2H2S + 3O2 → 2SO2 + 2H2O
\(n_{SO_2}=n_{H_2S}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(n_{NaOH}=\dfrac{50.1,28.25\%}{40}=0,4\left(mol\right)\)
Ta có : \(\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0,4}{0,4}=1\)
=> Tạo muối trung hòa
2NaOH + SO2 -----> Na2SO3 + H2O
\(C\%_{Na_2SO_3}=\dfrac{0,4.126}{0,4.64+\dfrac{50.1,28}{25\%}}.100=17,9\%\)

