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gii chi tit thi th i hc - m 502

Cu 1. Theo bi ra: nFe =1,12

= 0,02 (mol)56

; nCu =1,92

= 0,03 (mol)64

+

2 4H SO

H

n = 2.n = 2.0,4.0,5 = 0,4 (mol)

-33

NaNOHNOn = n = 0,4.0,2 = 0,08 (mol)

Cc PTHH xy ra:Fe + 4H+ + NO3

- Fe3+ + NO + 2H2O0,02 0,08 0,02 0,02 (mol)3Cu + 8H+ + 2NO3

- 3Cu2+ + 2NO + 4H2O0,03 0,08 0,02 0,03 (mol)H+ + NaOH Na+ + H2O(0,4 0,08 0,08) 0,24 (mol)Fe3+ + 3NaOH Fe(OH)3 + 3Na

+

0,02 0,06 (mol)Cu2+ + NaOH Cu(OH)2 + 2Na+

0,03 0,06 (mol)Suy ra:

NaOHn = 0,06 + 0,06 + 0,24 = 0,36 (mol) Vdd NaOH = 0,36/1 = 0,36 (lt) = 360 (ml). p n ng l C

Cu 2. Theo bi ra: nhh este =66,6

= 0,9 (mol)74

HCOOC2H5 + NaOH HCOONa + C2H5OHCH3COOCH3 + NaOH CH3COONa + CH3OH

Ta thy nancol = neste = 0,9 (mol)

2C2H5OH

o2 4H SO d, 140 C

C2H5-O-C2H5 + H2O2CH3OH

o2 4H SO d, 140 C CH3-O-CH3 + H2O

Ta thy: nnc =1

2nancol =

0,9= 0,45 (mol)

2

mnc = 0,45.18 = 8,1 (gam). p n ng l BCu 3. Vit cc PTHH (nu c)

Fe + H2SO4 (long, ngui) FeSO4 + H2Cl + 2FeCl2 2FeCl3H2S + CuCl2 CuS + 2HClH2S + FeCl2 khng xy ra (v FeS tan c trong HCl)

p n ng l DCu 4. Fe b n mn trc nn Fe ng vai tr cc m (Fe c tnh kh mnh hn, hay ni cch khcl Fe ng trc trong dy in ha)

Do , cc hp kim m Fe b n mn trc lCu Fe (I); Fe C(III); Sn Fe (IV)p n ng l C

Cu 5. Theo bi ra:2CO

7,84n = = 0,35 (mol)

22,4;

2H O

11,7n = = 0,65 (mol)

18

Ta c s phn ng xy ra:

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HCHO HCHO CO2 (0,35 mol)H2 CH3CHO H2O (0,65 mol)HCHO + H2

oNi, t CH3OHHCHO + O2 CO2 + H2O2CH3OH + O2 2CO2 + 4H2Op dng nh lut bo ton nguyn t cho nguyn t cacbon v hiro, ta c:

nC(HCHO) = 2C (C O )n = 0,35 (mol) = nHCHO

2 2H(H ) H(HCHO) H(H O)n + n = n

2H(H )n + 2.0,35 = 2.0,65

2H(H )

n = 0,6 (mol) 2H

0,6n = = 0,3 (mol)

2

Vy %2H /X

0,3.100%V = = 46,15%

0,3 + 0,35

p n ng l BCu 6. Hn hp Na2O (x mol) v Al2O3 (x mol)

Na2O + H2O 2NaOHx 2x (mol)

2NaOH + Al2O3 + 3H2O 2Na[Al(OH)4]2x x (tan)

Suy ra, hn hp ny tc dng va , to ra dung dch-Hn hp Cu (x mol) v FeCl3 (x mol):

Cu + 2FeCl3 2FeCl2 + CuCl2x 2x

Suy ra, Cu cn d (khng tan)-Hn hp BaCl2 (x mol) v CuSO4 (x mol):

