Đáp án:
\({C_\% }HCl = 40,98\% \)
Giải thích các bước giải:
\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}(1)\\
{H_2} + C{l_2} \to 2HCl(2)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{C{l_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{{H_2}}} < {n_{C{l_2}}} \Rightarrow \text{ $Cl_2$ dư } \\
C{l_2} + {H_2}O \to HCl + HClO(3)\\
{n_{C{l_2}}} \text{ dư }= 0,15 - 0,1 = 0,05\,mol\\
{n_{HCl(3)}} = {n_{HClO}} = {n_{C{l_2}}} \text{ dư }= 0,05\,mol\\
{n_{HCl(2)}} = 2{n_{{H_2}}} = 0,2\,mol\\
{n_{HCl}} = 0,2 + 0,05 = 0,25\,mol\\
{m_{{\rm{dd}}HCl}} = 0,25 \times 36,5 + 13,14 = 22,265g\\
{C_\% }HCl = \dfrac{{0,25 \times 36,5}}{{22,265}} \times 100\% = 40,98\%
\end{array}\)