2Al+3H2SO4--->Al2(SO4)3+3H2
x------1,5x
Mg+H2SO4----->MgSO4+H2
y-----y
n\(_{H2SO4}=\frac{39,2}{98}=0,4\left(mol\right)\)
Theo bài ra ta có pt
\(\left\{{}\begin{matrix}27x+24y=7,8\\1,5x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
%m\(_{Al}=\frac{0,2.27}{7,8}.100\%=69,23\%\)
%m\(_{Mg}=100-69,23=30,77\%\)