D. DAP AN VA HLTdNG DAN GIAI DE DAI HOC D A P A N V A H L / ( 3N G D A N G I A I M A D E
33 100.16 + So do hop thllc: 2NO3 (trong mu6i) ^ (trong oxit)
+ So do hop thllc: 2NO3 (trong mu6i) ^ (trong oxit)
+ Theo qui \Ac tS]
Chon dap dn B.
+ Theo qui t^c tang giam ^ m = 50 - - . n^^. .(2.62 - 16) = 44,6 gam 2 ^
B a i 17:
+ De sau k h i két thuc phan ufng khong c6 k i m loai t h i so mol Fe(N03)3 vCra du hoSc A p dung D L B T E => b > 2a
Chon đp an C. B a i 18:
+ V i de khong noi mi6c voi trong diT nen k h i dan san pham chay qua
binh diing dung dich nLfdc voi trong, ket thuc phan utog se sinh 2 muoi + Ta c6: m^ô + m„^o = 27,93 - 5,586 = 22,344 gam
Hay 44nc + 9H = 22,344 (1)
+ Theo de: 12nc + na = 4,872 (2)
+ TCr (1) va (2) nc = 0,336; n = 0,84 nc : n H = 4 : 10 -> C 4 H 1 0
Chon dap an C. B a i 19:
ncuisinh ra) = 0,02 mol, n K h i = 0,015 mol
C U S O 4 + 2NaCl ^ Cu + C I 2 + Na2S04
a mol —> a mol
+ V i n K h i = 0,015 mol nen C U S O 4 dir
C U S O 4 + H 2 O ^ . C u + H 2 S O 4 + ^ 0 2 b mol -> 0,5b . , . fa + b - 0,02 + Theo de ta c6 he <^ • [a + 0,5b = 0,015 => a = b = ỌOlmol -> n^^, = 0,02mol ^ [H"] = 0,01-> p H = 2 Chon dap an C. B a i 20: * Phaang phap n h a m
+ nnci = 0,15 mol, n c o , = 0,045 mol, nn^cô = 0,15 mol + n^^^cô = n H c i - n c o , = 0,105 mol ^ [NajCOa] = 0 , 2 I M
+ ^n.uco, = n^^co, + "^^co, " H H C I = 0,09 mol [NaHCOg] = 0,18M Chon dap an B.
B a i 21:
Ta CO p t pu: 3Cu + 8H* + 2 NÓ SCu^* + 2NO + 4H2O ncu = 0,05 mol, n„. = 0,12 mol, n^^ = 0,08 mol
0 12 3
+ Dd thay h e t triTdc -> ncu(phan .j„g) = -^-z^ = 0.045 m o l
o
n ^ ; o - = (0,08 - 0,12/4) = 0,05 m o l
NO;, (tao muoi)
+ Suy ra m^uoi = 0,045.64 + 0,05.62 + 0,02.96 = 7,9 gam Chon dap a n A .
B a i 22:
+ H C l la chat oxi hoa k h i va chi k h i phan iJng sinh H2. Chon dap a n C.
C a u 23:
* Phuong phap loai trCr
Nguyen tic: KhOf m a n h gap oxi-hoa manh
+ A ) Sai v i c6 muoi s&t t h i dung dich X phai c6 muoi k e m
+ B) Sai v i c6 muoi dong t h i dung dich X phai c6 muoi sSt
+ C) Sai v i CO sinh muoi sat I I I t h i dung dich X phai c6 muoi dong * Phiiong phap k i n h nghiem
+ Dung true o x i - hoa khuf de thay X gom: Mg(N03)2, Zn(N03)2, Fe(N03)2
Chon dap a n D.
B a i 24:
mp^ou = 6,4.71,875/100 = 4,6 gam
mnudc = 1,8 gam n^aac = 0,1 mol nn2 = 0,125 mol
4 6
Mruau = n = 92.n/3 (vdi n la so n h o m chuTc OH) • 0,125.2 - 0,1
D I thay A: CsHsCOIDa
Chon dap a n C.
B a i 25:
So dong phan c a c chat c6 cung so C nhiing d i cung cac nguyen tuf c6
so hoa t r i Idn hon t h i c6 so dong phan nhieu hon.
Vi C I C O hoa t r i I , O c6 hoa t h i I I , N c6 hoa t r i I I I n e n s o liTOng d o n g p h a n cua C 3 H 7 C I < CaHgO < C 3 H 9 N .
Chon dap an Ạ
B a i 26:
+ A p dung phuong phap tang giam k h o i liTOng: m^usi = 2,46 + 22.0,04 = 3,34 gam Chon dap d n D.
