B Phuomg trinh hoa hoc:
I Q H „ 0 , + A g , 0 )C,H„Q, + 2Ag i
T h e o ( l ) : S6'mol QH.^Og =^s6'mol Ag = - . ^ ^ ^ = 0,0725 (mol). 2 2 108
N6ng d6 mol cua dung djch glucoza la:
( Q H , A ) - = 0,725 (M).
5.24. Dap an diing la A .
Phucmg trinh phan ting trung hok:
CH3COOH + NaOH ^ CHjCOONa + H^O (1) Theo (1): S6' mol CHjCOONa = s6' mol CH3COOH = 0,125.0,4 = 0,05 (mol).
^ mcH^cooNa = 0.05.82 = 4,10 (gam).
5.25. Dap an diing la D. Phuomg trinh hoa hoc: Phuomg trinh hoa hoc:
CHjCOOH + q H j O H < = ^ ! ^ ^ CH3COOC2H5+H2O (1)
S6' mol cic chaft:
6 9 2
Theo (1): S6' mol este = s6 mol CH3COOH = 0,1 mol.
(Du rugru etylic ntn kh6'i lucmg este tinh theo axit axetic). HiSu su&t
phan ling la 60% ndn kh6'i lucmg este thu dugc la: m ^ , = 0 , 1 . 8 8 . ^ = 5,28 (gam).
5.26. Dap an diing la B.
Phucmg trinh hoa hoc: '•
Q H„ 0 , """^"^ >2C3H,OH + 2CQ, t (1)
9^ leo (1): S6' mol QHi^O, = - s6' mol CÔ = ^ 0,25 (mol). 2 2 22,4
Hieu suát 90% n&n khd'i lugng glucoza la:
100
m = 0,25.180.—= 50 (gam). 90
[S.27. C6ng thiic phan tut diing la Ạ
S6 mol cac chát:
Hcô = — = 0,15 (mol) ^ He = 0.15 mol va m^ = 1,8 gam. 44
2,7
= —^ = 0,15(mol)->n„ =0,3mol va m„ =0,3 gam. 18 ,
2 4
• -^rrio = 4 , 5 - ( l , 8 + 0,3) = 2 , 4 ( g a m ) ^ n o - — = 0,15(mol). 16
Phan tu chat him CO X CO C, H va Ọ (adsbygoogle = window.adsbygoogle || []).push({});
Dat CTPT cua X la C,H^O,. Ta c6: m ^xhV
X : y : z = He : HH : Ho = 0,15 : 0,3 : 0,15 = 1 : 2 : 1
CTPT cua X la (Clifi\ 60 ^ n = 2.
CTPT cua X la C2H4O2.
5 . 2 8 . Dap an dung la C. Phuomg trinh hoa hoc: Phuomg trinh hoa hoc:
(C,7H3,COO)3C3H, +3NaOH^3C,7H3,COONa + C3H,(OH)3 (1)
Theo (1): S6' kmol C3H,(OH)3 = ^ s6' kmol NaOH = = 0-01 (kmol).
Hieu suat phan utig la 80% nfin kh6'i lucmg glixerol thu dugc la: 80
m = 0,01.92.—= 0,736 (kg). 100
IỊ B A I T A P T L ; L U A N
5 . 2 9 . - Tfnh s6' mol cac chat ban d i u :
0 "56 1 44
= - 0,025 (mol); n,,^ - ^ = 0,02 (mol).
- Cac phuong trinh hoa hoc:
FeO + CO—J^Fe + CÔ (1)
CO, + Că0H)2 ^ CaCOj i + H^O ( 2 )
T h e o ( l ) v a ( 2 ) : n',o-nV.o ^n^^o, = y ^ = 0.01 (mol).
a) Nhu vay, chat ran thu dugc sau phan ling g6m Fe va FeO du:
. mp, =0,01.56 = 0,56 (gam).
m p , o = 0,01.72 = 0,72 (gam). b) Thé tich khi thu dugc sau phan ling:
v = v +v
= 0,01 + (0,025 - 0,01)].22,4 = 0,56 (lit). c) Phan ung kh6ng xay ra hoan toan vi FeO va CO d^u dụ
5.30. a) A va B c6 m6t chat tan trong H2SO4 loang:
A + H^SO, ^ A S O ^ + H , t (1) B + H2SO4: Kh6ng phan ling.
= 6,5 gam -> = - ^ . 1 0 0 % = 97,6%; = 2,4%. 6,659
- T h e o ( l ) : 0 ^ = 0 ^ , =0,1 (mol).
Nguyen tu kh6'i cua A : = ^ = 65 ^ A la kern (Zn).
- Phan dug hoa tan B: , y
B + 2H2SO4 d,n -> BSO4 + SO, t + 2H2O (2)
" B = "so, = ^ = 0.0025 (mol). (adsbygoogle = window.adsbygoogle || []).push({});
" 6 4 ' -> M„ = - ^ l i ^ = 64 ^ B la d6ng (Cu).
" 0,0025 . b) Tach hOn hgp Zn, Cu, CuO: