\(n_{H_2} = \dfrac{15,6-14}{2} = 0,8(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Mg + 2HCl \to MgCl_2 + H_2\)
Gọi \(n_{Al} = a \ mol;n_{Mg} = b\ mol\)
Ta có :
\(\left\{{}\begin{matrix}27a+24b=15,6\\1,5a+b=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,4\\b=0,2\end{matrix}\right.\)
Vậy :
\(\%m_{Al} = \dfrac{0,4.27}{15,6}.100\% = 69,23\%\\ \%m_{Mg} = 100\% - 69,23\% = 30,77\%\)