Đáp án:
a) 0,1375l
b) 23,3g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = 0,1 \times 1 = 0,1\,mol\\
{n_{KOH}} = 0,1 \times 1,5 = 0,15\,mol\\
{n_{NaOH}} = 0,1 \times 2 = 0,2\,mol\\
{n_{{H_2}S{O_4}}} = 0,1 + \dfrac{{0,15}}{2} + \dfrac{{0,2}}{2} = 0,275\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,275}}{2} = 0,1375l\\
b)\\
{n_{BaS{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,1\,mol\\
{m_{BaS{O_3}}} = 0,1 \times 233 = 23,3g
\end{array}\)