Ta có: \(n_{CO_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\\ n_{NaOH}=\dfrac{6,4}{40}=0,16\left(mol\right)\)
Ta có: \(T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,16}{0,07}\approx2,286\)
T >2 => Tạo muối NaHCO3.
PTHH: NaOH + CO2 -> NaHCO3
- nNaHCO3= nCO2= 0,07(mol)
=> m(muối)= mNaHCO3= 0,7.84= 58,8(g)