Đáp án:

\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 2,24l\\
b)\\
{C_{{M_{Ba{{(OH)}_2}}}}} = 0,75M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
2Ba + {O_2} \to 2BaO\\
Ba + 2{H_2}O \to Ba{(OH)_2} + {H_2}\\
BaO + {H_2}O \to Ba{(OH)_2}\\
{n_{Ba}} = \dfrac{{41,1}}{{137}} = 0,3mol\\
hh\,X:Ba(a\,mol);BaO(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
137a + 153b = 44,3
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,2\\
{n_{{H_2}}} = {n_{Ba}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
b)\\
{n_{Ba{{(OH)}_2}}} = {n_{Ba}} + {n_{BaO}} = 0,1 + 0,2 = 0,3mol\\
{V_{{H_2}O}} = \dfrac{{{m_{{H_2}O}}}}{1} = \dfrac{{400}}{1} = 400ml\\
{C_{{M_{Ba{{(OH)}_2}}}}} = \dfrac{{0,3}}{{0,4}} = 0,75M
\end{array}\)