a) Fe+2HCl--->FeCl2+H2
x---------------------------x(mol)
Mg+2HCl--->MgCl2+H2
y----------------------------y(mol)
n H2=4,48/22,4=0,2(mol)
Theo bài ra ta co hpt
\(\left\{{}\begin{matrix}56x+24y=8\\x+y=0.2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
%m Fe=0,1.56/8.100%=70%
%m Mg=100-70=30%