BaCl2 + CuSO4 BaSO4 + CuCl2To kt ta BaSO4

-Hn hp Ba (x mol) v NaHCO3 (x mol):Ba + 2H2O Ba(OH)2 + H2

x xBa(OH)2 + NaHCO3 BaCO3 + NaOH + H2O

x xTo ra kt ta BaCO3

Vy ch c 1 hn hp (Na2O/Al2O3) ha tan trong nc, to ra dung dch. p n ng l C

Cu 7. Theo bi ra: nX =6,72

0,3 (mol)22,4

M l CnH2n; N l CnH2n-2

Ta c: X12,4

M = = 41,30,3 n = 3. Vy M l C3H6 ; N l C3H4Gi x, y ln lt l s mol ca M, N

Ta c:x + y = 0,3 x = 0,2

42x + 40y = 12,4 y = 0,1

Vy trong X c 0,2 mol C3H6 v 0,1 mol C3H4. p n ng l DCu 8. S phn ng:

Este + NaOH mui + ancolTheo nh lut bo ton khi lng:

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meste + mNaOH = mmui + ancol 1,99 + mNaOH = 2,05 + 0,94 mNaOH = 1 (gam) nNaOH = 0,025 (mol) = neste (n chc)

Suy ra:este mui 3

1,99 2,05M = = 79,6; M = = 82 (CH COONa)

0,025 0,025

Vy 2 este l CH3COOCH3 (74); CH3COOC2H5 (88). p n ng l D

Cu 9. Amino axit X c dng: (H2N)a R (COOH)bX + HCl d Mui Y:(HOOC)b R (NH2)a + aHCl (HOOC)b-R-(NH3Cl)aMX m1 = MY = MX + 36,5.aX + NaOH d Mui Z(H2N)a R (COOH)b + bNaOH (H2N)a R (COONa)b + bH2OMX m2 = MZ = MX + 22bTheo bi ra: m2 m1 = (MX + 22b) (MX + 36,5.a) = 7,5

22b 36,5.a = 7,5 a = 1; b = 2Nh vy trong X c 1 nhm NH2 (tc l phn t X c 1 nguyn t N) v c 2 nhm COOH(phn t X c 4 nguyn t oxi)Trong cc phn t cho, ch c C5H9O4N l tha mn. p n ng l BCu 10. Theo bi ra: nKOH(110) = 0,11.2 = 0,22 (mol)

nKOH(140) = 0,14.2 = 0,28 (mol)Khi cho 110 ml dung dch KOH vo X th KOH thiu:ZnSO4 + 2KOH Zn(OH)2 + K2SO4

0,22 0,11 (mol)Suy ra: a = 0,11. 99 = 10,89 (gam)Khi cho 140 ml dung dch KOH vo X th ZnSO4 ht, KOH d ha tan mt phn Zn(OH)2ZnSO4 + 2KOH Zn(OH)2 + K2SO4

x 2x x

Zn(OH)2 + 2KOH K2[Zn(OH)4]y 2y

Ta c:2x + 2y = 0,28 x = 0,125

99.(x - y) = 10,89 y = 0,015

Vy4ZnSO= 161.x = 161.0,125 = 20,125 (gam) . p n ng l A

Cu 11. Cc hirocacbon: etilen, xiclopropan (vng no 3 cnh), stiren u lm mt mu dung dchbrom. Cn xiclohexan (vng no 6 cnh) khng tc dng vi dung dch brom

p n ng l CCu 12. PTHH xy ra:

CO (d) + CuO (r)ot Cu (r) + CO2

CO + Al2O3(r) khng xy raKhi 1 mol CuO (80g) phn ng th khi lng gim 16 gamVy khi khi lng oxit gim 9,1 8,3 = 0,8 gam th s mol CuO phn ng bng 0,05 (mol)Vy: mCuO(ban u) = mCuO(p/) = 0,05.80 = 4 (gam)p n ng l D

Cu 13. Theo bi ra:2CO

8,96n = = 0,4 (mol)

22,4;

2H O

7,2n = = 0,4 (mol)

18

V2 2H O CO

n = n v ete to ra t ancol n chc CTPT ete dng CnH2nO

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6 12 6C H O

m = 0,075. 180 = 13,5 (gam)

