\(n_{KOH}=0,1.0,1=0,01\left(mol\right)\)
\(n_{NaOH}=0,1.0,2=0,02\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=0,1.0,3=0,03\left(mol\right)\)
\(n_{BaCO_3}=\dfrac{3,94}{197}=0,02\left(mol\right)\)
=> Có tạo muối Ba(HCO3)2
Bảo toàn nguyên tố Ba: \(0,03.1=0,02.1+n_{Ba\left(HCO_3\right)_2}.1\)
=> \(n_{Ba\left(HCO_3\right)_2}=0,01\)
Ba(OH)2 + 2CO2 → Ba(HCO3)2
CO2 + Ba(OH)2 → BaCO3 + H2O
2KOH + CO2 → K2CO3 + H2O
2NaOH + CO2 → Na2CO3 + H2O
=> \(\Sigma n_{CO_2}=0,01.2+0,02+\dfrac{0,01}{2}+\dfrac{0,02}{2}=0,055\left(mol\right)\)
=> \(V_{CO_2}=0,055.22,4=1,232\)
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