giải chi tiết đề 192
TRANSCRIPT
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gii chi tit thi th i hc - m 192Cu 1. p n D
Cu 2. p n DCu 3. p n D
Cu 4. p n C
C l NaHCO3; D l Na2CO3; E l Ca(OH)2; F l CaCl2
V: CaCO3 ot CaO + CO2
CO2 d + NaOH NaHCO3CO2 + 2NaOH Na2CO3 + H2O
NaHCO3 + NaOH Na2CO3 + H2ONa2CO3 + HCl NaHCO3 + NaClNaHCO3 + Ca(OH)2 d 2CaCO3 + NaOH + H2ONa2CO3 + CaCl2 CaCO3 + 2NaCl
Cu 5. p n B
Do s mol Na+ u nh hn s mol HCO3- v Cl- nn trong dung dch cc ion Ca2+ v Mg2+ u tn ti
c hai dng mui bicacbonat v mui clorua. V vy nc thuc loi cng ton phn
Cu 6. p n DHF phn li: HF H
+ + F-T pH = 2 [H+] = 10-2
Ta bit:+ -
a+
[H ].[F ]= K
[H ]
Gi C l nng dung dch HF, khi cn bng, theo ta c:+ 2
-4-2
[H ]= 6,6.10
C - 10
V 10-2 nh so vi C nn coi C 10-2 C
Hay-2 2
-4[10 ] = 6,6.10
C
C =-2 2
-4
[10 ]= 0,152 (M)
6,6.10
Cu 7. p n C
t cng thc ca ru no l CnH2n+2-a(OH)aPhn ng t chy:
n 2n+2-a a 2 2 2
3n + 1 - aC H (OH) + O nCO + (n + 1)H O
2
Theo ta c:3n + 1 - a
= 3,52
3n + 1 a = 7
Hay n =6 + a
3(vi a n)
Nghim ph hp a = 3 n = 3 ru no l:
C3H5(OH)3 hay CH2OH-CHOH-CH2OH (glixerin)Cu 8. p n Dt cng thc phn t ca A l R(OH)n(COOH)mPhn ng ca A:
R(OH)n(COOH)m + NaHCO3 R(OH)n(COONa)m + mCO2 + mH2O
R(OH)n(COOH)m + (n + m)Na R(ONa)n(COONa)m + 2(n + m)
H2
t a l s mol ca A, t trn ta c: nCO2 = ma
2H
n + mn = a
2
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Theo c: ma =n + m
2. a = a nghim hp l l n = 1; m = 1
Vy A l R(OH)(COOH)Cng theo : MA = 45.2 = 90 gamSuy ra: MR+ 17 + 45 = 90 MR= 28 gamHay R l C2H4 Cng thc ca A: C2H4(OH)(COOH)Theo cho bit: Oxi ha A bng CuO cho B tc dng c vi dung dch AgNO3/NH3 nn A l ru
bc mt, do cng thc cu to ca A l:HO-CH2-CH2-COOHCu 9. p n B
3 4Fe O
4,06n = 0,0175 (mol)
232
Phn ng ca Fe3O4 vi H2SO4:
2Fe3O4 + 10H2SO4 ot 3Fe2(SO4)3 + SO2 + 10H2O
Mol: 0,0175 0,02625Vy nng ca mui trong dung dch X l:
Fe (SO )2 4 3M
0,02625C = = 0,0525 (M)
0,5
Cu 10. p n D
S mol NO = 4,48 = 0,2 (mol)22,4
t M l kim loi (ng thi l nguyn t khi), t phn ng oxi ha kh, kim loi m viHNO3, theo ta c:
NO3- + 3e NO
mol 0,2 0,6 0,2M - ne +
mol19,2
M
19,2
M.n
cn bng electron, s mol electron gc NO3- nhn phi bng s mol electron kim loi M
nhng, nn theo trn ta c:19,2
M. n = 0,6 19,2.n = 0,6M
Ch c gi tr n = 2 v M = 64 l thch hp
S mol Cu = 19,2 = 0,3 (mol)64
iu ch cht rn (CuO) t Cu theo quy trnh:
Cu + 3+HNO Cu(NO3)2 NaOH Cu(OH)2ot CuO
Mol 0,3 0,3Vy: mCuO = 0,3.80 = 24 (gam)
Cu 11. p n D
Cc phn ng xy ra:(NH4)2CO3 + 2HCl 2NH4Cl + CO2 + H2O
