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
C2H2 2CO2 +H2O
0,2 0,4 ^ 0,2 (mol)
H2 H2O
0,3 > 0,3 (mol)
=> m b i n h t a n g = nicoj + mj,^o = 0,4.44 + 0,5.18 = 26,6gam Chon dap a n D.
B a i 26: Ta c6 = = 3,5 > 2 ^ H6n hop andehit don thiJc
" A n d e h i t 0,05
phai chufa H C H O , do hai andehit la dong dSng ke tiep, andehit con lai la CH3CHỌ
Chon dap an B .
B a i 31: So do phan iJng:
HaN - (CH2)4 - CH(NH2) - COOH (x mol) .NaOH (0,2 mol)
'CIH3N - (CH2), - CmNHgCl) - COOH (x mol) NaCl (0,2 mol)
Ta c6: 0,4 = 0,2 + 2.x => x = 0,1 (mol)
=>m = 146.0,1 = 14,6gam. Chon dap an D.
B a i 38: H o n hop etyl isobutyrat, axit 2 - m e t y l propanoic, metyl butanoat phan ling vdi hon hop N a O H , K O H (chung la M O H ) tao h6n hop muoi dong phan
•V- - i - H C K O ^ m o l ) ( V i f a d u ) ^ ^
CgH^COOC^Hs
• CgH^COOH + M O H (0,03mol) -> C3H7COOM (0,03mol) + ...
[C3H7COOCH,, = ^ niMuoi = + m^^. + m^. = 3,62gam = ^ niMuoi = + m^^. + m^. = 3,62gam Chon dap an B. B a i 49: Amino axit X: (H2N)„R(C00H)„ (H2N)nR(C00H)™ + m N a O H ^ (H2N)„R(C00Na)n, + mHzO a > a (mol) (H2N)nR(C00H)„ + n H C l -> (C1H3N)„R(C00HU + nHsO a > a (mol)
Phucrng phap tSng giam k h o i iMng.
22m.a = 22,92 - 17,64 => m.a = 0,24 Ta c6: 36,5.n.a = 22,02 - 17,64 ạn = 0,12
Vay — = i C h i dap an D la dung,
m 2
B a i 56: MO + H2SO4 -> MSO4 + H2O
0,07 ^ 0 , 0 7 ^ 0,07 (mol)
Ta c6: 0,07.(M + 16) = 5,6 M = 64.M la C u
=^ ncuso,x..,o = "cuso, = 0,07mol ^ 17,5 = 0,07.(160 + 18.x)
=> X = 5 Chon dap an Ạ D A P A N V A H l / C i N G D A N G I A I M A D E 112 C a u D A C a u D A C a u D A C a u D A 1 C 16 A 31 B 46 D 2 D 17 D 32 A 47 C 3 A 18 B 33 B 48 C 4 A 19 C 34 D 49 A 5 B 20 A 35 C 50 B 6 C 21 D 36 A 51 B 7 A 22 A 37 B 52 B 8 B 23 B 38 B 53 C 9 A 24 C 39 B 54 C 10 A 25 D 40 D 55 A 11 D 26 B 41 A 56 B 12 D 27 B 42 A 57 C 13 B 28 B 43 C 58 B 14 B 29 A 44 D 59 C 15 B 30 C 45 A 60 C
Hi£c/ng dan gidi
B a i 7: Qua r a t nhieu p h a n uTng, Cu nhiidng electron, N*^ n h a n electron tao N*^ r o i nhLforng l a i cho oxi trot l a i N*^. Vay suót qua t r i n h phan ufng chi CO Cu nhtfdng electron va O2 nhan electron
Cu ^ Cu^* + 2.e O2 + 4.e 20"^
0,2 <t- 0,4 0,1 -> 0,4
=> mcu = 12,8 gam. Chon dap an Ạ
B a i 12: Phan i J n g : + O H " ^ H2O
Trifdc pir: 0,04Vx 0,03VY Pir: 0,03VY 0,03VY Sau pi/: (0,04Vx - 0,03VY) (sau phan ufng p H = 2 axit du)
Theo de, ta c6: sau phan ufng IH*] = 0,01M ^ (0,04Vx - 0,03VY) = 0,01.(Vx + V Y )
VY 3 •
Chon dap a n D.
B a i 14: m dat gia t r i Idn n h a t k h i Fe dir trong phan ufng vdi HNO3 tiep tuc phan ufng vdi Fê^ diia hét ve Fê*
Fe + 4HNO3 ^ Fe(N03)3 + N O + 2H2O
0,1 <- 0,4 0,1 (mol)