 25,15 gam. HCOOCH=CH2 3 30

AgNO NHt

/ 4Ag

HCOOCH=CHR 3 30AgNO NH

t/ 4Ag

nAg = 4nanehit = 1 mol mAg = 108 gam Theo nh lut bo ton khi lng: m + mNaOH = mmui + manehit m = 17 + 12,4 - 40.0,25 = 19,4 gam p n C Cu 39: Ha tan ht hn hp X gm FeS2, Cu2S v Cu trong V ml dung dch HNO3 1M. Sau khi cc phn ng xy ra hon ton thu c dung dch Y ch cha hai mui sunfat v 5,376 lt NO (sn phm kh duy nht, ktc). Gi tr ca V l

A. 960. B. 240. C. 120. D. 480. Gii

V dung dch thu c ch cha hai mui sunfat nn NO 3 ht

3HNOn = 3NOn = nNO = 0,24 mol

V = 240 ml p n B


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/18 - M thi 359

Cu 40: Hn hp M gm 4 axit cacboxylic. Cho m gam M phn ng ht vi dung dch NaHCO3, thu c 0,1 mol CO2. t chy hon ton m gam M cn dng va 0,09 mol O2, sinh ra 0,14 mol CO2. Gi tr ca m l

A. 5,80. B. 5,03. C. 5,08. D. 3,48. Gii

nCOOH = 2COn = 0,1 mol Theo nh lut bo ton nguyn t:

2 2 2COOH O CO H O

1n n n n2

2H On 2(0,1 + 0,09 - 0,14) = 0,1 mol

m = O (axit ) C Hm m m = 32.0,1 + 12.0,14 + 2.0,1 = 5,08 gam p n C

B. PHN RING: Th sinh ch c chn lm 1 trong 2 phn (Phn I hoc Phn II) Phn I. Theo chng trnh Chun (10 cu: T cu 41 n cu 50)

Cu 41: Thc hin phn ng nhit nhm hn hp gm Al v Cr2O3 trong iu kin khng c khng kh. Sau mt thi gian thu c 21,95 gam hn hp X. Chia X thnh hai phn bng nhau. Cho phn 1 vo lng d dung dch HCl long nng, thu c 3,36 lt H2 (ktc). Ha tan phn 2 vo lng d dung dch NaOH c nng, thu c 1,68 lt H2 (ktc). Bit cc phn ng ca phn 1 v phn 2 u xy ra hon ton. Hiu sut phn ng nhit nhm l

A. 30,0% B. 60,0%. C. 75,0%. D. 37,5%. Gii

Gi a, b ln lt l s mol ban u ca Al v Cr2O3. Ta c: 27a + 152b = 21,95 (1) 2Al + Cr2O3

0t Al2O3 + 2Cr 2x x x 2x X gm: x mol Al2O3, 2x mol Cr, (a - 2x) mol Al, (b - x) mol Cr2O3. Phn 2 + NaOH: Cr khng phn ng. Al2O3 + 2NaOH + 3H2O 2Na[Al(OH)4] Cr2O3 + 2NaOH + 3H2O 2Na[Cr(OH)4] 2Al + 2NaOH + 6H2O 2Na[Al(OH)4] + 3H2 0,05 0,075 nAl = a - 2x = 0,05.2 = 0,1 (2) Phn 1 + HCl: Cr2O3 + 6HCl 2CrCl3 + 3H2O Al2O3 + 6HCl 2AlCl3 + 3H2O 2Al + 6HCl 2AlCl3 + 3H2 0,05 0,075 Cr + 2HCl CrCl2 + H2 0,075 0,075 nCr = 2x = 2.0,075 = 0,15 x = 0,075 mol Kt hp (1)(2) a = 0,25 mol v b = 0,1 mol

V 2 3Cr On

0 125 0 11

, , ban uAl ban un

2 Tnh hiu sut phn ng theo Cr2O3

H = 0 075 100 750 1, . % %

, p n C

Cu 42: Qu trnh xy ra trong pin in ha Zn Cu v qu trnh xy ra khi in phn dung dch CuSO4 vi anot bng Zn c c im chung l

A. anot xy ra s kh H2O v catot xy ra s oxi ha ion Cu2+. B. anot xy ra s oxi ha H2O v catot xy ra s kh ion Cu2+. C. anot xy ra s oxi ha Zn v catot xy ra s kh ion Cu2+. D. anot xy ra s kh Zn v catot xy ra s oxi ha ion Cu2+.

