$\(\begin{array}{l}2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\ Mg + 2HCl \to MgC{l_2} + {H_2}\\a)\\n{H_2} = \frac{{8,96}}{{22,4}} = 0,4mol\\hh:Al(a\,mol),Mg(bmol)\\27a + 24b = 7,8\\1,5a + b = 0,4\\⇒a = 0,2\,b = 0,1\\\% mAl = 69,23\% \\ \% mFe = 30,77\% \\b)\\m{\rm{dd}}HCl = 292g\\m{\rm{dd}}spu = 7,8 + 292 -0,4 \times 2 = 299g\\C\% AlC{l_3} = 8,93\% \\C\% MgC{l_2} = 3,18\% \\c)\\2Al + 3C{l_2} \to 2AlC{l_3}\\Mg + C{l_2} \to MgC{l_2}\\nAl = \frac{{3,06 \times 69,23\% }}{{27}} = 0,07846mol\\nMg = 0,039\\m = 14,18g\end{array}\)$