$\begin{array}{l} \% {m_{Mg}} = 37,2\% \\ \% {m_{Al}} = 62,8\% \end{array}$
Ta có PTHH sau
$Mg+2HCl->MgCl_2+H_2$
$2Al+6HCl->2AlCl_3+3H_2$
Gọi $a$ là số mol $Mg$, $b$ là số mol $Al$
$\left\{ \begin{array}{l} 24a + 27b = 3,87\\ a + 1,5b = \dfrac{{4,368}}{{22,4}} = 0,195 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = 0,06\\ b = 0,09 \end{array} \right.$
$\%m_{Mg}=0,06.\dfrac{24}{3,87}.100\%=37,2%$
$\%m_{Al}=100\%-37,2\%=62,8\%$
