giải chi tiết đề 502
TRANSCRIPT
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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