$CuO+H_2\overset{t^o}\to Cu+H_2O$
$Fe_2O_3+3H_2\to 2Fe+3H_2O$
Gọi $n_{CuO}=a;n_{Fe_2O_3}=b$
Ta có :
$m_{hh}=80a+160b=16$
$m_{hhkl}=64a+112b=11,2$
Ta có hpt :
$\left\{\begin{matrix}
80a+160b=16 & \\
64a+112b=11,2 &
\end{matrix}\right.$
$⇒\left\{\begin{matrix}
a=0 (??)& \\
b=0,1 &
\end{matrix}\right.$
b/
$n_{H_2bđ}=0,1.3.160\%=0,48mol$
$⇒V_{H_2bđ}=0,48.22,4=10,752l$