Đáp án:
\(C{M_{Ba{{(OH)}_2}}} = \dfrac{{0,13}}{{0,5}} = 0,26M\)
\({V_{{H_2}S{O_4}}} = \dfrac{{42,47}}{{1,14}} = 37,25ml\)
Giải thích các bước giải:
\(\begin{array}{l}
BaO + {H_2}O \to Ba{(OH)_2}\\
{n_{BaO}} = 0,13mol\\
\to {n_{Ba{{(OH)}_2}}} = {n_{BaO}} = 0,13mol\\
\to C{M_{Ba{{(OH)}_2}}} = \dfrac{{0,13}}{{0,5}} = 0,26M\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,13mol\\
\to {m_{{H_2}S{O_4}}} = 12,74g\\
\to {m_{{H_2}S{O_4}}}dd = \dfrac{{12,74 \times 100}}{{30}} = 42,47g\\
\to {V_{{H_2}S{O_4}}} = \dfrac{{42,47}}{{1,14}} = 37,25ml
\end{array}\)