Gii Qu trnh xy ra trong pin in ha Zn Cu v qu trnh in phn dung dch CuSO4 vi anot bng Zn c im chung l:


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/18 - M thi 359

anot: Xy ra s oxi ha Zn. Zn Zn2+ + 2e catot: Xy ra s kh cation Cu2+. Cu2+ + 2e Cu p n C Cu 43: Khng nh no sau y l ng ?

A. Dung dch ancol etylic trong nc tn ti 3 loi lin kt hiro. B. Axit fomic khng lm mt mu nc brom. C. Khi tc dng vi hiro, xeton b kh thnh ancol bc I tng ng. D. Glixerol tan v hn trong nc v c v ngt. Theo SGK hoc dng phng php loi tr p n D

Cu 44: Thy phn hon ton m gam pentapeptit M mch h, thu c hn hp X gm hai - amino axit X1, X2 (u no, mch h, phn t c mt nhm NH2 v mt nhm COOH). t chy hon ton hn hp X trn cn dng va 2,268 lt O2 (ktc), ch thu c H2O, N2 v 1,792 lt CO2 (ktc). Gi tr ca m l

A. 2,295. B. 1,935. C. 2,806. D. 1,806. Gii

t 2n 2n 1C H O N l cng thc chung ca X1 v X2. Ta c:

2 2 2n 2n 1C H O N CO O

4 4n (1 5n n ) = (1 5 0 08 0 10125) = 0,025mol3 3

, , . , ,

x = n 2n 1O N2

C H1 n 0 0125mol2

,

22n 2n 1

CO

C H O N

nn

n

= 3,2 mX = (14.3,2 + 47).0,025 = 2,295 gam

M + 4H2O nX1 + mX2 a 4a na ma

2n 2n 1C H O Nn

= (n + m)a = 0,025 a = 0,005 mol

2H On = 4.0,005 = 0,02 mol

m = mX - 2H Om = 2,295 - 18.0,02 = 1,935 gam p n B

Cu 45: Ha tan ht 2,24 gam bt Fe vo 120 ml dung dch HCl 1M, thu c dung dch X. Cho X tc dng vi lng d dung dch AgNO3, sau khi kt thc cc phn ng thu c V lt kh NO (sn phm kh duy nht, ktc) v m gam cht rn. Gi tr ca m v V ln lt l

A. 17,22 v 0,224. B. 1,08 v 0,224. C. 18,3 v 0,448. D. 18,3 v 0,224. Gii

Fe + 2H+ Fe2+ + H2 0,04 0,08 0,04 Ag+ + Cl AgCl 0,12 0,12 3Fe2+ + NO 3

+ 4H+ 3Fe3+ + NO + 2H2O 0,03 0,04 0,01 Ag+ + Fe2+ Ag + Fe3+ 0,01 0,01 m = mAg + mAgCl = 143,5.0,12 + 108.0,01 = 18,3 gam; VNO = 0,01.22,4 = 0,224 lt p n D Cu 46: Thc hin cc th nghim sau:

(1) Cho Fe2O3 vo dung dch HI. (2) Cho Cu vo dung dch FeCl3. (3) Cho dung dch HCl vo dung dch Fe(NO3)2. (4) Sc kh SO2 vo dung dch KMnO4. (5) Sc kh CO2 vo dung dch NaOH. (6) Sc kh O2 vo dung dch KI. S th nghim c phn ng oxi ha - kh xy ra l A. 5. B. 2. C. 3. D. 4.

Gii Bao gm cc th nghim: (1), (2), (3), (4).


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/18 - M thi 359

Fe2O3 + 6HI 2FeI2 + I2 + 3H2O Cu + 2FeCl3 CuCl2 + 2FeCl2 3Fe2+ + NO 3

+ 4H+ 3Fe3+ + NO + 2H2O 5SO2 + 2KMnO4 + 2H2O K2SO4 + 2H2SO4 + 2MnSO4 CO2 + 2NaOH Na2CO3 + H2O Hoc: CO2 + NaOH NaHCO3 p n D Cu 47: Hn hp X gm hai axit cacboxylic. Cho m gam X phn ng vi lng d dung dch AgNO3 trong NH3 un nng, kt thc cc phn ng thu c 21,6 gam Ag. Cho ton b lng X trn phn ng va vi 200 ml dung dch NaOH 1M, thu c dung dch Y. C cn dung dch Y cn li 13,5 gam cht rn khan. Cng thc ca hai axit cacboxylic l

