$n_{HCl}= v_{dd}.C_M=0,2.0,5=0,1(mol)$
$Fe+2HCl->FeCl_2+H_2$
x 2x
$MgO+2HCl->MgCl_2+H_2O$
y 2y
Ta có hpt:
$\left \{ {{56x+40y=2,32} \atop {2x+2y=0,1}} \right.$
<=> $\left \{ {{x=0,02} \atop {y=0,03}} \right.$
Ta có $n_{Fe}=n_{H2}=0,02$
$=>V_{H2}=0,02.22,4=0,448l$