(NH4)2CO3 + 2NaOH Na2CO3 + 2NH3 + 2H2O
Al2O3 + 6HCl 2AlCl3 + 3H2O
Al2O3 + 2NaOH 2NaAlO2 + H2O
Fe3O4 + 8HCl FeCl2 + 2FeCl3 + 4H2O
Ca(HCO3)2 + 2HCl CaCl2 + 2H2O + 2CO2
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Ca(HCO3)2 + 2NaOH CaCO3 + Na2CO3 + 2H2O
2Al + 6HCl 2AlCl3 + 3H2
2Al + 2NaOH + 2H2O 2NaAlO2 + 3H2
HCl + NaOH NaCl + H2O
8Al + 3Fe3O3ot 4Al2O3 + 9Fe
Cu 12. p n D
Cn bng phn ng oxi ha kh, kt qu:3K2S + K2Cr2O7 + 7H2SO4 = 3S + Cr2(SO4)3 + 4K2SO4 + 7H2O
Cu 13. p n D
Gi x l khi lng bari cn lyPhn ng ha tan bari vo nc:Ba + 2H2O Ba(OH)2 + H2
gam 137 171 2gam x y z
T trn cho ta: y =171 2
.x; z = .x137 137
1 lt nc nng 1000 gam; nn khi lng dung dch sau khi ha tan:
mdd = 100 + x - 2x137
Vy C% =171/137.x.100
= 4,93%100 + x - 2x/137
Gii phng trnh trn c x = 41,3 gamCu 14. p n B
Kh nng kh ca kim loi c c trng bng th in cc ca chng. Kim loi c th in cc cngnh, tnh kh ca kim loi cng mnh. Trong cc kim loi cho trong , th in cc ca chng c spxp nh dn:
Cu > Ni > Fe > Ca > KVy tnh kh ca cc kim loi trn c sp xp tng dn:
Cu < Ni < Fe < Ca < KCu 15. p n C
So vi cng thc tng qut ca axit CnH2n+2-2aOm cc cng thc vit sau khng phi l axit:CH2O, v s nguyn t oxi lC2H2O2, v n = 2, a = 2 (v l)C2H5O4, v s nguyn t hiro lC6H9O3, v s nguyn t H v O u l
Cu 16. p n C
COOH COOH4
o
dd KMnO
t 2Br (1 mol)
bt Fe
CH3 Br
CH2Cl CH2OH 2Cl (1 mol)
askt o
NaOH
t
COOCH2
BrCu 17. p n B
Cu 18. p n D
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Dng dung dch HCl (hoc dung dch H2SO4 long) nhn bit: Khi nh HCl vo tng l- L no c hin tng si bt kh l dung dch Na 2CO3, v:
Na2CO3 + 2HCl 2NaCl + CO2 + H2O
- L no vn c khi lc, l dung dch Na2CO3, v:C6H5ONa + HCl C6H5OH + NaCl
- L no c mi gim, l dung dch CH3COONa, v:CH3COONa + HCl CH3COOH + NaCl
- L no khng c hin tng g l NaNO3Cu 19. p n B
Khi in phn dung dch:Na2SO4 khng tham gia in phn nn pH khng thay iCuCl2 c tham gia in phn nhng pH khng thay i
NaCl tham gia in phn v lm pH tngCu(NO3)2 tham gia in phn v lm pH gim
Cu 20. p n D
Cu 21. p n C
Phng trnh in li: FeSO4 Fe2+ + SO4
2
HCl H+ + Cl- anot (+) catot (-)
2Cl- - 2e Cl2 (1) 2H+ + 2e H2 (3)
2H2O 4e O2 + 4H+ (2) Fe2+ + 2e Fe (4)
T (1), (3): 2HCl H2 + Cl2 0,06 0,03
Thi gian hon thnh qu trnh (1), (3) theo cng thc Faraday m =1 A
. .ItF n
l
0,06.36,5 =1 36,5
. .1,34.t26,8 1
t = 1 gi 12 pht
T (2) v (4) ta c: 2FeSO4 + 2H2O dpdd
2Fe + O2 + 2H2SO4Gi a l s mol cht rn bm catot, theo (5) v cng theo cng thc Faraday c:
56.a =1 56
. .1,34.0,826,8 2
