Giải thích các bước giải:
\(\begin{array}{l}
C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O(1)\\
{n_{C{O_2}}} = 0,1mol\\
{n_{BaC{O_3}}} = 0,1mol\\
\to {n_{Ba{{(OH)}_2}}} = {n_{C{O_2}}} = 0,1mol\\
\to x = C{M_{Ba{{(OH)}_2}}} = \dfrac{{0,1}}{{0,1}} = 1M\\
C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O(2)\\
2C{O_2} + Ba{(OH)_2} \to Ba{(HC{O_3})_2}(3)\\
{n_{C{O_2}}} = 0,15mol\\
{n_{Ba{{(HC{O_3})}_2}}} = 0,038mol\\
\to {n_{C{O_2}(3)}} = 2{n_{Ba{{(HC{O_3})}_2}}} = 0,076mol\\
\to {n_{C{O_2}(2)}} = 0,074mol\\
\to {n_{Ba{{(OH)}_2}}} = {n_{C{O_2}(2)}} + \dfrac{1}{2}{n_{C{O_2}(3)}} = 0,112mol\\
\to x = C{M_{Ba{{(OH)}_2}}} = \dfrac{{0,112}}{{0,1}} = 1,12M\\
\to C{M_{Ba{{(OH)}_2}}} = 1 + 1,12 = 2,12M
\end{array}\)