Đáp án:
\(\begin{array}{l}
a)\\
\% Al = 17,2\% \\
\% Zn = 82,8\% \\
b)\\
{V_{HCl}} = 154ml
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{0,784}}{{22,4}} = 0,035mol\\
hh:Al(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
27a + 65b = 1,57\\
\dfrac{3}{2}a + b = 0,035
\end{array} \right.\\
\Rightarrow a = 0,01;b = 0,02\\
{m_{Al}} = 0,01 \times 27 = 0,27g\\
\% Al = \dfrac{{0,27}}{{1,57}} \times 100\% = 17,2\% \\
\% Zn = 100 - 17,2 = 82,8\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,07mol\\
{n_{HC{l_{cd}}}} = \dfrac{{0,07 \times 110}}{{100}} = 0,077mol\\
{V_{HCl}} = \dfrac{{0,077}}{{0,5}} = 0,154l = 154ml
\end{array}\)
