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gii chi tit thi th i hc - m 602

Cu 1. Cc cht khi t chy hon ton cho2 2CO H O

n = n l

(3). Xicloankan CnH2n(5). Anken CnH2n

(6). Ancol khng no (c mt lin kt i C=C), mch h CnH2nOa(8). Anehit no, n chc, mch h CnH2nO(9). Axit no, n chc, mch h CnH2nO2

Vy dy l 3, 5, 6, 8, 9. p n ng l CCh : Hp cht dng CnH2nOx th khi t chy hon ton u to ra:

2 2CO H On = n

Cu 2. X phi l axit 3 hiroxipropanoic. Tht vyHO-CH2-CH2-COOH + Na NaO-CH2CH2COONa + H2

a (mol) a (mol)HO-CH2-CH2-COOH + NaHCO3 HOCH2CH2COONa + CO2 + H2O

a (mol)

p n ng l C

Cu 3. Theo bi ra:2

-3-3

O

33,6.10n = = 1,5.10

22,4(mol)

PTHH: 2H2O2 2MnO 2H2O + O2

3.10-3 (mol) 1,5.10-3 (mol)Suy ra:

2 2

-3

H O (p/)n = 3.10 (mol)

Tc trung bnh ca phn ng (tnh theo H2O2)

2 2 2 2

2 2

-3H O H O (P / ) -4

H O

C n 3.10v = = = = 5,0.10

t t.V 60.0,1

(mol/(l.s))

p n ng l CCu 4. Vit phng trnh ion rt gn ca cc phn ng cho:(1). SO4

2- + Ba2+ BaSO4 (2). SO4

2- + Ba2+ BaSO4 (3). SO4

2- + Ba2+ BaSO4 (4). 2H+ + SO4

2- + BaSO3 BaSO4 + SO2 + H2O(5). 2NH4

+ + SO42- + Ba2+ + 2OH- BaSO4 + 2NH3 + 2H2O

(6). SO42- + Ba2+ BaSO4

Cc phn ng u c cng phng trnh ion rt gn: (1), (2), (3), (6)p n ng l B

Cu 5. Theo bi ra: nHCl = 0,2.0,1 = 0,02 (mol); nNaOH =40.4

= 0,04 (mol)100.40 Theo bi ra:+ 0,02 mol amino axit X tc dng va vi 0,02 mol HCl trong X c 1 nhm NH2+ 0,02 mol amino axit X tc dng va vi 0,04 mol NaOH trong X c 2 nhm COOHt cu to X dng H2N-R-(COOH)2Ta c: H2N-R(COOH)2 + HCl ClH3N-R(COOH)2

0,02 mol 0,02 mol 0,02 (mol)

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Suy ra:3 2ClH NR(COOH)

3,67M = = 183,5

0,02

R + 35,5 + 17 + 90 = 183,5 R = 41 (C3H5)Vy X l H2NC3H5 (COOH)2. p n ng l D

Cu 6. Photpho trng (P4) c cu trc tinh th phn t ( nt mng l cc phn t P4)-Nc (H2O rn) thuc loi tinh th phn t (nt mng l cc phn t H2O)-Kim cng c cu trc tinh th nguyn t ( nt mng l cc nguyn t cacbon)- th rn, mui n (NaCl) tn ti dng tinh th ion (nt mng l cc ion Na+, Cl- lun phin

nhau trn cc nh ca hnh lp phng)p n ng l B

Cu 7. Theo bi ra: nKOH = 0,1.0,4 = 0,04 (mol); nancol =-3336.10

= 0,015 (mol)22,4

Ta c s :Hn hp X (no, n chc) + KOH 1 mui + 1 ancol Trong khi X c 1 este RCOOR v 1 axit RCOOHSuy ra: nRCOOR = nancol = 0,015 (mol)

nRCOOH

= nKOH

- nancol

= 0,04 0,015 = 0,025 (mol)Gi thit, cc hp cht trong X u mch h t c cng thc chung l CnH2nO2 (este v

axit no, h, n chc, c cng cng thc chung)Phng trnh t chy: CnH2nO2 2+O nCO2 + nH2O

2 2CO H O

n = n

Theo bi ra, khi lng bnh ng Ca(OH)2 tng chnh bng tng khi lng ca CO2 v H2O chp th vo, do :

2 2CO H O44.n + 18.n = 682

M2 2CO H O

n = n

Suy ra:2 2CO H O

n = n = 0,11 (mol)