a = 0,02 mol
Vy mFe = 56.0,02 = 1,12 (gam)V kh anot =
2 2Cl OV = V = 22,4.(0,03 + 0,01) = 0,896 (lt)
Cu 22. p n D
MA = 2,25.32 = 72 gamGi cng thc phn t cht A l CxH2yOy (x, y nguyn dng), ta c:12x + 2y + 16y = 72 hay 2x + 3y = 12
Lp bng
x 1 2 3 4y 3,3 2,7 2 1,3
Ch cp x = 3, y = 2 tha mn Cng thc phn t ca A l C3H4O2Vy cng thc cu to c th c ca A:
CH2=CH-COOH; HCOOCH=CH2OHC-CH2-CHO; CH3-CO-CHO
Cu 23. p n B
nNaOH = 0,2.0,125 = 0,25 (mol); nNaOH = neste = 0,25 : 0,15
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este n chc v NaOH dGi cng thc este RCOOR (CxHyOz)
RCOOR + NaOH RCOONa + ROHMol 0,15 0,15 0,15
Cht rn gm RCOONa v NaOH d c khi lng:(R + 67)0,15 + 40.0,1 = 14,2 R = 1 (H)
Axit to este l HCOOHKhi lng bnh (1) tng l khi lng H2O
2H O
5,4n = = 0,3 (mol)
18
CO2 + Ba(OH)2 BaCO3 + H2Oa a
2CO2 + Ba(OH)2 Ba(HCO3)2Ba(HCO3)2 + 2NaOH BaCO3 + Na2CO3 + 2H2O
b b
Theo trn v :2CO
19,7 19,7n = a + 2b = + 2. = 0,3
197 197
2o
O
x y z 2 2t
yC H O xCO + H O
2
1 xy
2
0,1 0,3 0,3Gi tr tha mn: x = 3; y = 6Vy cng thc phn t ca este l C3H6O2
Cu 24. p n B
Gi R l k hiu tng qut, ng thi l nguyn t lng trung bnh ca hai kim loi, ta c:R + 2HCl R Cl2 + H2
2H
6,75n = = 0,3 (mol)
22,4
Vy R = 8,86 = 29,5330,3
Suy hai kim loi ha tr 2 thuc phn nhm chnh nhm II l Mg v Ca, vMg = 24 < 29,533 < 40 = Ca
Cu 25. p n C
Gi x l s mol Fe, 4x l s mol M trong hn hp kim loiFe + 2HCl FeCl2 + H2
Mol: x x2M + 2nHCl 2MCln + nH2
Mol: 4x 2nx2Fe + 3Cl2 2FeCl3
Mol: x 1,5x2M + nCl2 2MCln
Mol: 4x 2nx
T trn, theo ta c:
15,68x + 2nx = = 0,70
22,4
16,41,5x + 2nx = = 0,75
22,4
Gii h trn c x = 0,1S mol Cl2 phn ng vi M = 2nx = 0,7 0,1 = 0,6 (mol)
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Vy th tch Cl2 phn ng vi kim loi M l V = 0,6.22,4 = 13,44 (lt) S mol M = 4x = 4.0,1 = 0,4 (mol)Vy khi lng mol ca M =
108= 27 gam
0,4
M l AlCu 26. p n D
Cu 27. p n D
Dng dung dch H2SO4 long v dung dch NH3Khi cho dung dch H2SO4 long vo c 3 hp kim:- Trng hp no khng thy bt kh thot ra (khng c hin tng ha tan) l hp kim Cu Ag- Ly hai dung dch thu c ca hai trng hp cn li, ri cho tc dng vi dung dch NH3:
Trng hp no to kt ta khng tan trong NH3 d th hp kim ban u l Cu Al v:3Al + 3H2SO4 l Al2(SO4)3 + 3H2 Al2(SO4)3 + 6NH3 + 6H2O 2Al(OH)3 + 3(NH4)2SO4
Trng hp no c kt ta tan dn trong NH3 d th hp kim ban u l Cu Zn, v:Zn + H2SO4 l ZnSO4 + H2 ZnSO4 + 2NH3 + 2H2O Zn(OH)2 + (NH4)2SO4Zn(OH)2 + 4NH3 [Zn(NH3)4](OH)2 tan
Cu 28. p n DKim loi Me c th in cc cng nh th tnh kh ca kim loi cng mnh v tnh oxi ha ca cation
