Ta có: \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Na2O + CO2 ---to---> Na2CO3
Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\)
Vậy CO2 dư.
Theo PT: \(n_{Na_2CO_3}=n_{Na_2O}=0,1\left(mol\right)\)
=> \(m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)