\(Mg+2HCl-->MgCl2+H2\)
x-----------------------------------------x(mol)
\(2Al+6HCl-->2AlCl3+3H2\)
y-------------------------------------------1,5y(mol)
\(n_{H2}=\frac{7,728}{22,4}=0,345\left(mol\right)\)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}24x+27y=6,93\\x+1,5y=0,345\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,12\\y=0,15\end{matrix}\right.\)
\(\%m_{Mg}=\frac{0,12.24}{6,93}.100\%=41,56\%\)
\(\%m_{Al}=100-41,56=58,44\%\)