Mg + 2HCl → MgCl2 + H2 (1)
2Al + 6HCl → 2AlCl3 + 3H2 (2)
\(n_{H_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
a) Gọi x,y lần lượt là số mol của Mg và Al
Ta có: \(\left\{{}\begin{matrix}24x+27y=7,8\\x+1,5y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
Vậy \(n_{Mg}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(n_{Al}=0,2\left(mol\right)\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)
\(\%m_{Mg}=\frac{2,4}{7,8}\times100\%=30,77\%\)
\(\%m_{Al}=\frac{5,4}{7,8}\times100\%=69,23\%\)
b) Theo pT1: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\Rightarrow m_{MgCl_2}=0,1\times95=9,5\left(g\right)\)
Theo pt2: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\Rightarrow m_{AlCl_3}=0,2\times133,5=26,7\left(g\right)\)
Vậy \(m_{muối}=9,5+26,7=36,2\left(g\right)\)