# 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 Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.