Gọi 4a là mol Mg, 9a là mol Al.
$Mg+ 2HCl \to MgCl_2+ H_2$
$2Al+ 6HCl \to 2AlCl_3+ 3H_2$
$n_{H_2}= 0,04 mol$
$\Rightarrow 4a+13,5=0,04$
$\Leftrightarrow a= \frac{2}{875}$
$n_{Mg}= 4a= \frac{8}{875} mol$
$n_{Al}= 9a= \frac{18}{875} mol$
$\%m_{Mg}= \frac{\frac{8}{875}.24.100}{\frac{8}{875}.24+\frac{18}{875}.27}= 28,3\%$
$\%m_{Al}= 71,7\%$