[Hóa 10]Bài tập pưhh chương II

G

godrortol

H

hiepsh97

a)
$n_{H_2O}=0.225$
$n_Cu=0.08$
PT:
$2M+2H_2O \rightarrow 2MOH + H_2$
0.16-----0.16------0.16------0.08
$H_2+CuO \rightarrow Cu+H_2O$
0.08------------0.08

$n_{H_2O (pư)}=0.16 < 0.225$ => $H_2O$ dư => $M$ hết

$M_M= \frac{1.11}{0.16}=6.9375$

b)

$m_{MOH}=3.83$

Áp dụng ĐLBTKL:

$m_dd=m_M+m_{H_2O}-m_{H_2}=1.1+4.05-0.08.2=4.99$

=>C%=20.04%
 
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