Đáp án:
\(\begin{array}{l}
a)\\
{m_{Zn}} = 16,25g\\
{V_{{H_2}}} = 5,6l\\
b)\\
C{\% _{HCl(dư)}} = 1,16\% \\
C{\% _{CaC{l_2}}} = 4,52\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{m_{HCl}} = \dfrac{{182,5 \times 10\% }}{{100\% }} = 18,25g\\
\to {n_{HCl}} = 0,5mol\\
a)\\
{n_{Zn}} = {n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,25mol\\
\to {m_{Zn}} = 16,25g\\
\to {V_{{H_2}}} = 5,6l\\
b)\\
Ca{(OH)_2} + 2HCl \to CaC{l_2} + 2{H_2}O\\
{n_{HCl}} = 0,25mol\\
{m_{Ca{{(OH)}_2}}} = \dfrac{{130 \times 5\% }}{{100\% }} = 6,5g\\
\to {n_{Ca{{(OH)}_2}}} = 0,09mol\\
\to \dfrac{{{n_{Ca{{(OH)}_2}}}}}{1} < \dfrac{{{n_{HCl}}}}{2} \to {n_{HCl}}dư\\
\to {n_{HCl(pt)}} = 2{n_{Ca{{(OH)}_2}}} = 0,18mol\\
\to {n_{HCl(dư)}} = 0,07mol\\
\to {m_{HCl(dư)}} = 2,56g\\
{n_{CaC{l_2}}} = {n_{Ca{{(OH)}_2}}} = 0,09mol\\
\to {m_{CaC{l_2}}} = 9,99g\\
{m_{{\rm{dd}}}} = {m_{{\rm{ddHCl}}}} + {m_{{\rm{dd}}Ca{{(OH)}_2}}} = 221,25g\\
\to C{\% _{HCl(dư)}} = \dfrac{{2,56}}{{221,25}} \times 100\% = 1,16\% \\
\to C{\% _{CaC{l_2}}} = \dfrac{{9,99}}{{221,25}} \times 100\% = 4,52\%
\end{array}\)