Men+ cng yu v ngc liCu 29. p n A
Khi un si cc nc c cc phn ng:Ca2+ + 2HCO3
- CaCO3 + CO2 + H2OMg2+ + 2HCO3
- MgCO3 + CO2 + H2ONc cn li trong cc c: n(Na+) = 0,01 mol; n(Cl-) = 0,02 mol
n(Ca2+, Mg2+) = 0,03 - 0,025 = 0,005 (mol)Cu 30. p n C
Cu 31. p n CKhi so snh tan trong nc cn xem cht no c kh nng to lin kt hiro vi H 2O ln nht s c
tan ln nhtSo snh cc cht cho trong , C2H5OH c lin kt hiro vi H2O ln nht nn c tan trong nc
ln nhtCu 32. p n B
Axit axetic c s phn cc ca nhm > C = O nn H trong OH linh ng hn phenol, ancol. Cnphenol v ancol, do nhm C6H5- ht electron nn phenol c H trong OH linh ng hn trong ancolCu 33. p n C
Cn bng phn ng c:3CH3-CH=CH2 + 2KMnO4 + 4H2O = 3CH3-CH-CH2 + 2MnO2 + 2KOH
OH OHDo :
3 2 4 2CH CH=CH KMnO H On : n : n = 3 : 2 : 4
Cu 34. p n DTheo :
Nng ban u ca axit CH3COOH =3
= 0,2 (M)60.0,25
S phn li ca axit:
CH3COOH CH3COO- + H+
S mol: Ban u: 0,2Phn li: x x xCn bng 0,2 x x x
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Theo trn v ta c: -3x 1,4
= x = 2,8.100,2 100
Vy nng ca cc phn t v ion trong dung dch l:[CH3COOH] = 0,2 2,8.10
-3 = 0,1972 (M)[H+] = [CH3COO
-] = x = 2,8.10-3 (M)Cu 35. p n D
Na2CO3 l mui ca axit yu, baz mnh nn mi trng c pH > 7
NH4NO3 l mui ca axit mnh, baz yu nn mi trng c pH < 7K2SO4 l mui ca axit mnh, baz mnh nn mi trng c pH = 7Cu 36. p n D
Gi C l nng ban u ca axit, l in li:
CH3COOH H+ + CH3COO
-Mol Ban u C
in li C C CCn bng C - C C C
Theo trn v ta c: Ka =2
(C )
C(1 - a)
V axit yu nn nh, c th coi l 1 - 1
Do : 2(C )
C= 1,75.10-5
Hay
2-5(0,1. )
=1,75.100,1
=-5
-21,75.10= 1,32.10
0,1
Do p s D ngCu 37. p n D
p dng cng thc o o
o
P .V PV=
T T
Theo , ta c: o
1,761. .22,4
0,75.0,89688 =
273 273 + t C
toC = 136,5oC
Cu 38. p n C
t cng thc tng qut ca+ axit th nht l CnH2nO2+ axit th hai l CmH2mO2
Cng thc trung bnh ca 2 axit l n 2n 2C H O
(iu kin: n, m 0, nguyn; n > 0)n-1 2n-1 n-1 2n-1 2C H -COOH + KOH C H COOK + H O (1)
Gi x l s mol ca 2 axit ban uT (1): s mol 2 axit bng s mol mui v bng xT trn v theo , ta c:
(14 n + 32)x =13,4 (2)(14 n + 70)x = 21 (3)Gii (2) v (3) ta c: x = 0,2; n = 2,5
V hai axit l ng ng k tip m = n + 1T n = 2,5 suy ra n = 2, m = 3Vy cng thc cu to 2 axit l: CH3-COOH v CH3-CH2-COOH
Cu 39. p n BPhn ng ca oxit tan trong axit:
Fe2O3 + 3H2SO4 Fe2(SO4)3 + 3H2OMgO + H2SO4 MgSO4 + H2O
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ZnO + H2SO4 ZnSO4 + H2OTheo trn v ta c:
2 2 4H O H SO
n = n = 0,30.0,1 = 0,03 (mol)
