Gọi $x$, $y$ là số mol $FeO$, $ZnO$ ban đầu.
$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
$FeO+H_2\buildrel{{t^o}}\over\to Fe+H_2O$
$ZnO+H_2\buildrel{{t^o}}\over\to Zn+H_2O$
$\Rightarrow x+y=0,3$
$n_{Fe}=x (mol); n_{Zn}= y (mol)$
$Fe+H_2SO_4\to FeSO_4+H_2$
$Zn+H_2SO_4\to ZnSO_4+H_2$
$\Rightarrow n_{H_2}=x+y=0,3(mol)$
$\Rightarrow V_{H_2}=0,3.22,4=6,72l$