For the reaction 2SO2 + O2 -> 2 SO3, we first determine which is the excess reactant between SO2 and O2. We list down the molar mass of the reactants:
Molar mass of SO2 = 64.0638 g/mol
Molar mass of O2 = 32 g/mol
Using the stoichiometry of the reaction, we then calculate the amount of oxygen that will react with 90.0 g of SO2.
90.0 g SO2 x 1 mol SO2/64.0638 g x 1 mol O2/ 2 mol SO2 x 32 g O2/mol = 22.4776 g O2
Thus, we can conclude that O2 is the excess reactant while SO2 is the limiting reactant. Subtracting 22.4776 g O2 from the initial 100.0 g O2, we get 77.5224 g O2 left after the complete reaction of 90.0 g SO2.
