$n_{H_2}=7,84/22,4=0,35mol$
$a.CuO+H_2\overset{t^o}\to Cu+H_2O$
$ Fe_2O_3+3H_2\overset{t^o}\to 2Fe+3H_2O$
b.Gọi $n_{CuO}=a;n_{Fe_2O_3}=b(a,b>0)$
$\text{Ta có :}$
$m_{hh}=80a+160b=20g$
$n_{H_2}=a+3b=0,35mol$
$\text{Ta có hpt :}$
$\left\{\begin{matrix}
80a+160b=20& \\
a+3b=0,35 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,05 & \\
b=0,1 &
\end{matrix}\right.$
$⇒\%m_{CuO}=\dfrac{0,05.80}{20}.100\%=20\%$
$\%m_{Fe_2O_3}=100\%-20\%=80\%$