Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Al}} = 60\% \\
\% {m_{Cu}} = 40\% \\
b)\\
{C_\% }AlC{l_3} = 17,25\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Al}} = \dfrac{{0,15 \times 2}}{3} = 0,1\,mol\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
\% {m_{Al}} = \dfrac{{2,7}}{{4,5}} \times 100\% = 60\% \\
\% {m_{Cu}} = 100 - 60 = 40\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,3\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,3 \times 36,5}}{{14,6\% }} = 75g\\
{m_{{\rm{dd}}spu}} = 2,7 + 75 - 0,15 \times 2 = 77,4g\\
{C_\% }AlC{l_3} = \dfrac{{0,1 \times 133,5}}{{77,4}} \times 100\% = 17,25\%
\end{array}\)