V H = 90% nn6 12 6C H O

m (thc t) =13,5.100

90= 15,0 (gam)

p n ng l CCu 18. Hai ancol a chc, mch h, cng dy ng ng, t chy sinh ra

2 2H O COn > n Hai ancol

l cc ancol no mch h, a chc, dng:

n 2n + 2 aC H O 2+O n CO2 + ( n + 1)H2O

n

3

n + 1

4

Suy ra: 3( n + 1) = 4 n n = 3Vy trong X phi c mt ancol c s nguyn t cacbon trong phn t < 3; mt ancol c s nguynt cacbon trong phn t > 3. Trong s cc ancol cho, cp ancol tha mn l C2H4(OH)2 vC4H8(OH)2. p n ng l C

Cu 19. Theo bi ra:2H

2,24n = = 0,1 (mol)

22,4

2 4 2H SO Hn = n = 0,1 (mol)

2 4dd.H SO0,1.98.100

= = 98 (gam)10 S phn ng: Kim loi (Al, Zn) + H2SO4 (l) Mui + H2

Theo nh lut bo ton khi lng:mkim loi +

2 4dd.H SOm = m dd mui +

2H

3,68 + 98 = m dd mui + 0,1.2 m dd mui = 3,68 + 98 0,1.2 = 101,48 (gam). p n ng l ACu 20. PTHH xy ra:

CaOCl2 + 2HCl CaCl2 + Cl2 + H2O1 mol 1 mol2KMnO4 + 16HCl 5Cl2 + 2KCl + MnCl2 + 8H2O

1 mol 2,5 (mol)K2Cr2O7 + 14HCl 3Cl2 + 2KCl + 2CrCl3 + 7H2O1 mol 3 (mol)MnO2 + 4HCl Cl2 + MnCl2 + H2O1 mol 1 mol

Vy t K2Cr2O7 iu ch c nhiu Cl2 nht. p n ng l B

Cu 21. Theo bi ra: nAg =54

= 0,5 (mol)108

+ 0,25 mol anehit mch h X 0,5 mol Ag Anehit X dng RCHO (n chc)

+ 0,125 mol anehit X phn ng ht vi 0,25 mol H2

Anehit n chc X c gc R khng no (c 1 lin kt i C = C)Vy X c cng thc chung l CnH2n-1CHO. p n ng l A

Cu 22. Theo bi ra: nAl =12,42

= 0,46 (mol)27

; nY =1,344

= 0,06 (mol)22,4

Gi x, y ln lt l s mol N2O, N2

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Ta c:x + y = 0,06

x = 0,0344.x + 28.y

= 18.2 y = 0,03x + y

Cc qu trnh cho nhn electron:Al 3e Al3+0,46 1,38 mol

e (cho)n = 1,38 (mol) +5 0

22N + 10e N

0,3 0,03 (mol)+5 12N + 8e 2N

0,24 0,03.2 (mol) e (nhn)n = 0,3 + 0,24 = 0,54 < 1,38 Suy ra, c qu trnh: +5 -3

4 3N + 8e N (NH NO )

(1,38 0,54) 0,105 (mol)Cht rn gm c Al(NO3)3 (0,46 mol) v NH4NO3 (0,105 mol)

Vy: m = 0,46.213 + 0,105.80 = 106,38 (gam)p n ng l B

Cu 23. Theo bi ra:X YN O

940,8n = = 0,042 (mol)

22400

Mt khc:X Y 2N O /H

d = 22

X YN OM

= 222

X YN O

M = 44 (N2O)

Ta c: 2N+5 + 2.4e 2N+10,336 0,042.2 (mol)

e (nhn)n = 0,336 (mol)M ne Mn+a na (mol)

e (cho)n = n.a (mol) = 0,336 (mol)

Ta c: MM = Mm 3,024.n

= = 9na 0,336

Do : gi tr tha mn l n = 3 M = 27 (Al). p n ng l BCu 24. X + HCl XHClKhi 1 mol X phn ng th tng 36,5 gamDo khi tng 15 10 = 5 gam th s mol X bng:

nX =

5

36,5 . Vy MX =

10

= 745

36,5(C4H11N)

Cc ng phn ca C4H11N lCH3CH2CH2CH2NH2; (CH3)2CHCH2NH2; (CH3)3CNH2CH3CH2CH2NHCH3; (CH3)2CHNHCH3; CH3CH2NHCH2CH3(CH3)2NCH2CH3

