Đáp án:
\(\begin{array}{l}
{m_{Al}} = 5,4g\\
{m_{CuO}} = 6,6g\\
\% Al = 45\% \\
\% CuO = 55\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
{m_{CuO}} = 12 - 5,4 = 6,6g\\
\% Al = \dfrac{{5,4}}{{12}} \times 100\% = 45\% \\
\% CuO = 100 - 45 = 55\%
\end{array}\)