mmui sunfat = moxit +2 4H SO
m -2H O
m
= 2,18 + 0,03.96 0,03.18 = 5,21 (gam)Cu 40. p n C
K hiu chung 3 kim loi l MPhn ng ca kim loi vi HCl: M + 2HCl MCl2 + H2 Theo trn v ta c:
S mol kim loi nM =2H
3,024n = = 0,315 (mol)
22,4
T suy ra s mol mi kim loi trong hn hp l
nA = 0,315.4
= 0,06 (mol)9
Trong : (9 = 4 + 3 + 2)
nB = 0,315.3
= 0,045 (mol)9
; nC =2
0,315. = 0,03 (mol)9
t 3x l khi lng nguyn t ca A, th khi lng mol nguyn t ca B l 5x v ca C l 7xT , theo c: nA.3x + nB.5x + nC.7x = 4,92 gamThay cc gi tr nA; nB; nC vo phng trnh trn
x =4,92
= 80,615
Do : MA = 3x = 3.8 = 24 (gam) A l MgMB = 5x = 5.8 = 40 (gam) B l CaMC = 7x = 7.8 = 56 (gam) C l Fe
Cu 41. p n D
Ba(HCO3)2 + 2HNO3 Ba(NO3)2 + 2CO2 + 2H2OBa(HCO3)2 + Ca(OH)2 BaCO3 + CaCO3 + 2H2O
Ba(HCO3)2 + Na2SO4 BaSO4 + 2NaHCO3Ba(HCO3)2 + 2NaHCO3 BaSO4 + Na2SO4 + 2CO2 + 2H2OCu 42. p n C
t s hiu nguyn t ca A l Zn, ca B l Z + 1Theo ta c: Z + (Z + 1) = 31 Z = 15Vy s hiu nguyn t ca A l 15, ca B l 16Cu hnh electron ca A l 1s22s22p63s23p3
ca B l 1s22s22p63s23p4Cu 43. p n C
Phn ng tng hp NH3: N2 + 3H2 2NH3
Hng s cn bng: K =2
3
32 2
[NH ]
[N ].[H ]
Khi cn bng ha hc c thit lp, nu nhit phn ng tng, K gim (c ngha l NH3 gim),phn ng chuyn dch theo chiu nghch, thu nhit nn theo nguyn l L SatLi phn ng thun, tc phnng to NH3, l phn ng ta nhitCu 44. p n DCu 45. p n D
iu kin cht hu c c ng phn hnh hc:- Phn t c ni i- Cc nguyn t hay nhm nguyn t nh cacbon cha ni i phi c bn cht khc nhauDa vo , thy ch c hai cht sau l c ng phn hnh hc:
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CH3CH=CHC2H5 v CH3CH=CHCH3Cu 46. p n B
(C2H5O)n hay C2nH5nOnTa bit: ancol no c s H = 2C + 2Do c: 5n = 2.2n + 2 n = 2
Nn cng thc phn t ca ancol no l C4H10O2Cu 47. p n D
Phn t cc axit hu c c nhm C=O phn cc nn tnh axit ca chng mnh hn nc
Cng trong phn t cc axit , nu c nhm ht electron (nh cc halogen chng hn) s c tnh axitmnh hn cc axit c nhm y electron (nh cc gc R)
Do , cc cht cho trong , H2O c tnh axit yu nht, CH2ClCH2COOH c tnh axit mnh nhtCu 48. p n B
Cu 49. p n DGi V (lt) l th tch bnh cu:
nHCl = HClV V
; m = 36,5. (gam)22,4 22,4
;
2H O= V.100.1 = 1000 V (gam)
mdd
= mHCl
+2H O
m =36,5
+ 1000V22,4
Vy C% =36,5V/22,4
.100% = 0,163%36,5V/22,4 + 1000V
Cu 50. p n CTrong dung dch A c:
HF H+ + F- (1)
NaF Na+ + F- (2)
V c cn bng (2) nn cn bng (1) b chuyn dch sang tri, nn c th coi [HF] = 0,1; F- = 0,1Do , theo :
Ka =+ -
-4[H ].[F ]= 6,8.10
[HF]
[H+] =-4
0,1.6,8.10
0,1
pH = -lg (6,8.10-4) pH = 3,17