Vy c 7 ng phn. p n ng l B

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Cu 25. Hai kim loi phi l Ag (sinh ra) v Fe (d) Zn phn ng ht nn trong dung dch cmui Zn(NO3)2; AgNO3 phn ng ht (v Fe cn d). C 2 mui nn mui cn li phi l Fe(NO3)2.Vy 2 mui trong dung dch l Zn(NO3)2 v Fe(NO3)2p n ng l CCu 26. Thuc th c chn l Cu(OH)2:Gly Ala Gly + Cu(OH)2 mu tm

(Tripeptit c phn ng mu biure c trng)Gly Ala + Cu(OH)2 khng xy ra(ipeptit khng c phn ng mu biure). p n ng l A

Cu 27. Theo bi ra: nFe =3HNO

6,72= 0,12 (mol); n = 0,4.1 = 0,4 (mol)

56

Fe + 4HNO3 Fe(NO3)3 + NO + 2H2O0,1 0,4 (mol) 0,1

3 3Fe(NO )

n = 0,1 (mol) ; nFe (d) = 0,12 0,1 = 0,02 (mol)

Fe + 2Fe(NO3)3 3Fe(NO3)20,02 0,04 (mol)

3 3Fe(NO )n (cn li) = 0,1 0,04 = 0,06 (mol)

Cu + 2Fe(NO3)3 Cu(NO3)2 + Fe(NO3)20,03 0,06 (mol)

Vy m = 0,03.64 = 1,92 (gam). p n ng l ACu 28. t CTPT ca X l CxHyOz. Ta c:

x : y : z = C H Om m m 21 2 4

: : = : : = 1,75 : 2 : 0,25 = 7 : 8 : 112 1 16 12 1 16

CTGN ca X l C7H8O (cng l CTPT)Cc ng phn hp cht thm:

O-CH3 CH2-OH CH3 CH3 CH3; ; OH ; ;

OHOH

C 5 ng phn. p n ng l ACu 29. Cc cht v ion c c tnh oxi ha v tnh kh (c trng thi oxi ha trung gian): S; FeO;SO2; N2; HCl (5 cht/ion)

Cu 30. Cu(NO3)2(rn)ot CuO (rn) + 2NO2 + 2

1O

2

x (mol) 2x 0,5xKhi lng cht rn gim bng khi lng ca hn hp kh NO2 v O2:

2 2(NO .O )= 6,58 - 4,96 = 1,62 = 2x.46 + 0,5x.32 x = 0,015 (mol)

2NO2 + 21 O2

+ H2O 2HNO3

2x 0,5x 2x (mol)

3HNOn = 2x = 2.0,015 = 0,03 (mol)

[H+] = [HNO3] =0,03

= 0,1 (M)0,3

pH = -lg [H+] = 1. p n ng l D

Cu 31. Poli (metyl metacrylat) c iu ch tCH2=C(CH3)-COOCH3 (metyl metacrylat)

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-Nilon -6 c iu ch t: H2N-[CH2]5-COOH(Axit 6 aminohexanoic)

p n ng l CCu 32. X l phenol, tht vy

C6H5OH + NaOH C6H5ONa + H2OC6H5OH + 3Br2 C6H2Br3(OH) + 3HBr

C6H5OH + NaHCO3 khng phn ngp n ng l DCu 33. X c cu hnh electron lp ngoi cng l ns2np6 X nhm VIA Ha tr cao nht vioxi bng 6 (XO3), ha tr trong hp cht kh vi hiro bng:

8 - 6 = 2 (H2X)

Theo bi ra: %2X/H X

X.100= = 94,12%

X + 2 X = 32 (S)

Vy %3X(XO )

X.100= = 40%

X + 3.16. p n ng l B

Cu 34. Cc cht l:

C2H5OH + CuOot CH3CHO + H2O + Cu

C2H4 + 21

O2

oxt, t CH3CHO

C2H2 + H2Ooxt, t [CH2=CH-OH] CH3CHO

p n ng l CCu 35. Theo bi ra: nHCl = 0,2.1 = 0,2 (mol)

2 3Na COn = 0,1.1,5 = 0,15 (mol) ;

3KHCOn = 0,1.1 = 0,1 (mol)

