Lời giải:

1,

 \[\begin{array}{l}2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\a......................................1,5a(mol)\end{array}\]

  \[\begin{array}{l}Mg + 2HCl \to MgC{l_2} + {H_2}\\b.......................................b(mol)\end{array}\]

2, Đặt a = n_Al, b = n_Mg

\[{n_} = 0,4(mol)\]Ta có:

\[\left\{ \begin{array}{l}27a + 24b = 7,8\\1,5a + b = 0,4\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 0,2\\b = 0,1\end{array} \right.\]→ n_Al = 0,2 mol → m_Al = 5,4 (g)

→ %Al = 69,23%, %Mg= 30, 76%

3, m_muối = \[{m_{AlC{l_3}}} + {m_{MgC{l_2}}}\] = 0,2.133,5 + 0,1. 95 = 36,2 (g).