Đáp án:
\({V_{Ba{{(OH)}_2}}} = 250ml\)
Giải thích các bước giải:
\(\begin{array}{l}
2C{O_2} + Ba{(OH)_2} \to Ba{(HC{O_3})_2}(1)\\
C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O(2)\\
{n_{Ba{{(HC{O_3})}_2}}} = \dfrac{{19,425}}{{259}} = 0,075mol\\
{n_{C{O_2}(1)}} = 2{n_{Ba{{(HC{O_3})}_2}}} = 0,15mol\\
\Rightarrow {n_{C{O_2}(2)}} = {n_{C{O_2}}} - {n_{C{O_2}(1)}} = 0,3 - 0,15 = 0,15mol\\
\Rightarrow {n_{Ba{{(OH)}_2}}} = {n_{Ba{{(HC{O_3})}_2}}} + {n_{C{O_2}(2)}} = 0,075 + 0,15 = 0,225mol\\
{V_{Ba{{(OH)}_2}}} = \dfrac{n}{{{C_M}}} = \dfrac{{0,225}}{{0,9}} = 0,25l = 250ml
\end{array}\)