Khi nh t t dung dch HCl (H+) vo X th:Na2CO3 + HCl NaCl + NaHCO30,15 0,15 0,15 (mol)

Sau phn ng ny, lng HCl cn li l:nHCl (d) = 0,2 0,15 = 0,05 (mol)

Do , xy ra qu trnh:HCO3

- + H+ CO2 + H2O0,05 0,05 0,05 (mol)

Vy V =2CO

V = 0,05.22,4 = 1,12 (lt). p n ng l B

Cu 36. Ancol no, n chc, mch h: CnH2n+2OPhng trnh ha hc t chy:CnH2n+2O + 1,5nO2 nCO2 + (n+1)H2O1 (mol) 1,5 (mol) n (mol) (n + 1) (mol)

Theo PTHH ta thy:2 2O CO

1,5.Vn = 1,5.n = (mol)

22,4

Theo nh lut bo ton khi lng: mancol +2 2 2O CO H O

m = m + m

m +1,5.V.32 V.44

= + a22,4 22,4

m = a -V

5,6

p n ng l ACu 37. Trch cc mu th vo ng nghim

-Cho t t dung dch HCl vo cc mu th:+ Mu no to ra kh bay ln mu cha amoni hirocacbonat NH4HCO3

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HCl + NH4HCO3 CO2 + NH4Cl + H2O+ Mu to ra kt ta trng, sau kt ta b tan mu cha natri aluminat Na[Al(OH)4]:

HCl + Na[Al(OH)4] Al(OH)3 + NaCl + H2O3HCl + Al(OH)3 AlCl3 + 3H2O

+ Mu to ra vn c mu cha natri phenolat:HCl + C6H5ONa C6H5OH + NaCl

+ Mu to ra cht lng khng tan mu cha ancol etylic C2H5OH:HCl + C2H5OH

ot C2H5Cl + H2OKhng tan

+ Mu to ra 2 lp cht lng khng trn ln vo nhau mu cha benzen C6H6HCl + C6H6 khng tc dng

+ Mu to ra dung dch ng nht mu cha anilin C6H5NH2:HCl + C6H5NH2 C6H5NH3Cl (tan)

p n ng l B

Cu 38. Theo bi ra:2CO

0,448n = = 0,02 (mol)

22,4

nNaOH = 0,1.0,06 = 0,006 (mol); 2Ba(OH)n = 0,1.0,12 = 0,012 (mol) Suy ra: -

2NaOH Ba(OH)OHn = n + 2.n = 0,006 + 2.0,012 = 0,03 (mol)

2+2Ba(OH)Ba

n = n = 0,012 (mol)

Cc phn ng xy ra: (ion rt gn)CO2 + OH

- HCO3-

0,02 0,02 0,02 (mol)HCO3

- + OH- CO32- + H2O

0,01 (0,03 0,02) 0,01 (mol)CO3

2- + Ba2+ BaCO3 0,01 0,01 0,01 (mol)

Vy m = 3BaCO = 0,01.197 = 1,97 (gam) p n ng l DCu 39. Cc kim loi c tnh kh trung bnh v yu c th iu ch bng phng php in phndung dch mui ca chng

Cc kim loi c th l Fe, Cu, Agp n ng l B

Cu 40. Ion X2+: 1s22s22p63s23p63d6 Nguyn t X: 1s22s22p63s23p63d64s2

(X X2+ + 2e)Suy ra: X chu k 4 (v c 4 lp electron), nhm VIIIB (v l nguyn t d v c 8 electron

ha tr)p n ng l ACu 41. Xc nh cu to cc cht:

C2H2: CH CH (ankin)C2H4: CH2=CH2 (anken)CH2O: HCHO (anehit)CH2O2 (mch h): HCOOH (axit fomic)C3H4O2 (mch h, n chc, khng lm chuyn mu qu tm m):

HCOOCH=CH2 (vinyl fomat)

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Nh ta bit, cc loi hp cht ankin c ni ba u mch; anehit; axit v este ca axitfomic; u phn ng c vi dung dch AgNO3/NH3 to ra kt ta:

