Em tham khảo nha :
\(\begin{array}{l}
a)\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{Fe}} = \dfrac{3}{4}{n_{{H_2}}} = 0,225mol\\
{m_{Fe}} = 0,225 \times 56 = 12,6g\\
b)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = {n_{Fe}} = 0,225mol\\
{V_{{H_2}}} = 0,225 \times 22,4 = 5,04l
\end{array}\)