Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,1mol\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
{m_{A{l_2}{O_3}}} = 10 - 2,7 - 2,75 = 4,55g\\
\% Al = \dfrac{{2,7}}{{10}} \times 100\% = 27\% \\
\% Cu = \dfrac{{2,75}}{{10}} \times 100\% = 27,5\% \\
\% A{l_2}{O_3} = 100 - 27 - 27,5 = 45,5\%
\end{array}\)