Sục 13,44 lít CO2 (đktc) vào 200 ml dung dịch X gồm Ba(OH)2 1,5M và NaOH 1M.

* Hướng dẫn giải

$\begin{array}{l}+n_{Ba{\left(OH\right)}_2}=\mathrm{0,3};n_{NaOH}=\mathrm{0,2};n_{BaC{l}_2}=\mathrm{0,24};n_{KOH}=\mathrm{0,3.}\\ +\underset{\mathrm{0,6}}{\underbrace{C{O}_2}}+\underset{ddX}{\underbrace{\left\{\begin{array}{l}Ba{\left(OH\right)}_2:\mathrm{0,3}\\ NaOH:\mathrm{0,2}\end{array}\right\}}}\Rightarrow 1<\frac{n_{O{H}^{-}}}{n_{C{O}_2}}=\frac{\mathrm{0,8}}{\mathrm{0,6}}<2\stackrel{}{\to }\left\{\begin{array}{l}C{O_3}^{2-}\\ HC{O_3}^{-}\end{array}\right\}\\ \Rightarrow n_{C{O_3}^{2-}}=n_{O{H}^{-}}-n_{C{O}_2}=\mathrm{0,2}<n_{B{a}^{2+}}\Rightarrow n_{BaC{O}_3}=\mathrm{0,2.}\\ \Rightarrow \underset{ddY}{\underbrace{\left\{\begin{array}{l}N{a}^{+}:\mathrm{0,2}\\ B{a}^{2+}:\mathrm{0,1}\\ HC{O_3}^{-}:\mathrm{0,4}\end{array}\right\}}}+\left\{\begin{array}{l}BaC{l}_2:\mathrm{0,24}\\ KOH:\mathrm{0,3}\end{array}\right\}\stackrel{}{\to }\left\{\begin{array}{l}{n}_{C{O_3}^{2-}}=\mathrm{0,3}<n_{B{a}^{2+}}=\mathrm{0,34}\\ n_{BaC{O}_3}=\mathrm{0,3}\iff \boxed{\mathrm{59,1}gam}\end{array}\right.\end{array}$