$n_{H_2}=13,44/22,4=0,6mol$
$PTHH :$
$2Al+6HCl\to 2AlCl_3+3H_2$
$Mg+2HCl\to MgCl_2+H_2$
Gọi $n_{Al}=a;n_{Mg}=b$
$\text{Ta có :}$
$m_{hh}=27a+24b=12,6g$
$n_{H_2}=1,5a+b=0,6mol$
$\text{Ta có hpt :}$
$\left\{\begin{matrix}
27a+24b=12,6 & \\
1,5a+b=0,6 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,2 & \\
b=0,3 &
\end{matrix}\right.$
$⇒\%m_{Al}=\dfrac{0,2.27.100\%}{12,6}=42,86\%$
$\%m_{Mg}=100\%-42,86\%=57,14\%$