Gi s nguyn t cacbon trong mt phn t axit RCOOH l a, s nguyn t cacbon trong mtphn t este l (a + r)

V nguyn t cacbon c bo ton, nn ta c:

2C(RCOOH) C(RCOOR') C(O )n + n = n

0,025.a + 0,015.(a + r) = 0,11.1 5a + 3(a + r) = 22 8a + 3r = 22Ta c bng:

r 1 2 3 4 5a 19/8 2 13/8 10/8 7/8Kl Loi Tha Loi Loi Loi

Vy axit l CH3COOH; este l CH3COOC2H5. p n ng l D

Cu 8. Theo bi ra: nNaOH = a (mol);2H

22,4.an = = a (mol)22,4

Ta thy: nX = nNaOH =2H

n

trong phn t X c 1 nhm chc tc dng vi 1 NaOH; 2 nhm chc tc dng c vi 2NaDo cu to X c th l HO-CH2-C6H4-OH. Tht vy:HO-CH2-C6H4-OH + NaOH HO-CH2-C6H4-ONa + H2O

HO-CH2-C6H4-OH + 2Na NaO-CH2-C6H4-ONa + H2p n ng l A

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Cu 9. t Y trn ngn la n kh, thy ngn la c mu vng trong Y c nguyn t natriHai mui X, Y c th l KMnO4, NaNO3:2KMnO4 K2MnO4 + MnO2 + O2 2 (mol) 1 (mol)

NaNO3 2NaNO2 + O22 (mol) 1(mol)p n ng l A

Cu 10.

+ X (C3H7NO2) + NaOH H2NCH2COONa + Z

cht hu c Z l CH3OH. Tht vy:H2NCH2COOCH3 + NaOH H2NCH2COONa + CH3OH

+ Y (C3H7NO2) + NaOH CH2=CHCOONa + T + .

Kh T l NH3. Tht vy:CH2=CHCOONH4 + NaOH CH2=CHCOONa + NH3 + H2O

Vy Z, T ln lt l CH3OH v NH3. p n ng l ACu 11. Cc ipeptit to ra t alanin v glyxin:

Ala Gly; Gly Ala; Gly Gly; Ala Ala (4 ipeptit)p n ng l A

Cu 12. Theo bi ra:

2H O

0,351n = = 0,0195 (mol)

18;

2CO

0,4368n = = 0,0195 (mol)

22,4

2 2CO H O

n = n

Mt khc: X tc dng c vi Cu(OH)2Suy ra, X c th l mt anehit no, h, n chc. Vy X c th l C2H5CHOp n ng l BCu 13. S phn ng:

Al H2SO4 (l) Al2(SO4)3 Ba(OH)2 BaSO4 O2, to BaSO4Fe FeSO4 Fe(OH)2 Fe2O3

H2SO4(X) (Y) (Z)

Vy trong Z c BaSO4 v Fe2O3. p n ng l CCu 14.

a)4HCl + PbO2 PbCl2 + Cl2 + 2H2OCht kh

b) HCl + NH4HCO3 NH4Cl + CO2 + H2Oc) 2HCl + 2HNO3 2NO2 + Cl2 + 2H2OCht khd) 2HCl + Zn ZnCl2 + H2 C 2 phn ng trong HCl th hin tnh khp n ng l D

Cu 15. T visco l t nhn to-Phn ng iu ch cao su buna N t buta 1,3 ien v acrilonitrin(CH2=CH-CN) l phn ng trng hp-Poli (etylen terephtalat) c iu ch bng phn ng trng ngng:

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nHOOC-C6H4-COOH + nHO-CH2-CH2-OH axit terephtalic etilen glicol (-OC-C6H4-COO-CH2-CH2-O-)n + 2nH2O

Poli (etylen terephtalat)

-Trng hp stiren thu c polistiren:nC6H5-CH=CH2

ot ,p, xt (-CH(C6H5)-CH2-)n

Polistirenp n ng l C

Cu 16. iu kin cc cht (monome) tham gia phn ng trng hp l c lin kt bi trong phn t.Th d nh:

+ 1,1,2,2 tetraftoluen: nCF2=CF2 (-CF2-CF2-)n+ Propilen: nCH2=CHCH3 (-CH2-CH(CH3)-)n+ Stiren: nC6H5CH=CH2 (-CH(C6H5)-CH2-)n