A. HCOOH v HOOCCOOH. B. CH3COOH v HOOCCOOH. C. HCOOH v C2H3COOH. D. HCOOH v C2H5COOH.

Gii Axit c phn ng trng bc l HCOOH. HCOOH + 2[Ag(NH3)2]OH (NH4)2CO3 + 2NH3 + H2O + 2Ag 0,1 0,2 X + NaOH: nNaOH = 0,2 mol HCOOH + NaOH HCOONa + H2O 0,1 0,1 0,1 R(COOH)x + xNaOH R(COONa)x + xH2O a ax a nNaOH = 0,1 + ax = 0,2 ax = 0,1 mcht rn = 68.0,1 + (R + 67x)a = 13,5 Ra = 0 R = 0 Cng thc ca axit cn li l HOOCCOOH p n A Cu 48: Pht biu no sau y l sai khi so snh tnh cht ha hc ca C2H2 v CH3CHO ?

A. C2H2 v CH3CHO u lm mt mu nc brom. B. C2H2 v CH3CHO u c phn ng trng bc. C. C2H2 v CH3CHO u c phn ng cng vi H2 (xc tc Ni, t0). D. C2H2 v CH3CHO u lm mt mu dung dch KMnO4. Theo SGK p n B

Cu 49: Cho cc pht biu sau: (1) Amiloz c cu trc mch phn nhnh. (2) Fructoz lm mt mu nc brom. (3) Saccaroz khng b oxi ha bi dung dch AgNO3 trong NH3, un nng. (4) Xenluloz trinitrat c dng lm thuc sng. (5) Thy phn mantoz thu c glucoz v fructoz. (6) Saccaroz ch c cu to dng mch vng. S pht biu ng l A. 2. B. 4. C. 1. D. 3. Theo SGK bao gm: (3), (4), (6) p n D

Cu 50: Ha tan hn hp X gm 3,2 gam Cu v 23,2 gam Fe3O4 bng lng d dung dch H2SO4 long, thu c dung dch Y. Cho dung dch NaOH d vo Y thu c kt ta Z. Nung Z trong khng kh n khi lng khng i, thu c m gam cht rn. Bit cc phn ng u xy ra hon ton. Gi tr ca m l

A. 28,0. B. 26,4 C. 27,2. D. 24,0. Gii

Cch 1: Phng php thng thng. nCu = 0,05 mol;

3 4Fe On = 0,1 mol

Fe3O4 + 4H2SO4 FeSO4 + Fe2(SO4)3 + 4H2O 0,1 0,1 0,1 Cu + Fe2(SO4)3 CuSO4 + 2FeSO4 0,05 0,05 0,05 0,1


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/18 - M thi 359

2FeSO4 2NaOH 2Fe(OH)2 0

2O t, Fe2O3 0,2 0,1 Fe2(SO4)3 3NaOH 2Fe(OH)3

0t Fe2O3 0,05 0,05 CuSO4 2NaOH Cu(OH)2

02O t, CuO

0,05 0,05 m = 160.0,15 + 80.0,05 = 28 gam Cch 2: Bo ton nguyn t.

CuO Cun n 0 05mol;, 2 3 3 4Fe O Fe O3n n2

= 0,15 mol

m = 2 3CuO Fe O

m m = 160.0,15 + 80.0,05 = 28 gam p n A Phn II. Theo chng trnh Nng cao (10 cu: T cu 51 n cu 60)

Cu 51: Nhn xt no sau y l sai ? A. Cho dung dch K2Cr2O7 vo dung dch Ba(NO3)2 xut hin kt ta mu vng ti. B. Cr(OH)2 tan trong dung dch NaOH khi c mt O2. C. Trong mi trng kim, ion CrO 24

(mu vng) phn ng vi H2O sinh ra ion Cr2O 27 (mu da cam).

D. Trong mi trng axit H2SO4 long, ion Cr2O 27 oxi ha c H2S thnh S.

Gii A. 2Ba2+ + Cr2O 27

+ H2O 2BaCrO4 + 2H+ (vng ti) B. 4Cr(OH)2 + O2 + 2H2O 4Cr(OH)3 Cr(OH)3 + NaOH Na[Cr(OH)4] Ngoi ra Na[Cr(OH)4] c th b oxi ha thnh mui Na2CrO4 C. Ion Cr2O 27

khng tn ti trong mi trng kim.

