Em tham khảo nha :
\(\begin{array}{l}
M + 2{H_2}O \to M{(OH)_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{n_M} = {n_{{H_2}}} = 0,25mol\\
{M_M} = \dfrac{{10}}{{0,25}} = 40dvC\\
M:Canxi(Ca)\\
{n_{Ca{{(OH)}_2}}} = {n_{Ca}} = 0,25mol\\
{m_{Ca{{(OH)}_2}}} = 0,25 \times 74 = 18,5g\\
C{\% _{Ca{{(OH)}_2}}} = \dfrac{{18,5}}{{10 + 200 - 0,25 \times 2}} \times 100\% = 8,83\%
\end{array}\)