The balanced equation for the above reaction is as follows; 2S + 3O₂ --> 2SO₃ Stoichiometry of O₂ to SO₃ is 3:2 O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present. Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol 3 mol of O₂ forms 2 mol of SO₃ therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol Number of SO₃ moles formed - 0.0833 mol Answer is 4) 0.08 mol