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HNG DN GII CHI TIT MT S CU M 401

Cu 23.Trong hp cht Na2O2 c hin din ion O2

2Cu 27.Bo ton C cho s mol C2H2 trong X = 0,35 mol. Sau bo ton H cho s mol H2 trong

X = (1 - 0,35) = 0,65 mol. Vy %H = 65% (chn C)

Cu 28.Ch rng cc anehit khng b oxi ha bi dung dch Br2 trong CCl4 m ch b oxi ha bi dung dchBr2 trong nc (nc brom)Cu 29.Theo 100ml dung dch X khi c axit ha bng H2SO4 long lm mt mu va 250ml dungdch KMnO4 0,015M tc 0,00375 mol KMnO4. Ch c FeSO4 b oxi ha bi KMnO4.Gi a l s mol FeSO4. Bo ton electron cho: a = 5.0,00375 = 0,01875.

Do %FeSO4 =0,01875 . 152 . 100

= 57%5

Vy % tp cht = 100 - 57 = 43%Cu 30.

Gi M l phn t khi ca amino axit, ta c h:a(M + 36,5) = 30,7 a = 0,2

a(M + 22) = 27,8 M = 117

. Vy m = aM = 23,4

Cu 31.

Chn C. HCOOCH=CHOOCH + 2H2O+ oH ,t 2HCOOH + HOCH2CHO

Cu 32.

Ta c2 3Fe O

14,4n = = 0,09 mol

160nn

3 4Fe O

0,09.2n = = 0,06 mol

3

Ch rng 1 phn t Fe3O4 cho 1e, cn mt phn t K2Cr2O7 nhn 6e, ta c

2 2 7K Cr O

0,06n = = 0,01 mol

6. Vy V = 0,01 lt

Cu 33.Theo , A l C6H5COOCH=CH2 v B l CH3CHO

Cu 34.Gi a, b ln lt l s mol CH3OH b oxi ha thnh HCHO v HCOOH

Ch rng nHCHO = a; nHCOOH = b;2H O

n = a + b;3CH OH (d-)

n = c, ta c h:

12,964a + 2b = = 0,12

108 a = 0,025b a + b c 0,56

+ + = = 0,025 b = 0,012 2 2 22,4

c = 0,0050,224

b = = 0,01

22,4

Vy %CH3OH b oxi ha =(a + b)100

a + b + c= 87,5%

Cu 36.

Hm lng H2S trong khng kh0,3585.34

= = 0,017 mg/l3.239

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Cu 37.V X chy cho

2 2 2O CO H On : n : n = 1,5 : 2 : 1 nn X c dng (CHO)n. Ta phi c n chn v

230n < = 7,9

29. Nhng n = 2 hoc n = 4 u khng c cht X tha bi. Vy n = 6, tc X l C 6H6O6

ng vi cng thc cu to C6(OH)6Cu 38.

Theo a, b, c cng ln lt l s mol 3 kh trn c trong 1 mol hn hp.

Suy ra x =44a

44a + 28b + 2c

vx 44

k = =

a 44a + 28b + 2c

z =2c

44a + 28b + 2c

vz 2

t = =c 44a + 28b + 2c

Nhng 2 < 44a + 28b + 2c < 44 nn k > 1, cn t < 1

Cu 40.H2O2 to thnh O2 khi gp cc cht oxi ha mnh. C 4 trng hp l:

- H2O2 + Cl2(bo ha) O2 + 2HCl- 5H2O2 + 3H2SO4 + 2KMnO4 2MnSO4 + 5O2 + K2SO4 + 8H2O- H2O + 2Hg(NO3)2 O2 + Hg2(NO3)2 + 2HNO3- H2O2 + Ca(ClO)2 CaCl2 + 2H2O + 2O2

Cu 41.

Dng nc NH3 cho vo 5 mu- Mu to kt ta trng bn l AlCl3- Mu to kt ta trng ri tan l ZnCl2- Mu to kt ta xanh ri tan l CuCl2- Mu to s phn lp l C6H5NH3Cl (do xut hin anilin khng tan)- Mu to dung dch ng nht l NaCl

Cu 42.Phng trnh phn ng:

CuSO4 + 2NaCl Cu + Na2SO4 + Cl2mol 0,06 0,12 0,06

CuSO4 + H2O Cu + H2SO4 +1

2 O2mol 0,04 0,02

n kh =

2 2l On + n = 0,06 + 0,02 = 0,08 mol

Vkh = 22,4 . 0,08 = 1,792 (lt)

Cu 43.Ta phi chn nhng cht c O gn trc tip vo vng benzen i electron t do trn nguyn

t O tham gia lin hp vo vng benzen, t xut hin cc v tr giu electron phn ng th bromtheo c ch i electron d dng xy ra. Vy c 4 cht tha mn bi l o-crezol, m-crezol, p-crezol vmetyl phenyl ete.

