a)PTHH Fe +2HCl ---> FeCl\(_2\)+H\(_2\)(1)
Mg+ HCl ---> MgCl\(_2\)+H\(_2\)(2)
b) Ta có :\(n_{H_2}=\)\(\frac{11,2}{22,4}\)=0,5 mol
Gọi n\(_{Fe}=x\) => m\(_{Fe}=56x\)
\(n_{Mg}=y\) => m\(_{Mg}=24y\)
=> 56x +24y =21,6 (g)(*)
Mặt khác : theo pthh1
\(n_{H_2}=n_{Fe}=x\left(mol\right)\)
theo pyhh 2
\(n_{H_2}=n_{Mg}=y\left(mol\right)\)
Suy ra x +y=0,5 (**)
Từ (*)(**) ta có hệ pt
\(\left\{{}\begin{matrix}56x+24y=21,6\\x+y=0,5\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)
m\(_{Fe}=0,3.56=16,8\left(g\right)\)
m\(_{Mg}=21,6-16,8=4,8\left(g\right)\)
C) Theo pthh (1)(2)
tổng\(n_{HCl}=2Tổng\) \(n_{H_2}=1mol\)
m\(_{HCl}=36,5.1=36,5\left(g\right)\)
m\(_{ddHCl}=\frac{36,5.100}{7,3}=500\left(g\right)\)