Polistiren+ Vinyl clorua: nCH2=CHCl (-CH2-CH(Cl)-)n

Poli (vinyl clorua)

p n ng l BCu 17. Xenluloz (C6H10O5)n c cc tnh cht:-C dng si (si xenluloz): Phn t c phn t khi rt ln, mch khng phn nhnh v

khng xon-Tan trong nc Svayde (dung dch [Cu(NH3)4](OH)2-Phn ng vi HNO3 c/H2SO4 c:[C6H7O2(OH)3]n + 3nHNO3 () 2 4

H SO dac [C6H7O2(NO3)3]n + 3nH2O-B thy phn trong dung dch H+/to:

(C6H10O5)n + nH2O+ oH , t nC6H12O6

Xenluloz Glucoz Xenluloz c cc tnh cht 1, 3, 4, 6. p n ng l B

Cu 18. Theo bi ra:2H

0,224n = = 0,01 (mol)

22,4; ncht tan = 0,5.0,04 = 0,02 (mol)

+ Trng hp 1: Kim loi M l kim loi kim oxit l M2O

M + H2O MOH + 21

H2

x x 0,5x (mol)M2O + H2O 2MOH

y 2y (mol)

Theo bi ra ta c:0,5.x = 0,01

x + 2y = 0,02

M.x + (2M + 16)y = 2,9

Gii ra ta c x = 0,02; y = 0 loi+ Trng hp 2: M l kim loi kim th oxit l MO

M + 2H2O M(OH)2 + H2 x x x

MO + H2O M(OH)2y y

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Theo bi ra ta c:x = 0,01

x + y = 0,02

Mx + (M + 16).y = 2,9

x = 0,01

y = 0,01

M = 137 (Ba)

Vy kim loi M l Ba. p n ng l ACu 19. t cng thc ca anken l CnH2n. Gi x, y ln lt l s mol ca H2, CnH2n c trong hnhp X.

CnH2n + H2Ni CnH2n+2

Ban u: y x 0Phn ng: y y yCn li: 0 (x y) yTheo nh lut bo ton khi lng, ta c: mX = mY = 2x + 14n.y

Theo bi ra: Xm 2x + 14ny

= = 9,1.2 = 18,2x + y x + y

(1)

Ym 2x + 14ny= 13.2 = 26(x - y) + y x

(2)

T (1, 2)

x + y 26 10

= =x 18,2 7

10x = 7x + 7y 3x = 7y y =3x

7(3)

Thay (3) vo (2) ta c:

14.n.3x2x +

7 = 26x

n = 4

Suy ra: anken c CTPT l C4H8V anken ny c kh nng cng HBr cho sn phm hu c duy nht nn anken phi c cu to: CH3-CH=CH-CH3Tht vy: CH3-CH=CH-CH3 + HBr CH3-CH2-CH(Br)-CH3

Sn phm duy nhtp n ng l C

Cu 20. Theo bi ra: nNO =3,66

= 0,15 (mol)22,4

Gi x, y ln lt l s mol Fe3O4 v Cu phn ngTheo bi ra, ta c: 232x + 64y + 2,4 = 61,2 232x + 64 y = 58,8 (1)

V Cu (kim loi) cn d nn trong dung dch Y c hai loi mui l Cu(NO3)2 v Fe(NO3)2Do c cc qu trnh cho nhn electron xy ra l

Cu - 2e Cu2+y 2y y

+8/3

Fe + 23

e Fe+2

3x 3.2

3x 3x

+5 +2

N + 3e N0,45 0,15 (mol)

V e (cho) e (nhn)n = n nn: 2y = 2x + 0,45 (2)

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T (1), (2) ta c: x = 0,15; y = 0,375Vy: m =

3 2 3 2Cu(NO ) Fe(NO )+ m = 188.y + 180.3x

= 188.0,375 + 180.3.0,15 = 151,5 (gam)p n ng l B

Cu 21. Theo bi ra:2SO

3,248n = = 0,145 (mol)

22,4

C qu trnh nhn electron: S+6 + 2e S+40,29 0,145 (mol)

+ Trng hp 1: Gi s oxit st l FeO:Fe2+ - 1e Fe+30,29 (mol) 0,29 (mol) 0,29 (mol) mFeO = 0,29.72 = 20,88 (gam)

Suy ra: mmui =2 4 3Fe (SO )

0,29= .400 = 58 (gam)

2

+ Trng hp 2: Gi s oxit st l Fe3O4:3Fe8/3 _ 3.1/3e 3Fe+3

3.0,29 0,29 (mol)Suy ra:

3 4Fe O

3.0,29= .232 = 67,28 (gam)

3 20,88 (gam)

loi trng hp ny. p n ng l D

Cu 22. Theo bi ra:2H

4,48n = = 0,2 (mol)

