Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
\% Al = \dfrac{{5,4}}{{15}} \times 100\% = 36\% \\
\% Cu = 100 - 36 = 64\% \\
b)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,2mol\\
{m_{AlC{l_3}}} = 0,2 \times 133,5 = 26,7g\\
c)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6mol\\
{m_{HCl}} = 0,6 \times 36,5 = 21,9g\\
C{\% _{HCl}} = \dfrac{{21,9}}{{219}} \times 100\% = 10\% \\
d)\\
C{\% _{AlC{l_3}}} = \dfrac{{26,7}}{{5,4 + 219 - 0,3 \times 2}} \times 100\% = 11,9\%
\end{array}\)