Cu 44.X

42,65 58,5 . 0,3M = 36,5 = 89 (anilin)

0,2

Cu 45.D dng tm c

2 2NO SOn = n = 0,1 mol

t cng thc X l FexOy (a mol). Bo ton electron cho h:a(56x + 16y) = 26 ax = 0,355

a(3x 2y) = 1 . 0,1 + 2 . 0,1 ay = 0,3825

Bo ton Fe cho nFe = ax = 0,355 mol nn m = 0,355 . 56 = 19,88 gam

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Cu 48.Gi a, b ln lt l s mol ASO4 v BSO4 mi phn.

Th nghim 1 cho: a + b17,475

= = 0,075 mol233

Th nghim 2, gi s 2 mui u b in phn theo phn ng:

2MSO4 + 2H2O d.p.d.d 2M + O2 + 2H2SO4 (1)

Th ta phi c s mol kh thot ra anot = s mol O2a + b 0,56

= 0,0375 0,0252 22,4

iu ny chng t ch c mt mui b in phn nh (1), mt mui khng b in phn (tc thc chtl in phn nc). Gi s ch c ASO4 b in phn, ta c h:

22a(A + 96) + b(B + 96) = = 11

a = 0,0252

a + b = 0,075 b = 0,05

a = 0,025 0,025A + 0,05B = 3,8(*)

(*) A + 2B = 152. Ch c A = 24 v B = 64 l ph hp. Vy hai kim loi cn tm l Mg v Cu

Cu 50.

Ankin CnH2n-2 c (2n - 2) nguyn t H. Nhng nguyn t H ny lin kt trc tip vi cc nguynt C bng (2n - 2) lin kt . Mt khc gia n nguyn t C c (n - 1) lin kt . Vy tng s lin kt trong ankin CnH2n-2 l (2n - 2 + n - 1) = 3n - 3

Cu 53.Nu 0,1 mol Cu(NO3)2 phn ng ht s to mY > 8 gam. Theo th mY = 6,4 gam chng t

Cu(NO3)2 cn d. Vy Al v Zn phn ng ht.Gi a, b, c ln lt l s mol Al, Zn v Cu ban u. Ch rng rn Y ch l CuO v rn Z ch l Al2O3,ta c h:

27a + 65b + 64c = 3,76a = 0,02

6,41,5a + b + c = = 0,08 b = 0,02

80 c = 0,03a 1,02

= = 0,012 102

Vy %mZn =0,02 . 65 . 100

= 34,57%3,76

Cu 54.t cng thc trung bnh 2 anehit l CxHyO v a l s mol 2 anehit, ta c h:

a(12x + y + 16) = 11,36

24,64ax = = 0,56 a = 0,24

44

ay 7,2= = 0,4

2 18

Nhng nAg =69,12

0,64108

> 2 .0,24 chng t c mt anehit l HCHO

Gi b, c ln lt l s mol HCHO v anehit cn li, ta c h:

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30b + Mc = 11,36 b = 0,08

b + c = 0,24 c = 0,16

4b + 2c = 0,64 M = 56

Vy hn hp gm HCHO v CH2=CH-CHO

Cu 56.S dung dch lm mt mu nc brom l fomon; glucoz; mantoz; axit fomic v axetanehit.Cu 57.

Ta c phn ng:

2HBr (k) H2 (k) + Br2 (k)Ban u: 3,2 0 0 (mol)Phn ng: 2a a a (mol)Lc cn bng: (3,2 - 2a) a a (mol)

Vy 7 7 32 2C 22

a a.

[H ].[Br ] V VK = = 4,6.10 = 4,6.10 a = 2,16.10[HBr] 3,2 2a

V

Do %HBr b phn hy = 3

2,16.10 .2.100= 0,135%

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Cu 58.Theo , A c cng thc C2nH4nO3n. Cng thc ny ch c 1 nn A ch c mt nhm COOH.

Vy cng thc A c th vit C2n-1Hn+1(COOH)(OH)3n-2.Ta phi c: s nhm OH s C gc 3n - 2 2n - 1 n 1. Vy n = 1. Do A c cng thcl C2H4O3Cu 59.

C 7 hp cht lng tnh l: Al(OH)3; KHCO3; NH4NO2; HCOONH3CH3; Cr2O3; (CH3COO)2Pbv Sn(OH)2Cu 60.

Ha tan hon ton rn Y vo dung dch HNO3 phi c dung dch cha 0,1 mol Cu(NO3)2 v0,2 mol Al(NO3)3. Bo ton N cho

3 3 2 3 3 2HNO Cu(NO ) Al(NO ) NO NOn = 2n + 3n + n + n

= 2 . 0,1 + 3 . 0,2 + 0,2 = 1 mol nn3dd.HNO

V = 0,5 lt