22,4;

2CO

26,4n = = 0,6 (mol)

44

t Y l R1COOH (CaHbO2) (x mol)Z l R2(COOH)2 (CaHcO4) (y mol)

S cc phn ng:

R1COOH + Na R1COONa + 21

2

x 0,5x (mol)R2(COOH)2 + 2Na R2(COONa)2 + H2

y yTheo bi ra, ta c: 0,5x + y = 0,2 (1)

CaHbO2 2+O aCO2

x a.x (mol)CaHcO4 2

+O aCO2y a.y (mol)

Theo bi ra, ta c: ax + ay = 0,6 a(x + y) = 0,6 a =0,6

x + y(2)

T (1) x + y > 0,2. Thay vo (2) suy ra: a < 0,6 = 30,2

V axit a chc, nn a > 1. Vy a = 2Suy ra: Y l CH3COOH; Z l HOOC-COOH (3)T (1), (2), (3) ta c x = 0,2; y = 0,1

Vy: Z/X0,1.90.100%

%m = = 42,86 %(0,1.90 + 0,2.60)

. p n ng l C

Cu 23. X, Y ln lt l: HOCH2-CHO

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V HO-CH2-CH2-CHO. Tht vy:-Cc cht ny c % khi lng oxi ln lt l 53,33% v 43,24% (nh bi ra)-u tc dng vi Na (nhm OH)-u c phn ng trng gng (nhm CHO)p n ng l A

Cu 24. Theo bi ra:2CO

3,967n = = 0,1775 (mol)

22,4

;2CO

6,38n = = 0,145 (mol)

44

t cng thc chung ca 2 este l n 2 n 2C H O (este no, mch h, n chc).Phng trnh phn ng t chy este:

n 2 n 2C H O + 23n - 2

O2

n CO2 + n H2O

3n - 2

2 n

0,1775 (mol) 0,145 (mol)

Suy ra: 0,1775. n = 0,145.3n - 2

2

0,1775. n = 0,2175. n - 0,145 0,04 n = 0,145 n = 3,625Vy CTPT 2 este l C3H6O2 v C4H8O2. p n ng l ACu 25. Cc PTHH xy ra cc th nghim:

(I) Fe + H2SO4 (l) FeSO4 + H2(II) SO2 + Br2 + 2H2O H2SO4 + 2HBr(III) CO2 + H2O + NaClO HClO + NaHCO3 C 3 phn ng ha hc xy ra: p n ng l B

Cu 26. Nhit si ph thuc vo:+ Lin kt hiro gia cc phn t+ Phn t khi ca chtDy sp xp nhit si tng dn l

CH3CHO < C2H5OH < HCOOH < CH3COOH. p n ng l B

Cu 27. Theo bi ra:2 3H Al(OH)

3,36 39n = = 0,15 (mol); n = = 0,5 (mol)

22,4 78

S phn ng xy ra:Al Al NaAl(OH)4

ot +NaOH 2CO Al(OH)3 Fe3O4 Fe Fe (rn)

Al2O3 H2

Al + NaOH + 3H2O NaAl(OH)4 + 23

H2

0,1 (mol) 0,15V nguyn t nhm c bo ton nn:nAl (ban u) =

3Al(OH)n = nAl (d) + nAl (p/ nhit nhm)

0,5 = 0,1 + nAl (p/ nhit nhm) nAl (p/ nhit nhm) = 0,5 0,1 = 0,4 (mol)

8Al (d) + 3Fe3O4ot 9Fe + 4Al2O3

0,4 0,15 (mol)Suy ra:

3 4Fe On (ban u) = 0,15 (mol)

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Vy m = mAl (ban u) +3 4Fe O

m = 0,5.27 + 0,15.232 = 48,3 (gam)

p n ng l B

Cu 28. Theo bi ra:2O X

1,6n = n = = 0,05 (mol)

32

MX =3,7

= 740,05

X 2+ O CO21

(mol)74

0,7

(mol)22,4

Suy ra, s nguyn t cacbon trong 1 phn t X bng nC(X) >0,7.74

= 2,31322,4.1

V X phn ng c vi NaOH v dung dch AgNO3 trong NH3 nn trong phn t X c 2nguyn t oxi

Gi s phn t X c 2 nguyn t oxi X c cng thc phn t C3H6O2. Cu to ca X thamn s l HCOOC2H5

Nu phn t X c 3,4 nguyn t oxi: khng c CTPT, CTCT no tha mn

p n ng l ACu 29. Theo bi ra:

