Đáp án:
\(\begin{array}{l}
a)\\
{m_{CaC{l_2}}} = 11,1g\\
b)\\
{m_{HCl}} = 7,3g\\
c)\\
{C_\% }CaC{l_2} = 5\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Ca{(OH)_2} + 2HCl \to CaC{l_2} + 2{H_2}O\\
{n_{Ca{{(OH)}_2}}} = \dfrac{{7,4}}{{74}} = 0,1\,mol\\
{n_{CaC{l_2}}} = {n_{Ca{{(OH)}_2}}} = 0,1\,mol\\
{m_{CaC{l_2}}} = 0,1 \times 111 = 11,1g\\
b)\\
{n_{HCl}} = 2{n_{Ca{{(OH)}_2}}} = 0,2\,mol\\
{m_{HCl}} = 0,2 \times 36,5 = 7,3g\\
c)\\
{m_{{\rm{dd}}spu}} = 7,4 + 214,6 = 222g\\
{C_\% }CaC{l_2} = \dfrac{{11,1}}{{222}} \times 100\% = 5\%
\end{array}\)