$\text{Dẫn CO qua hỗn hợp}$
$\text{thì MgO không bị khử}$
$3CO + Fe_2O_3\hspace{0,2cm} \underrightarrow{t^{o}}\hspace{0,2cm} 2Fe + 3CO_2$
$n_{Fe_2O_3}=a; n_{MgO}=b$
$\Rightarrow 160a+40b=12 \hspace{0,5cm}(1)$
$\Rightarrow 2a.56+40b=9,12 (2)$
$\text{Từ (1), (2)}\Rightarrow a=b=0,06$
$Fe + 2HCl \rightarrow FeCl_2 + H_2 \uparrow$
$n_{H_2}=n_{Fe}=2n_{Fe_2O_3}=0,12 (mol)$
$\Rightarrow V_{H_2}=0,12.22,4=2,688 (l)$