Sulfur dioxide was passed through 60 g of a solution with a mass fraction of sodium hydroxide of 11%. This formed sodium sulfite.

Sulfur dioxide was passed through 60 g of a solution with a mass fraction of sodium hydroxide of 11%. This formed sodium sulfite. Calculate the volume of the reacted gas.

1) Let’s compose the reaction equation:
2NaOH + SO2 ↑ = Na2SO3 + H2O
2) Calculate the mass and amount of sodium hydroxide substance contained in the solution:
m (NaOH) = (m (solution) ∙ ω) / 100% = (60 ∙ 11) / 100 = 6.6 g;
n (NaOH) = m (NaOH) / M (NaOH) = 6.6 / 40 = 0.165 mol.
3) Determine the volume of sulfur dioxide that has reacted:
according to the reaction equation
n (SO2) = 0.5 ∙ n (NaOH) = 0.5 ∙ 0.165 = 0.0825 mol;
V (SO2) = n (SO2) ∙ Vm = 0.0825 ∙ 22.4 = 1.85 l
Answer: 1.85 liters.