Em tham khảo nha :
\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = \dfrac{{13}}{{65}} = 0,2mol\\
{m_{HCl}} = \dfrac{{200 \times 28,5}}{{100}} = 57g\\
{n_{HCl}} = \dfrac{{57}}{{36,5}} = 1,56mol\\
\dfrac{{0,2}}{1} < \dfrac{{1,56}}{2} \Rightarrow HCl\text{ dư}\\
{n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
{n_{HC{l_{pu}}}} = 2{n_{Zn}} = 0,4mol\\
{m_{HC{l_{pu}}}} = 0,4 \times 36,5 = 14,6g\\
{m_{HC{l_d}}} = 57 - 14,6 = 42,4g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,2mol\\
{m_{ZnC{l_2}}} = 0,2 \times 136 = 27,2g\\
C{\% _{HC{l_d}}} = \dfrac{{42,4}}{{200 + 13 - 0,2 \times 2}} \times 100\% = 19,95\% \\
C{\% _{ZnC{l_2}}} = \dfrac{{27,2}}{{200 + 13 - 0,2 \times 2}} \times 100\% = 12,8\%
\end{array}\)
