a,
$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$\Rightarrow n_{Fe}=n_{H_2}=0,1(mol)$
$\%m_{Fe}=\dfrac{0,1.56.100}{28,8}=19,4\%$
$\%m_{Fe_3O_4}=80,6\%$
b,
$n_{Fe_3O_4}=\dfrac{28,8-0,1.56}{232}=0,1(mol)$
$Fe_3O_4+8HCl\to 2FeCl_3+FeCl_2+4H_2O$
Theo PTHH, $n_{FeCl_2}=n_{Fe}+n_{FeCl_2}=0,2(mol)$
Bảo toàn e: $n_{FeCl_2}=5n_{KMnO_4}$
$\Rightarrow n_{KMnO_4}=0,04(mol)$
$\to V=0,04l=40ml$