2Br X

48 13,44n = = 0,3 (mol); n = = 0,6 (mol)

160 22,4

2 2Ag C

36n = = 0,15 (mol)

240

Gi x, y, z ln lt l s mol ca CH4, C2H4 v C2H2 trong 8,6 gam X. Suy ra, s mol ca CH4,C2H4 v C2H2 trong 13,44 lt kh X l kx, ky v kz

Theo bi ra, ta c:16x + 28y + 26z = 8,6 (1)

k(x + y + z) = 0,6 (2)

* X + Br2:

C2H4 + Br2 C2H4Br2y yC2H2 + 2Br2 C2H2Br4

z 2zTa c: y + 2z = 0,3 (3)* X + AgNO3/NH3 (Ag2O/NH3):C2H2 + Ag2O 3

NH Ag2C2 + H2Okz kz

Ta c: kz = 0,15 (4)T (1, 2, 3, 4) ta c: x = 0,2; y = 0,1; z = 0,1

Vy % 4CH /X0,2.100%

V = = 50%0,2 + 0,1 + 0,1 (Cht kh c t l th tch cng l t l v s mol). p n ng l A

Cu 30. V tr cc nguyn t trong bng tun hon:

Nhm IA IIA IVA VAChu k 2 NChu k 3 Na Mg Si PChu k 4 K

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Theo quy lut bin i bn knh nguyn t trong chu k v trong nhm, ta c:Na > Mg > Si > P (Trong chu k)

K > Na (Trong nhm)P > N (Trong nhm)

Suy ra: K > Na > Mg > Si > P > N K > Mg > Si > Np n ng l D

Cu 31. Theo bi ra: nFe = 2,24 = 0,04 (mol)56

,

3 3 2AgNO Cu(NO )n = 0,2.0,1 = 0,02 (mol); n = 0,2.0,5 = 0,1 (mol)

Cc phng trnh phn ng xy ra:Fe(d) + 2AgNO3 Fe(NO3)2 + 2Ag 0,01 0,02 0,02 (mol)Fe + Cu(NO3)2(d) Fe(NO3)2 + Cu 0,03 0,03 0,03 (mol)

Cht rn thu c gm Ag (0,02 mol) v Cu (0,03 mol)mY = mAg + mCu = 0,02.108 + 0,03.64 = 4,08 (gam)

p n ng l B

Cu 32. Theo bi ra: nX =3

367,2.10 = 3.10 (mol)22,4

nX =2,24

= 0,1 (mol)22,4

;3CaCO

2n = = 0,02 (mol)

100

Cc phng trnh ha hc xy ra:

Al2O3dpnc 2Al + 2

3O

2 (*)

C + O2 CO22C + O2 2COCO2 + Ca(OH)2(d) CaCO3 + H2O

Trong hn hp kh c th c cc kh sau:CO2 (x mol); CO (y mol) v O2 (z mol)Theo bi ra, ta c: x + y + z = 3000 (1)44x + 28y + 32z

= 16.2 44x + 28y + 32z = 96000x + y + z

(2)

Khi 0,1 mol X + Ca(OH)2 d 0,02 mol CaCO3 (0,02 mol CO2)3000 mol X z mol CO2

z = 600T (1), (2), (3) ta gii ra c: x = 600; y = 1800; z = 600 (3)Suy ra, tng s mol kh O2 sinh ra theo phng trnh (*) (p dng bo ton nguyn t oxi):

2Oy 1800n = x + + Z = 600 + + 600 = 2100 (mol)2 2

V H = 100% mAl =2100.2.27

= 75.600 (gam) = 75,6 (kg)1,5

p n ng l DCu 33.

Theo bi ra:3 2Cu(NO )

n = 0,8.0,2 = 0,16 (mol) ;2 4H SO

n = 0,8.0,25 = 0,2 (mol)

Suy ra: -3NO

n = 0,16.2 = 0,32 (mol) ; 2+ +Cu Hn = 0,16 (mol); n = 0,2.2 = 0,4 (mol)

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Fe + -3 (d)NO + 4H+

Fe3+ + NO + 2H2O

0,1 0,1 0,4 0,1 0,1 (mol)Sau phn ng, thu c hn hp bt kim loi (Fe v Cu) nn xy ra cc phn ng:Fe + 2Fe3+ 3Fe2+0,05 0,1 (mol)Fe (d) + Cu

2+ Fe2+ + Cu 0,16 0,16 0,16 (mol)

Cht rn sau phn ng c 0,16 mol Cu;m

- (0,1 + 0,05 + 0,16)56

mol Fe

Theo bi ra, ta c: 0,16.64 + m (0,1 + 0,05 + 0,16).56 = 0,6m 0,4m = 0,31.56 0,16.64 = 7,12 m = 17,8 (gam)V = VNO = 0,1.22,4 = 2,24 (lt). p n ng l C

Cu 34.

