Em tham khảo nha:

\(\begin{array}{l}
a)\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
CaO + {H_2}O \to Ca{(OH)_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Ca}} = {n_{{H_2}}} = 0,1\,mol\\
{m_{Ca}} = 0,1 \times 40 = 4g\\
{m_{CaO}} = 9,6 - 4 = 5,6g\\
\% {m_{Ca}} = \dfrac{4}{{9,6}} \times 100\%  = 41,67\% \\
\% {m_{CaO}} = 100 - 41,67 = 58,33\% \\
b)\\
{n_{CaO}} = \dfrac{{5,6}}{{56}} = 0,1\,mol\\
{n_{Ca{{(OH)}_2}}} = 0,1 + 0,1 = 0,2\,mol\\
{m_{Ca{{(OH)}_2}}} = 0,2 \times 74 = 14,8g
\end{array}\)

Đáp án + Giải thích các bước giải:

$a\big)$

$Ca+2H_2O\to Ca(OH)_2+H_2$

$CaO+H_2O\to Ca(OH)_2$

$b\big)$

Theo PT: $n_{Ca}=n_{H_2}=\frac{2,24}{22,4}=0,1(mol)$

$\%m_{Ca}=\frac{0,1.40}{9,6}.100\%=41,67\%$

$\%m_{CaO}=100-41,67=58,33\%$

$c\big)$

Theo PT: $n_{CaO}=\frac{9,6-4}{5,6}=0,1(mol)$

`\sum n_{Ca(OH)_2}=n_{Ca}+n_{CaO}=0,2(mol)`

$\to m_{Ca(OH)_2}=0,2.74=14,8(g)$