D. 3H2S + Cr2O 27 + 8H+ 3S + 2Cr3+ + 7H2O

p n C Cu 52: Cho 11,25 gam hn hp bt gm Zn v Cu vo 250 ml dung dch AgNO3 1M. Sau khi cc phn ng xy ra hon ton, thu c dung dch X v 30,2 gam cht rn. Khi lng ca Cu trong hn hp ban u l


V mAg max = 0,25.108 = 27 gam < 30,2 gam nn AgNO3 ht v hn hp kim loi cn. Zn + 2Ag+ Zn2+ + 2Ag x1 2x1 2x1 Cu + 2Ag+ Cu2+ + 2Ag y1 2y1 2y1 m tng = (216 - 65)x1 + (216 - 64)y1 = 30,2 - 11,25 151x1 + 152y1 = 18,95 (1)

Agn = 2(x1 + y1) = 0,25 x1 + y1 = 0,125 (2)

Gii h (1)(2) ta c:

11

x 0 05moly 0 075mol

,,

Do y1 > 0 nn nZn ban u = x1 = 0,05 mol mCu ban u = 11,25 - 65.0,05 = 8 gam p n A Cu 53: Hn hp M gm axit axetic v anehit X. t chy hon ton m gam M cn dng va 0,13 mol O2, sinh ra 0,1 mol CO2 v 0,1 mol H2O. Cho ton b lng M trn vo lng d dung dch AgNO3 trong NH3 un nng, sau khi kt thc cc phn ng thu c 0,04 mol Ag. Cng thc ca X l

A. C3H7CHO. B. HCHO. C. CH3CHO. D. C2H5CHO. Gii


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/18 - M thi 359

2COn = 2H On X l anehit no, n chc, mch h (CnH2nO hay RCHO) Gi a, b ln lt l s mol CH3COOH v CnH2nO. Ta c:

a + 0,5b = 2 2 2CO H O O

1n n n2

= 0,02

M + AgNO3/NH3: Xt hai trng hp sau: Nu RCHO HCHO CH3COOH + NH3 CH3COONH4 RCHO + 2[Ag(NH3)2]OH RCOONH4 + 3NH3 + 2Ag + H2O b 2b nAg = 2b = 0,04 b = 0,02 mol a = 0,01 mol

2COn 2a nb 2 0 01 0 02n 0 1. , , , n = 4 (C3H7CHO)

Nu RCHO l HCHO CH3COOH + NH3 CH3COONH4 HCHO + 4[Ag(NH3)2]OH (NH4)2CO3 + 6NH3 + 4Ag + 2H2O b 4b nAg = 4b = 0,04 b = 0,01 mol a = 0,015 mol

2COn 2a b 2 0 015 0 01 0 04 0 1. , , , , (loi)

p n A Cu 54: Cho s chuyn ha:

X HBr (1:1) Y 0dd NaOHlong

t Z 0

CuOt

T 3 30dd AgNO NH

t/ CH3CH2COONH4

Trong Y, Z, T l cc sn phm chnh. Cht X l A. propen. B. xiclopropan. C. propin. D. eten.

Gii

CH2

H2C CH2 + HBr CH3CH2CH2Br (Y) CH3CH2CH2Br + NaOH

0t CH3CH2CH2OH + NaBr (Z) CH3CH2CH2OH + CuO

0t CH3CH2CHO + Cu + H2O (T) CH3CH2CHO + 2[Ag(NH3)2]OH CH3CH2COONH4 + 3NH3 + 2Ag + H2O p n B Cu 55: Cho hn hp kh gm N2 v H2 vo bnh kn, chn khng (dung tch khng i), c cha sn cht xc tc. Sau khi nung nng bnh mt thi gian ri a v nhit ban u, thy p sut trong bnh gim 18,4% so vi p sut ban u. T khi ca hn hp kh thu c sau phn ng so vi H2 l 6,164. Hiu sut phn ng tng hp NH3 l

A. 29,67%. B. 60%. C. 70,33%. D. 40%. Gii

V hiu sut phn ng khng ph thuc vo s mol ban u ca hn hp kh N2 v H2 nn n gin ta coi nhh kh ban u = 1 mol. t a, b ln lt l s mol N2 v H2 ban u. Ta c: nt = a + b = 1 (1) Phn ng tng hp NH3: N2 + 3H2 0

xt

t 2NH3

B: a b 0 [ ]: (a x) (b 3x) 2x nS = (a x) + (b 3x) + 2x = (a + b) - 2x = 1 2x V, T = const:

S S

t t

1 0 184P n 1 2xP n 1 1

, x = 0,092 ns = 0,816 mol


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/18 - M thi 359

mt = mS = 0,816.2.6,164 = 10,06 gam hay 28a + 2b = 10,06 (2) Gii h (1)(2) ta c: a = 0,31 mol; b = 0,69 mol.

