\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(x\) \(2x\) \(x\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(y\) \(2y\) \(y\)
\(n_{H_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
\(\Rightarrow x+y=0,4\)
Có \(24x+56y=12\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0,4\\24x+56y=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,325\\y=0,075\end{matrix}\right.\)
\(\Rightarrow m_{Mg}=0,325.24=7,8\left(g\right)\Rightarrow\%m_{Mg}=65\%\)
\(\Rightarrow m_{Fe}=56.0,075=4,2\left(g\right)\Rightarrow\%m_{Fe}=35\%\)
Có \(n_{HCl}=2x=2.0,325=0,65\left(mol\right)\)
\(n_{HCl}=2y=2.0,075=0,15\left(mol\right)\)
Có 0,15<0,65=> chọn 0,15(mol)
\(\Rightarrow V_{HCl}=\frac{0,15}{2}=0,075\left(l\right)\)
