a)nNa2O = 6.2/62=0.1mol
Na2O + H2O -> 2NaOH
(mol) 0.1 0.2
a) để tạo ra muối axit thì: \(\dfrac{nNaOH}{nSO2}\le1\)
\(\dfrac{0.2}{nSO2}\le1\)=>\(nSO2\ge\dfrac{0.2}{1}=0.2mol\)
\(VSO2\ge0.2\cdot22.4=4.48\left(l\right)\)
b) Để tạo muối trung hòa thì \(\dfrac{nNaOH}{nSO2}\ge2\)
=> \(nSO2\le\dfrac{0.2}{2}=0.1mol\)
\(VSO2\le0.1\cdot22.4=2.24\left(l\right)\)