Đáp án:

$a\big)↓↓↓$ 

$b\big)m_{Ca(OH)_2}≈16,12(g)$

Giải thích các bước giải:

$a\big)$

$Ca+2H_2O\to Ca(OH)_2+H_2↑$

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

$b\big)$

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

Theo PT: $n_{Ca}=n_{H_2}=0,1(mol)$

$\to m_{Ca}=0,1.40=4(g)$

$\to m_{CaO}=10,6-4=6,6(g)$

$\to n_{CaO}=\frac{6,6}{56}=\frac{33}{280}(mol)$

Theo PT: $\sum n_{Ca(OH)_2}=n_{Ca}+n_{CaO}=0,1+\frac{33}{280}=\frac{61}{280}(mol)$

$\to m_{Ca(OH)_2}=\frac{61}{280}.74≈16,12(g)$

Đáp án:

 16,12g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
CaO + {H_2}O \to Ca{(OH)_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Ca}} = {n_{{H_2}}} = 0,1\,mol\\
{m_{CaO}} = 10,6 - 0,1 \times 40 = 6,6g\\
{n_{CaO}} = \dfrac{{6,6}}{{56}} = \dfrac{{33}}{{280}}\,mol\\
{m_{Ca{{(OH)}_2}}} = (\dfrac{{33}}{{280}} + 0,1) \times 74 = 16,12g
\end{array}\)