Consider the oxidation of sulfur dioxide to sulfur trioxide: 2 SO ₂( g ) + O ₂( g ) 2 SO ₃( g ) If 12.8 g of SO ₂ is reacted with 4.8 g of O ₂ and the percentage yield of the reaction is 75 % , the actual mass of SO ₃ produced is:

Step-by-step solution

The balanced chemical equation is: 2 SO ₂( g ) + O ₂( g ) 2 SO ₃( g ) Molar mass of SO ₂ = 64 g mol ⁻¹ Molar mass of O ₂ = 32 g mol ⁻¹ Number of moles of SO ₂ = 12.8 64 = 0.2 mol Number of moles of O ₂ = 4.8 32 = 0.15 mol To identify the limiting reagent, divide the moles by their respective stoichiometric coefficients: For SO ₂: 0.2 2 = 0.1 For O ₂: 0.15 1 = 0.15 Since 0.1 From the stoichiometry, 2 moles of SO ₂ produce 2 moles of SO ₃ . Theoretical moles of SO ₃ = 0.2 mol Theoretical mass of SO ₃ = 0.2 80 = 16 g