S + O2 => SO2 (1)
SO2 + 2NaOH => Na2SO3 + H2O (2)
SO2 + NaOH => NaHSO3 (3)
nS = \(\frac{a}{32}\left(mol\right)\)
nNaOH = \(\frac{b}{40}\left(mol\right)\)
(1) => nSO2 = nS = \(\frac{a}{32}\left(mol\right)\)
+ Trường hợp 1 : X : NaOH dư , Na2SO3
=> \(\frac{a}{b}=< \frac{2}{5}\)=> xảy ra (2)
(2) => nNaOH phản ứng = 2.nSO2 = \(\frac{a}{16}\left(mol\right)\)
nNaOH dư = \(\frac{b}{40}-\frac{a}{16}\left(mol\right)\)
(2) => nNa2SO3 = 2.nSO2 = \(\frac{a}{16}\left(mol\right)\)
+ Trường hợp 2 : X : NaHSO3 , Na2SO3
=> \(\frac{2}{5}< \frac{a}{b}< \frac{4}{5}\)=> xảy ra (2),(3)
Gọi nNa2SO3 = x (mol) , nNaHSO3 = y (mol)
(2),(3) => nNaOH = 2x + y = \(\frac{b}{40}\) (mol) (I)
(2),(3) => nSO2 = x + y = \(\frac{a}{32}\) (mol) (II)
(I),(II) => x = \(\frac{b}{40}-\frac{a}{32}\) , y = \(\frac{a}{16}-\frac{b}{40}\)
+ Trường hợp 3 : X : NaHSO3
=> \(\frac{a}{b}>=\frac{4}{5}\)=> xảy ra (3)
(3) => nNaHSO3 = 2.nNaOH = \(\frac{b}{20}\left(mol\right)\)
