\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ Đặt:n_{Mg}=s\left(mol\right);n_{Fe}=i\left(mol\right)\left(s,i>0\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}24s+56i=12,8\\s+i=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}s=0,3\\i=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,3.24}{12,8}.100=56,25\%\\ \Rightarrow\%m_{Fe}=100\%-56,25\%=43,75\%\)
Mg+2HCl->FeCl2+H2
x------------------------x mol
Fe+2HCl->MgCl2+H2
y-------------------------y mol
ta có :
\(\left\{{}\begin{matrix}24x+56y=12,8\\x+y=0,4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0,3\\y=0,1\end{matrix}\right.\)
=>%m Mg=\(\dfrac{0,3.24}{12,8}.100\)=56,25%
=>%m Fe=43,75%
