By passing 2.8 liters of sulfur (IV) oxide through an excess of sodium hydroxide, 15.12 g of sodium

By passing 2.8 liters of sulfur (IV) oxide through an excess of sodium hydroxide, 15.12 g of sodium sulfite were obtained. Determine the yield of the reaction product in% of the theoretically possible.

SO2 + 2NaOH = Na2SO3 + H2O

Find the amount of sulfur dioxide substance:

n (SO2) = V (SO2) / VM = 2.8 / 22.4 = 0.125 mol;

According to the stoichiometry of the reaction:

ntheor (Na2SO3) = n (SO2) = 0.125 mol;

npract (Na2SO3) = m (Na2SO3) / M (Na2SO3) = 15.12 / 126 = 0.12 mol;

η = npract (Na2SO3) / ntheor (Na2SO3) = 0.12 / 0.125 = 0.96 = 96%.

Answer: the reaction yield was 96%.