B a i 27:
' Axeton, benzen, xiclobutan khong l a m m a t mau dung dich thuoc t i m
du a nhiet do caọ
Chon dap a n Ạ
B a i 28:
n^ô = 0,08 mol, n,,^^ - 0,064 mol
S 6 H 2.n„^o 2.0,064 8
D l thay: 10 + 0,15.40 = 16 -> E la este vong -> Chon dap a n C. la phu hop
Chon dap a n C.
B a i 29:
So m g K O H can de trung hoa 10 k g chat beo A l a :
10000.7 = 70000 m g = 0,07 k g
Suy r a N a O H can de trung hoa chat beo A la:
^ = 0,00125 (Kmol)
56
N a O H can dung de phan iJng v d i 10 k g A l a :
M 2 0 - ^ =
0,035 (Kmol)40 1000
Ap dung d i n h luat bao toan k h o i lUgfng t a c6
n i A + m N a O H ( p h a n iJiig v6i A) = T^xh phbng + IHgiixerol + n i H 2 0
^ m^aph6ng = 10 + 0,035.40 - O,0M25.18 - 92.1/3.(0,035 - 0,00125) = 10,3425 gam Chon dap a n A .
B a i 30:
+ Glu CO 2 n h o m cacboxyl va 1 n h o m aminọ
+ Lys CO 1 n h o m cacboxyl va 2 n h o m aminọ Chon dap a n B.
B a i
31:
M = ^^'^ ^ 49,67 -> dap an B la phu hap
36^5 Chon dap an B.
B a i 32:
M a t do electron cang cao t h i t i n h bazcf cang manh. Chon dap an Ạ
B a i 33:
Luu y:
+ 1 mol mantozor tien hanh phan uTng trang bac sinh 2 mol A g
+ 1 mol mantozot thuy phan hoan toan sau do lay dung dich t h u dxiac
tien hanh phan ufng trang bac sinh 4 mol Ag.
nmantozo = 0,1 mol
+ Suy ra: mAg = 108.(0,1.0,5.2 + 0,1.0,5.4) = 32,4 gam Chon dap an B.
B a i 34:
nButa-l,3-clien = nBrom = 0,125 mol
45,75-0,125.54 ^07=^^1
nstiren =
—' —
= 0,375 mol 104Ti le mat xich butadien va stiren trong cao su buna-S la
T= « i ? 5 , 1 = 1 : 3
0,375 3 Chon dap an Ạ
B a i 35:
Axetilen, metyl axetat, axeton khong tham gia phan ilng trang guang.
Chon dap an B. B a i 36:
+ Phenol + brom-> ket tua t r ^ n g
+ Stiren + brom -> mat mau dung dich brom + Ancol benzylic + brom-> khong c6 hien tirong Chon dap dn Ạ
Bai 37:
PhU0ng phap ducfng cheo + phucfng phap 3 d6ng + Theo de dx/He = 3,75 nen:
n. n 4.3,75-28 = 1 -> n,,^ - n^y,^ = x(mol) -> C2H6 4.3,75-2 H2 + C2H4 Ban dau: x x Phan urng: x . H % x . H % x . H % . Con lai: x . ( l - H % ) x . ( l - H % ) x . H %
+ M a t khac: dy/He = 5 nen:
x . ( l - H % ) . 2 + x . ( l - H % ) . 2 8 + x.H%.30 x.(2 - H%)
^ H % - 50%
* Phuctng phap B T K L + t i i chon liTOng chat + D L B T K L : nx.4.3,75 = nỴ4.5 -> 3nx = 4nY
-> Chon nx = 4 mol va ny = 3 mol
+ So mol k h i giam chinh la so mol H2 phan ufng n = 5.4
I I j p h a n ijfng = 1 mol
Hi ban dau ~ ^ C j l l ^ ban dau ~ ^
+ Suf dung dudng cheo: n -> H % = - = 50%
2
Chon dap an B.
B a i 38:
+ CH3OH + C O - ^ C H a C O O H
+ C2H5OH + O2 > CH3COOH + H2O + 2CH3CHO +O2 2CH3COOH
Chon dap an D.
B a i 39:
CH4, C2H5OH, CgHeCbenzen) khong phan iJng dugc vdi nudc brom.
B a i 40:
PhUdng phap tif chon laong chat
CO2 : H 2 O = 11 : 12 ^ chon n,,o,^ = 11 mol va n„^o = 12 mol
Suy ra nx = 12-11 = 1 mol
