$n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1(mol)$
$n_{HCl}=1(mol)$
$Fe_3O_4+8HCl\to 2FeCl_3+FeCl_2+4H_2O$
$\Rightarrow n_{Fe^{2+}}=0,1(mol)$
Sau phản ứng, dd có $1$ mol $Cl^-$
$\mathop{Fe}\limits^{+2}\to \mathop{Fe}\limits^{+3}+e$
$\mathop{Cl}\limits^{-1}\to \mathop{Cl}\limits^0+e$
$\mathop{Mn}\limits^{+7}+5e\to\mathop{Mn}\limits^{+2}$
Bảo toàn e:
$n_{Cl^-}+n_{Fe^{2+}}=5n_{KMnO_4}$
$\Rightarrow n_{KMnO_4}=0,22(mol)$
$\to V=0,44l$