+ 3NaOH + Cr(NO3)3 Cr(OH)3 + 3NaNO3NaOH (d) + Cr(OH)3 Na[Cr(OH)4] (tan)

+ CO2 + Ca(OH)2 CaCO3 + H2O

CO2(d) + CaCO3 + H2O Ca(HCO3)2(tan)+ 4HCl + Na[Al(OH)4] Al(OH)3 + NaCl + 4H2O

3HCl (d) + Al(OH)3 AlCl3 + 3H2O (tan)+ 3NH3(d) + 3H2O + AlCl3 Al(OH)3 + 3NH4Clp n ng l D

Cu 35.

+ 2KClO3o

2MnO , t 3O2 + 2KCl100

122,5

1,5.100(mol)

122,5

+ 2KMnO4ot O2 + K2MnO4 + MnO2

100

158

0,5.100(mol)

158

+ 2KNO3ot O2 + 2KNO2

100

101

0,5.100(mol)

101

+ 2AgNO3ot 2Ag + 2NO2 + O2

100

170

0,5.100

170

Suy ra, cht to lng O2 ln nht l KClO3. p n ng l CCu 36.

Ta c:2CuCln = 0,5.0,1 = 0,05 mol ; nNaCl = 0,5.0,5 = 0,25 (mol) kh ht ion Cu2+ th thi gian in phn cn:

(Cu2+ + 2e Cu , y l qu trnh kh xy ra u tin)

tCu = Cum .n.F 0,05.64.2.96500

= = 1930A.I 64.5

(giy) < 3860 (giy)

Thi gian in phn H2O ( catot) l 3860 1930 = 1930 (giy)2H2O + 2e 2OH

- + H2

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Suy ra: -OHI.t 5.1930

n = = = 0,1 (mol)n.F 1.96500

Phng trnh ha hc ha tan Al trong dung dch thu c:

Al + OH- + 3H2O [Al(OH)4]- +

2

3

2

0,1 0,1 (mol)

Vy m = mAl (max) = 0,1.27 = 2,70 (gam)p n ng l CCu 37.

+) NaCl + KOH khng xy ra+) Na2CO3 + Ca(OH)2 2NaOH + CaCO3 +) 2NaCl + 2H2O

dpnc, mn 2NaOH + Cl2 + H2 +) Cu(OH)2 + NaNO3khng xy ra+) NH3 + Na2CO3 khng xy ra+) Na2SO4 + Ba(OH)2 BaSO4 + 2NaOH Cc th nghim iu ch c NaOH l II, III, VI. p n ng l A

Cu 38. Cc cht tc dng c vi Na: c nhm OH trong phn t-Cc cht tc dng vi Cu(OH)2 (to thng) c 2 nhm OH 2 nguyn t cacbon cnh nhau

trong phn t Cc cht va tc dng vi Na v Cu(OH)2 l cht c 2 nhm OH cnh nhau trong phn t.Cc cht l

a) HO-CH2-CH2-OHc) HOCH2-CH(OH)-CH2OHd) CH3-CH(OH)-CH2OH

p n ng l D*Ch : Ch nhng ancol a chc c 2 nhm OH cnh nhau (tr ln) mi tc dng vi Cu(OH)2(nhit thng) to thnh dung dch phc cht c mu xanhCu 39.

Theo bi ra:2 4H SOn = 0,1.0,05 = 0,005 (mol) ; nHCl = 0,1.0,1 = 0,001 (mol)nNaOH = 0,1.0,2 = 0,02 (mol);

2Ba(OH)n = 0,1.0,1 = 0,01 (mol)

Suy ra: +2 4HCl H SOH

n = n + 2.n = 0,01 + 2.0,005 = 0,02 (mol)

-2NaOH Ba(OH)OH

n = n + 2n = 0,02 + 0,01.2 = 0,04 (mol)

Qu trnh xy ra: H+ + OH- H2OBan u: 0,02 0,04 (mol)Phn ng: 0,02 0,02 (mol)Cn li: 0 0,02 (mol)

- 0,02OH = = 0,1 (M)0,1 + 0,1

pOH = -lg.0,1 = 1 pH = 14 1 = 13,0

p n ng l BCu 40.

