\(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ n_{H_2\left(đktc\right)}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+56b=5,2\\a+b=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{5,2}.100\approx46,15\%\\\%m_{Fe}\approx53,85\%\end{matrix}\right.\)
=> Chọn A