Em tham khảo nha:

\(\begin{array}{l}
a)\\
Mg + C{l_2} \to MgC{l_2}\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
Zn + C{l_2} \to ZnC{l_2}\\
{n_{C{l_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
BTKL:\\
m + {m_{C{l_2}}} = {m_m} \Rightarrow m = 12,5 - 0,1 \times 71 = 5,4g\\
b)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,128}}{{22,4}} = 0,095\,mol\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,19\,mol\\
BTKL:\\
{m_X} + {m_{HCl}} = {m_2} + {m_{{H_2}}}\\
 \Rightarrow {m_2} = 5,4 + 0,19 \times 36,5 - 0,095 \times 2 = 12,145g\\
hh:Mg(a\,mol),Fe(a\,mol),Al(c\,mol),Zn(d\,mol)\\
T{N_1}:\\
{n_{C{l_2}}} = 0,1\,mol \Rightarrow a + 1,5b + 1,5c + d = 0,1(1)\\
T{N_2}:\\
{n_{{H_2}}} = 0,095 \Rightarrow a + b + 1,5c + d = 0,095(2)\\
(1) - (2) \Rightarrow 0,5b = 0,1 - 0,095 \Rightarrow b = 0,01\\
\% {m_{Fe}} = \dfrac{{0,01 \times 56}}{{5,4}} \times 100\%  = 10,37\% 
\end{array}\)