Em tham khảo nha :
\(\begin{array}{l}
P1:\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
P2:\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
Cu + C{l_2} \to CuC{l_2}\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
hh:Fe(a\,mol),Cu(b\,mol),Al(c\,mol)\\
\left\{ \begin{array}{l}
56a + 64b + 27c = 70\\
a + \dfrac{3}{2}c = 0,6 \times 2\\
\dfrac{3}{2}a + b + \dfrac{3}{2}c = 0,95 \times 2
\end{array} \right.\\
\Rightarrow a = 0,6;b = 0,4;c = 0,4\\
{m_{Fe}} = 0,6 \times 56 = 33,6g\\
{m_{Cu}} = 0,4 \times 64 = 25,6g\\
\% Fe = \dfrac{{33,6}}{{70}} \times 100\% = 48\% \\
\% Cu = \dfrac{{25,6}}{{70}} \times 100\% = 36,6\% \\
\% Al = 100 - 48 - 36,6 = 15,4\%
\end{array}\)