Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Al}} = 29,03\% \\
\% {m_{Cu}} = 70,97\% \\
b)\\
{C_M}AlC{l_3} = {C_M}HCl = 0,5M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Al}} = \dfrac{{0,15 \times 2}}{3} = 0,1\,mol\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
\% {m_{Al}} = \dfrac{{2,7}}{{9,3}} \times 100\% = 29,03\% \\
\% {m_{Cu}} = 100 - 29,03 = 70,97\% \\
b)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1\,mol\\
{n_{HCl}} = 0,2 \times 2 = 0,4\,mol\\
{n_{HCl}} = 0,4 - 0,15 \times 2 = 0,1\,mol\\
{C_M}AlC{l_3} = {C_M}HCl = \dfrac{{0,1}}{{0,2}} = 0,5M
\end{array}\)