Đáp án:
\(\begin{array}{l}
a){V_{{H_2}}} = 4,48l\\
b)C{\% _{Ca{{(OH)}_2}}} = 14,8\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
a)\\
{n_{Ca}} = 0,2mol\\
\to {n_{{H_2}}} = {n_{Ca}} = 0,2mol\\
\to {V_{{H_2}}} = 4,48l
\end{array}\)
\(\begin{array}{l}
b)\\
{n_{Ca{{(OH)}_2}}} = {n_{Ca}} = 0,2mol\\
\to {m_{Ca{{(OH)}_2}}} = 14,8g\\
{m_{{\rm{dd}}}} = {m_{Ca}} + {m_{{H_2}O}} - {m_{{H_2}}} = 100g\\
\to C{\% _{Ca{{(OH)}_2}}} = \dfrac{{14,8}}{{100}} \times 100\% = 14,8\%
\end{array}\)
