a)\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(Fe+2HCl--.FeCl2+H2\)
x-----------------------------------x(mol)
\(Mg+2HCl-->MgCl2+H2\)
y-------------------------------------y(mol)
theo bài ta có hpt
\(\left\{{}\begin{matrix}56x+24y=9,2\\x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
\(\%m_{Fe}=\frac{0,1.56}{9,2}.100\%=60,87\%\)
\(\%m_{Mg}=100-60,87=39,13\%\)
b) \(n_{HCl}=2n_{H2}=0,5\left(mol\right)\)
\(V_{HCl}=V_{dd}=\frac{0,5}{2}=0,25\left(l\right)\)
\(C_{M\left(FeCl2\right)}=\frac{0,1}{0,25}=0,4\left(M\right)\)
\(C_{M\left(Mgcl2\right)}=\frac{0,15}{0,25}=0,6\left(M\right)\)