Đáp án:
\(\begin{array}{l}
b)\\
{m_{Cu}} = 9,1g\\
\% Al = 37,24\% \\
\% Cu = 62,76\% \\
c)\\
{C_{{M_{HCl}}}} = 4M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
{m_{Al}} = n \times M = 0,2 \times 27 = 5,4g\\
{m_{Cu}} = 14,5 - 5,4 = 9,1g\\
\% Al = \dfrac{{5,4}}{{14,5}} \times 100\% = 37,24\% \\
\% Cu = 100 - 37,24 = 62,76\% \\
c)\\
{n_{HCl}} = 3{n_{Al}} = 3 \times 0,2 = 0,6mol\\
{C_{{M_{HCl}}}} = \dfrac{n}{V} = \dfrac{{0,6}}{{0,15}} = 4M
\end{array}\)