\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b.Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}24x+56y=13,6\\95x+127y=34,9\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{13,6}.100=17,65\%\%\\\%m_{Fe}=82,35\%\end{matrix}\right.\)