\(Đặt:a=n_{Mg};b=n_{Al}\left(mol\right)\\ n_{SO_42^-}=0,38.0,5=0,19\left(mol\right)\\ n_{Cl^-}=0,5.1=0,5\left(mol\right)\\ n_{H_2}=\dfrac{8,736}{22,4}=0,39\left(mol\right)\\ n_{H^+\left(trong.axit\right)}=0,19.2+0,5=0,78\left(mol\right)\\ Vì:n_{H_2}=\dfrac{n_{H^+}}{2}\\ \Rightarrow Chỉ.tác.dụng.với.HCl\\ \Rightarrow\left\{{}\begin{matrix}24a+27b=7,74\\a+1,5b=0,39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,12\\b=0,18\end{matrix}\right.\)
=> Kim loại đã tan hết.
\(Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{muối}=m_{MgCl_2}+m_{AlCl_3}=0,12.95+0,18.133,5=35,43\left(g\right)\)
