S + O2 → SO2
Moles of S:
11/32.06 = 0.3431 mol
Moles of O2:
44/15.999 = 2.75 mol
Sulfur will be the limiting reagent.
Oxygen will be the excess reagent
2.75 - 0.3431 = 2.4071 mols of oxygen leftover
2.4071 * 15.999 = 38.511
38.511 grams of oxygen will be leftover
Reaction
S + O₂ ⇒ SO₂
mole S = 11 : 32 g/mol = 0.34375
mole O₂ = 44 : 32 g/mol = 1.375
S as a limiting reactant (smaller) and converted all to sulfur dioxide
mole O₂ (unreacted) = 1.375 - 0.34375 = 1.03125
mass O₂ (unreacted) = 1.03125 x 32 = 33 g
