What is the maximum mass of S8 that can be produced by combining 80.0 g of each reactant? 8SO2+16H2S⟶3S8+16H2O

Answer 1

(80.0g SO₂) * (1 mol SO₂ / 64.1g SO₂) * (3 mol S₈ / 8 mol SO₂) = 0.468 mol S₈.

(80.0g H₂S) * (1 mol H₂S / 34.1g H₂S) * (3 mol S₈ / 16 mol H₂S) = 0.440 mol S₈. So H₂S is the limiting reactant, and maximum mass is: (0.440 mol S₈) * (256.5g S₈ / 1 mol S₈) = 113g S₈.