$n_{Fe}=\dfrac{14}{56}=0,25(mol)$
$n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1(mol)$
Bảo toàn $O$: $n_{H_2O}=4n_{Fe_3O_4}=0,4(mol)$
Đặt $x$ là số mol $H_2$
Bảo toàn $H$:
$n_{H_2SO_4}=n_{H_2}+n_{H_2O}=x+0,4(mol)$
BTKL:
$14+23,2+98(x+0,4)=88,4+2x+0,4.18$
$\to x=0,2$
$\to V_{H_2}=0,2.22,4=4,48l$