Sơ đồ phản ứng:
\(CO_2+\left\{{}\begin{matrix}KOH:0,4\left(mol\right)\\NaOH:0,3\left(mol\right)\\K_2CO_3:0,4\left(mol\right)\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}K^+:1,2\left(mol\right)\\Na^+:0,3\left(mol\right)\\CO_3^{2-}:a\left(mol\right)\\HCO_3^-:b\left(mol\right)\end{matrix}\right.+H_2O\)
\(Ba^{2+}+CO_3^{2-}\rightarrow BaCO_3\downarrow\)
Ta có: \(n_{BaCO_3}=\dfrac{39,4}{197}=0,2\left(mol\right)=n_{CO_3^{2-}}=a\)
Bảo toàn điện tích: \(n_{K^+}+n_{Na^+}=2n_{CO_3^{2-}}+n_{HCO_3^-}\) \(\Rightarrow n_{HCO_3^-}=b=1,1\left(mol\right)\)
Bảo toàn Cacbon: \(n_{CO_2}=a+b-n_{K_2CO_3}=0,9\left(mol\right)\) \(\Rightarrow V_{CO_2}=0,9\cdot22,4=20,16\left(l\right)\)
