PTHH:
$\rm 2Al + 3Fe_3O_4 \xrightarrow{t^o} Al_2O_3 + 9FeO(1)$
$\rm 2Al + 3FeO \xrightarrow{t^o} 2Al_2O_3 + 3Fe(2)$
$\rm 2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2(3)$
$\rm Fe + 2HCl \rightarrow FeCl_2 + H_2(4)$
$\rm FeO + 2HCl \rightarrow FeCl_2 + H_2O (5)$
$\rm Fe_3O_4 + 8HCl \rightarrow FeCl_2 + 2FeCl_3 + 4H_2O(6)$
$\rm Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O(7)$
$\rm \xrightarrow{\text{BTNT O}} n_{H_2O} = 4n_{Fe_3O_4} = 0,16 (mol)$
$\rm \xrightarrow{\text{BTNT H}} n_{HCl} = 2n_{H_2} + 2n_{H_2O} = 0,62 (mol)$
Ta có:
$\rm n_{(-Cl)} = n_{HCl} = 0,62 (mol)$
$\rm n_{Fe} = 3n_{Fe_3O_4} = 0,12 (mol)$
$\rm \Rightarrow m = m_{\text{muối khan}} = m_{Al} + m_{Fe} + m_{(-Cl)} = 0,12.27 + 0,12.56 + 0,62.35,5 = 31,97 (g)$