*Trng hp 1: X l flo, Y l clo:Gi x, y ln lt l s mol ca NaF, NaClTheo bi ra, ta c: 42x + 58,5y = 6,03 (1)NaF + AgNO3 khng xy raNaCl + AgNO3 AgCl + NaNO3

y y (mol)

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Ta c: 143,5.y = 8,61 y = 0,06T (1, 2) ta c: x = 0,06; y = 0,06

Vy NaF/hh0,06.42.100%

%m = = 41,8 (%)6,03

p n ng l D

* Trng hp 2: Hai mui NaX v NaY u to kt ta vi AgNO3t cng thc chung ca 2 mui l Na X

Na X + AgNO3 Ag X + NaNO3Khi 1 mol Na X 1 mol Ag X khi lng tng 85 gamVy khi tng (8,61 6,03) th s mol Na X l:

NaX

8,61 - 6,03n = = 0,03 (mol)

85

Suy ra:X

6,03M = = 201

0,03

(Khng c halogen no c trong t nhin m c M > 201 loi trng hp ny)Cu 41. Theo bi ra: nKOH = 0,1.1,5 = 0,15 (mol);

3 4H POn = 0,2.0,5 = 0,1 (mol)

V 1 32 R > 15 (CH3)Do , gi tr tha mn ca R, R ln lt l R = CH2 (14); R = C2H5 (29)

Vy X l H2N-CH2-COOC2H5H2N-CH2-COOC2H5 + NaOH H2N-CH2-COONa + C2H5OH0,25 0,25 0,25 (mol)

Sau phn ng, cht rn thu c gm:NaOH: 0,3 0,25 = 0,05 (mol); Mui: 0,25 (mol)Vy m = mNaOH (d) + mmui = 0,05.40 + 0,25.97 = 26,25 (gam)p n ng l A

Cu 48. Theo bi ra:2O

17,92n = = 0,8 (mol)

22,4

t cng thc chung ca hai anehit l n 2nC H O (anehit no, mch h, n chc)* X + H

2:

n 2n 2 n 2n + 2C H O + H C H O m (g) ( m + 1) (gam)Ta c: m +

2H= m + 1

2H= 1 (g)

2Hn = 0,5 (mol) = nanehit

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* X + O2:

n 2nC H O+n

2

3 - 1O

2

2nCO + 2nH O

1 (mol) n3 - 1

2

mol

0,5 (mol) 0,8 (mol)Ta c: 0,8.1 = 0,5. n

3 - 1

2

3 n - 1 = 3,2 n = 1,4

Vy m = 0,5. (14. n + 16) = 0,5. (14.1,4 + 16) = 17,8 (gam)p n ng l A

Cu 49. X + Br2 1 : 1 XBr2

%mBr=160.100

= 74,08X + 160

16000 = 74,08.X + 160.74,08 X = 56 (C4H8)

V X + HBr 2 sn phm hu c X c cu to: CH2=CH-CH2-CH3

(But 1 en)p n ng l ACu 50. ng dng khng phi ca ozon l iu ch oxi trong phng th nghim. ( iu ch oxitrong phng th nghim ngi ta nhit phn KClO3 (xc tc MnO2, KMnO4,). p n ng l DCu 51. Trong dung dch glucoz tn ti c dng mch h (vng , - glucoz) v dng mchh:

Vng - glucoz CH2(OH)(CHOH)4CHO vng glucoz- dng mch h (CH2(OH)-[CHOH]4CHO) c 5 nhm OH k nhau-Khi glucoz dng mch vng, ch c nhm OH C1 (OH) hemiaxetal) tc dng vi

CH3OH (xc tc HCl) to ra metyl glicozit-Glucoz (dng mch h c nhm CHO) tc dng c vi nc brom:CH2(OH)-(CHOH)4CHO + Br2 + H2O CH2(OH)(CHOH)4COOH + 2HBrp n ng l C

Cu 52. chua (do H+) ca t tng ln khi bn NH4NO3 (mui to t axit mnh HNO3 v bazyu NH3):

NH4NO3 NH4+ + NO3

-

NH4+ H

+ + NH3p n ng l C

Cu 53. Theo bi ra:2NO

1,344n = = 0,06 (mol)

22,4

Gi x, y ln lt l s mol ca Cu, Al trong hn hp X

Theo bi ra, ta c: 64x + 27y = 1,23 (1)Cc qu trnh xy ra:

