Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Al}} = 22,88\% \\
\% {m_{Zn}} = 77,12\% \\
b)\\
{C_\% }AlC{l_3} = 14,63\% \\
{C_\% }ZnC{l_2} = 20,87\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{16,24}}{{22,4}} = 0,725\,mol\\
hh:Al(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
27a + 65b = 29,5\\
1,5a + b = 0,725
\end{array} \right.\\
\Rightarrow a = 0,25;b = 0,35\\
\% {m_{Al}} = \dfrac{{0,25 \times 27}}{{29,5}} \times 100\% = 22,88\% \\
\% {m_{Zn}} = 100 - 22,88 = 77,12\% \\
b)\\
{m_{{\rm{dd}}spu}} = 29,5 + 200 - 0,725 \times 2 = 228,05g\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,25\,mol\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,35\,mol\\
{C_\% }AlC{l_3} = \dfrac{{0,25 \times 133,5}}{{228,05}} \times 100\% = 14,63\% \\
{C_\% }ZnC{l_2} = \dfrac{{0,35 \times 136}}{{228,05}} \times 100\% = 20,87\%
\end{array}\)