CH CH + Ag2O 3NH AgC CAg + H2OBc axetilua

HCHO + 2Ag2O 3NH CO2 + 4Ag + H2O

HCOOH + Ag2O3NH

2Ag

+ CO2 + H2OHCOOCH=CH2 + Ag2O 3NH HOCOOCH=CH2 + 2Ag C 4 cht to kt tap n ng l B

*Ch : - Phn bit vi hin tng trng gng (to kt ta bc kim loi Ag), cc cht Ag,AgC CAg, AgC C-R, u l kt taCu 42. Phn ng ha hc xy ra khi cho Ba(OH)2 ti d vo cc dung dch trn:

+ Ba(OH)2 + (NH4)2SO4 BaSO4 + 2NH3 + 2H2O+ Ba(OH)2 + FeCl2 Fe(OH)2 + BaCl2+ 3Ba(OH)2 + 2Cr(NO3)2 2Cr(OH)3 + 3Ba(NO3)2+ 2Cr(OH)

3+ Ba(OH)

2 Ba[Cr(OH)

4]

2(tan)

+ Ba(OH)2 + K2CO3 BaCO3 + 2KOH+ 3Ba(OH)2 + 2Al(NO3)3 2Al(OH)3 + 3Ba(NO3)2+ 2Al(OH)3 + Ba(OH)2 Ba[Al(OH)4]2 (tan)C 3 ng nghim c kt ta. p n ng l D

Cu 43. Theo bi ra:2H

5,6n = = 0,25 (mol)

22,4

Gi x, y ln lt l s mol ca Al, Sn c trong hn hp ban uTa c: 27x + 119y = 14,6 (1)X + HCl (d):

Al + 3HCl AlCl3 + 1,5H2

x 1,5x (mol)Sn + 2HCl SnCl2 + H2 y y (mol)

Ta c: 1,5x + y = 0,25T (1) v (2) ta c: x = 0,1; y = 0,1 (2)

X + O2: 4Al + 3O2ot 2Al2O3

x3

x4

(mol)

Sn + O2ot SnO2

y y (mol)

Suy ra: 2O 3 3n = x + y = .0,1 + 0,1 = 0,175 (mol)4 4

Vy2O

V = 0,175.22,4 = 3,92 (lt). p n ng l A

*Ch : Sn + 2HCl SnCl2 + H2

Sn + O2ot SnO2

Cu 44. Cacbohirat (Cn(H2O)m) l hp cht hu c tp chc, c nhiu nhm OH (nhm chc caancol). p n ng l D

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Cu 45. Th t trong cc dy in ha :Mg2+ Zn2+ Cu2+ Ag+Mg Zn Cu Ag

Dung dch thu c cha 3 ion kim loi 3 ion phi l Mg2+, Zn2+ v Cu2+ (d)V Cu2+ cn d Mg v Zn phn ng ht

Cc qu trnh phn ng xy ra:Mg - 2e Mg2+1,2 2,4 (mol)Zn - 2e Zn2+x 2x (mol)Ag+ + 1e Ag 1 1 (mol)Cu2+ + 2e Cu 2 4 (mol)

V Cu2+ cn d nn: 2,4 + 2x < 1 + 4 x < 1,3Trong cc gi tr cho th x = 1,2 < 1,3 tha mn yu cu bi ton

p n ng l D

Cu 46. Theo bi ra:2CO NaOH

11,2n = = 0,5 (mol); n = 0,5.1 = 0,5 (mol)

22,4

Ta thy:2CO

n = NaOHn = 0,5 (mol)

Trong phn t axit c s nhm COOH bng s nguyn t cacbon trong phn t. Vy 2axit phi l HCOOH v HOOC-COOH

p n ng l D

Cu 47. Theo bi ra: nX =10,3

= 0,1 (mol)103

X + NaOH Y + Z

+ X: C4H9NO2: mch h+ Y: Nng hn khng kh MY > 29; lm giy qu tm m chuyn mu xanh (suy ra Y l amin)+ Z: Cht tan, lm mt mu dung dch Br2 Z l mui natri ca axit cacboxylic khng no,

n chc, mch hVy X phi l CH2=CH-COOH3NCH3:CH2=CH-COOH3NCH3 + NaOH CH2=CH-COONa + CH3NH2 + H2O

(X) (Z) (Y)0,1 mol 0,1 (mol)

Vy m = mZ = 0,1.94 = 9,4 (gam). p n ng l C

Cu 48. Xt cn bng: 2NO2(k) N2O4(k) , H ?

