Câu 9 ạ e cảm ơn( dò kết quả thui)

Em tham khảo nha :

\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = \dfrac{{32,5}}{{65}} = 0,5mol\\
{n_{HCl}} = 2{n_{Zn}} = 1mol\\
{m_{HCl}} = 36,5 \times 1 = 36,5g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{36,5 \times 100}}{{18,25}} = 200g\\
{n_{{H_2}}} = {n_{Zn}} = 0,5mol\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,5mol\\
{m_{ZnC{l_2}}} = 0,5 \times 136 = 68g\\
C{\% _{ZnC{l_2}}} = \dfrac{{68}}{{32,5 + 200 - 0,5 \times 1}} \times 100\%  = 29,37\% 
\end{array}\)