V 22

HN

nn

3 nn tnh hiu sut phn ng theo s mol H2.

3 0 092 100 400 69. , . % %

,

p n D Cu 56: Thc hin cc th nghim sau:

(1) Cho kim loi Mg ti d vo dung dch FeCl3. (2) Cho kim loi Na vo dung dch CuSO4. (3) Cho AgNO3 vo dung dch Fe(NO3)2. (4) Nhit phn AgNO3. (5) Cho kh CO i qua ng ng bt Al2O3 nung nng. Cc th nghim thu c kim loi khi kt thc cc phn ng l A. (1), (2), (3), (4). B. (1), (3), (4). C. (2), (5). D. (1), (3), (4), (5).

Gii (1) Mg + 2FeCl3 MgCl2 + 2FeCl2 Mg d + FeCl2 MgCl2 + Fe (2) 2Na + 2H2O + CuSO4 Cu(OH)2 + Na2SO4 + H2 (3) AgNO3 + Fe(NO3)2 Fe(NO3)3 + Ag (4) 2AgNO3

0t 2Ag + 2NO2 + O2 (5) Al2O3 + CO

0t Khng xy ra khi nung nng. Cu 57: Cho 0,1 mol este n chc X tc dng va vi 200 ml dung dch NaOH 1M un nng, thu c 19,8 gam hn hp hai mui. S ng phn cu to tha mn iu kin trn ca X l

A. 5. B. 2. C. 4. D. 3. Gii

nNaOH : nX = 1 : 2 X c dng RCOOAr RCOOAr + 2NaOH RCOONa + ArONa + H2O 0,1 0,2 0,1 0,1 R + Ar = 198 - 67 - 39 = 92 R = 1 (H-) v Ar = 91 (C7H7-) hoc R = 15 (CH3-) v Ar = 77 (C6H5-) Cng thc cu to ph hp ca X l

CH3COO HCOO CH3

HCOO

CH3

HCOO

CH3 p n C Cu 58: Cho cc chuyn ha sau:

X + H2O 0xt t, Y

Y + Br2 + H2O Axit gluconic + HBr Axit gluconic + NaHCO3 Z + Natri gluconat + H2O Z + H2O nh sngclorophin X + E

Cc cht X, Y ln lt l A. xenluloz, glucoz. B. tinh bt, glucoz. C. tinh bt, fructoz. D. saccaroz, glucoz.

Gii C6H10O5)n + nH2O

0H t, nC6H12O6 tinh bt glucoz CH2OH[CHOH]4CHO + Br2 + H2O CH2OH[CHOH]4COOH + 2HBr CH2OH[CHOH]4COOH + NaHCO3 CH2OH[CHOH]4COONa + CO2 + H2O (Z)


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/18 - M thi 359

6nCO2 + 5nH2O nhsngclorophin (C6H10O5)n + 6nO2

p n B Cu 59: Cho 0,1 mol amino axit M phn ng va vi 80 ml dung dch HCl 1,25M. C cn cn thn dung dch to thnh thu c 17,35 gam mui khan. Bit M l hp cht thm. S ng phn cu to tha mn iu kin trn ca M l

A. 4. B. 2. C. 1. D. 3. Gii

nM : nHCl = 1 : 1 M cha 1 nhm NH2 R(NH2)(COOH)x + HCl R(NH3Cl)(COOH)x Theo nh lut bo ton khi lng: mM = mmui - mHCl = 17,35 - 36,5.0,1 = 13,7 gam

M = 13 70 1

,,

= 137 gam/mol R + 45x + 16 = 137 R = 121 - 45x

Nu x = 1 R = 76 (C6H4) M l H2NC6H4COOH Nu x = 2 R = 31 (loi)

COOHNH2

COOH

NH2

COOH

NH2 p n D Cu 60: Cho dy cht: Al, Al(OH)3, NaHCO3, (NH4)2CO3, CH3COONH4, CH3COOC2H5, Na2HPO3. S cht trong dy c tnh lng tnh l

A. 7. B. 5. C. 4. D. 6. Gii

Bao gm: Al(OH)3, NaHCO3, (NH4)2CO3, CH3COONH4. p n C

----------- HT ---------- Chc cc em t kt qu cao hn na trong k thi th sp ti!


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