Em tham khảo nha :
\(\begin{array}{l}
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
CaO + {H_2}O \to Ca{(OH)_2}\\
{n_{{H_2}}} = \dfrac{{0,224}}{{22,4}} = 0,01mol\\
{n_{Ca}} = {n_{{H_2}}} = 0,01mol\\
{m_{Ca}} = 0,01 \times 40 = 0,4g\\
\% Ca = \dfrac{{0,4}}{1} \times 100\% = 40\% \\
{m_{CaO}} = 1 - 0,4 = 0,6g\\
{n_{CaO}} = \dfrac{{0,6}}{{56}} = 0,0107mol\\
{n_{{H_2}O}} = 2{n_{Ca}} + {n_{CaO}} = 0,0307mol\\
{m_{{H_2}O}} = 0,0307 \times 18 = 0,5526g
\end{array}\)