(Nu ) (khng mu)Khi h nhit ca bnh m mu nu nht dn khi h nhit cn bng trn chuyndch theo chiu thun (chiu t tri sang phi, chiu chuyn thnh N2O4 khng mu)

Theo nguyn l chuyn dch cn bng, ta suy ra c chiu thun l chiu ta nhit ( H < 0).p n ng l D

Cu 49. Theo bi ra:2O

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

22,4

Ancol X (no, mch h, a chc) dng CnH2nOa (a n)

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CnH2nOa + 23n + 1 - a

O2

nCO2 + (n + 1)H2O

1 (mol)3n + 1 - a

2mol

0,2 (mol) 0,8 (mol)

Suy ra: 0,2.3n + 1 - a

2= 0,8.1 0,3n + 0,1 0,1a = 0,8

3n + 1 a = 8 3n a = 7V a > 1 n 3

a n n 3,5Suy ra: n = 3, do X dng C3H8Oa+ Nu a = 2: X tr thnh C3H8O2

C3H8O2 + 4O2 3CO2 + 4H2O1 (mol) 4 (mol)0,2 (mol) 0,8 (mol) (Tha mn bi)

+ Nu a = 3: X tr thnh C3H8O3C3H8O3 + 3,5O2 3CO2 + 4H2O1 (mol) 3,5 (mol)0,2 (mol) 0,7 (mol) 0,8 (mol) (Loi)

V X tc dng c vi Cu(OH)2 nn X phi c 2 nhm OH 2 nguyn t cacbon cnh nhau.Do cu to ca X:

CH3-CH(OH)-CH2(OH): propan 1,2- iolX + Cu(OH)2:2C3H8O2 + Cu(OH)2 (C3H7O2)2Cu + 2H2O0,1 mol 0,05 (mol) phc mu xanh

Vy m = 0,05.98 = 4,9 (gam). p n ng l ACu 50. Phn ur c cng thc l (NH2)2CO:

2NH3 + CO2

ot ,p cao

(NH2)2CO + H2O-Phn hn hp cha ng thi 3 nguyn t nit, photpho v kali, c gi l phn NPK-Phn ln cung cp photpho cho cy di dng ion photphat (PO43-)Cn phn m cung cp nit cho cy di dng ion nitrat (NO3

-) v ion amoni (NH4+)

-Amophot l hn hp cc mui NH4H2PO4 v (NH4)2HPO4 (thu c vi axit photphoricH3PO4)

Cn hn hp (NH4)2HPO4 v KNO3 l nitrophotka (phn hn hp)Vy ni dung ng l B. p n ng l B

Cu 51. Phng trnh tng hp NH3:

N2(k) + 3H2(k) 2NH3(k)Ban u: 0,3M 0,7M 0Phn ng: x 3x 2xCn bng (0,3 x) (0,7 3x) 2xTheo bi ra:

2HV (sau p/) = 50%Vhh

2H hh

n = 50%n

0,7 3x =1

.(0,3 - x + 0,7 - 3x + 2x)2

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0,7 3x =1

(1 - 2x)2

1,4 6x = 1- 2x 4x = 0,4 x = 0,1Vy trng thi cn bng c:

[N2] = 0,3 x = 0,3 0,1 = 0,2 M[H2] = 0,7 3x = 0,7 0,3 = 0,4M

[NH3] = 2x = 2.0,1 = 0,2MSuy ra: KC =

22

3

3 3

2 2

NH (0,2)= = 3,125

0,2.(0,4)N . Hp n ng l D

Cu 52. Ta c:2+ 2+

o o o

pin Zn - Cu Cu /Cu Zn /ZnE = E - E = 1,1 (1)

+ 2+

o o o

pin Cu - Ag Ag /Ag Cu /CuE = E - E = 0,46 (2)

Vi +o

Ag /AgE = 0,8V. T (2) 2+oCu /CuE = +0,34V

T (1) 2+o

Zn /ZnE = 0,34 1,1 = - 0,76 (V)

