\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Mg}=y\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x x ( mol )
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}56x+24y=0,83\\x+y=0,2\end{matrix}\right.\)
Nghiệm âm, bạn xem lại đề nhé!