Cu - 2e Cu2+x 2x

Al - 3e Al3+y 3y

N+5 + 1e N+40,06 0,06 (mol)

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Ta c: 2x + 3y = 0,06 (2)T (1), (2) ta c: x = 0,015; y = 0,01Sc kh NH3 d vo dung dch Cu(NO3)2, Al(NO3)3 th ch c kt ta ca Al(OH)3

Al(NO3)3 + 3NH3 + H2O Al(OH)3 + 3NH4NO3y y

m =3Al(OH)

= 0,01.78 = 0,78 (gam)

%mCu/X = 64.x.100% 64.0,015.100%= = 78,05%1,23 1,23

p n ng l D

Cu 54. Theo bi ra: nAg =54

= 0,5 (mol)108

Thng thng: RCH2OH+CuO RCHO 2 3+Ag O/NH 2Ag

0,2 (mol) 0,4 (mol)Nhng theo bi ra: nAg = 0,5 > 0,4 (mol) c anehit fomic HCHOSuy ra, trong X c CH3OH v C2H5OH (hai ancol no, n chc, mch h, k tip nhau trong

dy ng ng)

CH3OH HCHO 4Ag x 4x

C2H5OH CH3CHO 2Ag y 2y

Theo bi ra ta c:x + y = 0,2

4x + 2y = 0,5

x = 0,05

y = 0,15

Vy m =3 2 5CH OH C H OH

+ m = 0,05.32 + 0,15.46 = 8,5 (gam)

p n ng l BCu 55. Pin c sut in ng chun ln nht khi pin to ra t cp in cc cch xa nhau nht trongdy in ha

Do , trong cc pin cho th pin Zn Cu (v tr ca cp Zn2+

/Zn v Cu2+

/Cu cch xa nhaunht). c S chun ln nht2+ 2+

o o o

pin Cu /Cu Zn /ZnE (Zn - Cu) = E -E = + 0,34 - (-0,76) = 1,1 V

p n ng l BCu 56. Cc phn ng ha hc xy ra trong s trn:

CH3-CH(OH)-CH2-CH3o

2 3H SO d, 170 C CH3-CH=CH-CH3 + H2OButan 2 ol (X)

CH3-CH=CH-CH3 + HBr CH3-CH2-CH(Br)-CH3(X) (Y)

CH3-CH2-CH(Br)-CH3 + Mgete CH3-CH(MgBr)CH2-CH3

(Y) (Z)p n ng l D

Cu 57. Theo bi ra:2H

6,4n = = 0,04 (mol)

160; nNaOH = 0,04.0,75 = 0,03 (mol)

Gi x, y, z ln lt l s mol ca CH2=CH-COOH; CH3COOH v CH2=CH-CHO trong hnhp X

Theo bi ra, ta c: x + y + z = 0,04 (1)

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* X + dung dch Br2:CH2=CH-COOH + Br2 CH2Br-CHBr-COOH

x xCH2=CH-CHO + 2Br2 + H2O CH2Br-CHBr-COOH + 2HBr

z 2zTa c: x + 2z = 0,04 (2)

* X + NaOH:CH2=CH-COOH + NaOH CH2=CH-COONa + H2Ox x

CH3COOH + NaOH CH3COONa + H2Oy y

Ta c: x + y = 0,03 (3)T (1), (2), (3) ta gii ra c: x = 0,02 ; y = 0,01; z = 0,01Vy

2CH =CH-COOH= 72.x = 72.0,02 = 1,44 (gam)

p n ng l DCu 58. Nc cng toan l hn hp axit HCl v HNO3 c kh nng ha tan vng

Al + 3HCl + HNO3

AlCl3 + NO

+ 2H2O0,02 0,06 (mol) 0,02 (mol)Vy nHCl = 0,06 (mol); nNO = 0,02 (mol). p n ng l C

*Ch : Khng xt qu trnh to phc ca AuCl3:

AuCl3 + HCl H[AuCl4] Axit tetracloroauricCu 59. Qu trnh in li ca mui tan trong nc:

CH3COONa CH3COO- + Na+

0,1M 0,1MQu trnh phn li ca axit axetic:

CH3COOH CH3COO- + H+

Ban u 0,1M 0,1M 0Phn li x x xCn bng (0,1 x) (0,1 + x) x

Ta c:

+ -

3 -5

a

3

H CH COO x(0,1 + x)K = = = 1,75.10

CH COOH 0,1 - x

Gi thit x

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Nn: manilin =186.30

= 55,8 (gam)100

p n ng l B