\(a)\)

\(2Al + 6HCl \to 2AlCl_3^{} + 3H_2^{}\)

\(Mg + 2HCl \to MgCl_2^{} + H_2^{}\)

\(Zn + 2HCl \to ZnCl_2^{} + H_2^{}\)

\(2Al + 6HCl \to 2AlCl_3^{} + 3H_2^{}\)

\(Mg + 2HCl \to MgCl_2^{} + H_2^{}\)

\(Zn + 2HCl \to ZnCl_2^{} + H_2^{}\)

\(b)\)

Gọi số mol lần lượt của kim loại là x,y,z

\(P1:\)

\(2Al + 6HCl \to 2AlCl_3^{} + 3H_2^{}\)

\(\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{2}x\)

\(Mg + 2HCl \to MgCl_2^{} + H_2^{}\)

\(\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\)

\(Zn + 2HCl \to ZnCl_2^{} + H_2^{}\)

\(\,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z\)

\(\frac{3}{2}x + y + z = 0,4\,mol\)

\(27x + 24y + 65z = 12,8(1)\)

\(\frac{3}{2}x + y + z = 0,4\,mol(2)\)

\(P2:\)

\(V_{O_2^{}(Phan\,\,Ung)}^{} = \frac{{44,8}}{5} = 8,96\,lit\)

\(n_{O_2^{}}^{} = \frac{{8,96}}{{22,4}} = 0,4\,mol\)

\(4Al + 3O_2^{} \to 2Al_2^{}O_3^{}\)

\(\,\,\,x\,\,\,\,\,\,\,\,\frac{3}{4}x\)

\(2Mg + O_2^{} \to 2MgO\)

\(\,\,\,y\,\,\,\,\,\,\,\,\frac{1}{2}y\)

\(2Zn + O_2^{} \to 2ZnO\)

\(\,\,\,z\,\,\,\,\,\,\,\,\frac{1}{2}z\)

\(k.(x + y + z) = 0,7(3)\)

\(k(\frac{3}{4}x + \frac{1}{2}y + \frac{1}{2}z) = 0,4(4)\)

\(Lay\,(3)\,chia\,(4):\)

\(\frac{{k.(x + y + z)}}{{k(\frac{3}{4}x + \frac{1}{2}y + \frac{1}{2}z)}} = \frac{{0,7}}{{0,4}}\)

\( \to \frac{{\frac{3}{4}x + \frac{1}{2}y + \frac{1}{2}z}} = \frac{7}{4}\)

\( \to (x + y + z).4 = (\frac{3}{4}x + \frac{1}{2}y + \frac{1}{2}z).7\)

\( \to  - 1,25x + 0,5y + 0,5z = 0(5)\)

\(Tu\,(1),(2),(5)\)

\( \to x = 0,1\,mol;y = 0,15\,mol;z = 0,1\,mol\)

\(\% m_{Al}^{} = \frac{{0,1.27}}{{12,8}}.100 = 21,09\% \)

\(\% m_{Mg}^{} = \frac{{0,15.24}}{{12,8}}.100 = 28,125\% \)

\(\% m_{Zn}^{} = \frac{{0,1.65}}{{12,8}}.100 = 50,78125\% \)