Vy2+

o

Zn /ZnE v2+

o

Cu /CuE ln lt l - 0,76 V v +0,34 Vp n ng l CCu 53. Khi nung PbS trong khng kh xy ra phn ng:

PbS (r) + 23

O2

PbO (r) + SO2

Khi 1 mol PbS (239 gam) phn ng ht th khi lng gim 239 223 = 16 gamVy a (gam) PbS phn ng, khi lng gim m 0,95m = 0,05 m (gam)

a =0,05.m.239

= 0,7469m (gam)16

Vy % khi lng PbS chy l: % mPbS =0,7469m.100%

= 74,69%m

p n ng l ACu 54. C6H5NH2 + HONO

ot C6H5OH + N2 + H2OAnilin axit nitr

Sn phm to ra khng c mui iazoni (phn ng to mui iazoni xy ra khi thc hin nhit thp, 0 5oC).

C6H6 + Br2(dd) khng xy raBenzenC2H5NH2 + HONO C2H5OH + N2 + H2OEtylamin axit nitr bt kh

Ch cc ancol a chc c 2 nhm OH 2 nguyn t cacbon cnh nhau mi phn ng vi

Cu(OH)2 to ra phc mu xanh lam:CH2(OH)-CH2-CH2OH + Cu(OH)2 khng xy raCH2(OH)-CH(OH)-CH3 + Cu(OH)2 phc mu xanh lam

p n ng l C*Ch : - Amin bc mt (R-NH2) tc dng vi axit nitr (HONO) nhit thng cho ancol hoc

phenol v gii phng kh nit (N2)-Anilin v cc amin thm bc mt tc dng vi axit nitr nhit thp (0 5 oC) cho mui

iazoni:

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C6H5NH2 + HONO + HClo0 - 5 C C6H5N2

+Cl- + 2H2OMui iazoni

Cu 55. Cc hp cht c nhm CHO (chc anehit) th c kh nng tham gia phn ng trnggng (trng bc) nh glucoz, mantoz, axit fomic, anehit axetic. p n ng l A*Ch : - Trong dung dch, gc - glucoz ca mantoz c th m vng to ra nhm CHO. Do dung dch mantoz tham gia phn ng trng bc

-Axit fomic va c nhm COOH va c cu to ca nhm CHOCu 56. Cc cht gy nghin nh thuc phin, cn sa, heroin, cocain, moocphin, seduxen,.. p nng l C*Ch : Cc thuc nh penicilin, ampicilin, erythromicin, l cc loi thuc khng sinhCu 57. X (C5H8O2) + NaOH mui + sn phm hu c

Suy ra X phi l este n chc: RCOORRCOOR + NaOH RCOONa + ROH

nRCOONa = nX =5

100= 0,05 (mol)

Suy ra MRCOONa =3,4

= 680,05

R + 44 + 23 = 68 R = 1 (H)

Nh vy X c dng HCOOR (HCOOC4H7)V sn phm hu c cn li khng lm mt mu dung dch brom, do X phi c cu to:

HCOOC(CH3)=CH-CH3. Tht vy:HCOOC(CH3)=CH-CH3 + NaOH HCOONa + [CH3-CH=C(CH3)OH]

CH3-CH2-COCH3p n ng l B

Cu 58. Cc PTHH xy ra trong s trn:C6H5OH + (CH3CO)2O xt C6H5OOCCH3 + CH3COOHPhenol anhirit axetic Phenyl axetat

C6H5OOCCH3 + 2NaOH(d) C6H5ONa + CH3COONa + H2ONatri phenolatVy X l anhirit axetic (CH3CO)2O, Y l natri phenolat C6H5ONa

p n ng l BCu 59. Cc phn ng xy ra trong s :

CH3CH2Cl + KCN CH3CH2CN + KCl(X)

CH3CH2CN + 2H2O+

3H O CH3CH2COOH + NH3(Y)

p n ng l DCu 60. Hon thnh cc PTHH (nu c):

Cu + Pb(NO3)2 khng xy raCu + HCl khng xy ra2Cu + 4HCl + O2 2CuCl2 + 2H2OCu + H2SO4 (l) khng xy ra

p n ng l C