What is the maximum mass of S8 that can be produced by combining 83.0 g of each reactant?

Question

What is the maximum mass of S8 that can be produced by combining 83.0 g of each reactant?

8SO2+16H2S⟶3S8+16H2O

1. mass of S8: ? g

Answer 1

8SO₂ + 16H₂S ==> 3S₈ + 16H₂O .. balanced equation

Any time we are given the amounts of BOTH reactants (as is the case here), we must first determine which reactant is limiting. One way to do this is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value comes out less represents the limiting reactant.

For SO₂ (molar mass 64.1 g/mol): 83.0 g x 1 mol / 64.1 g = 1.29 mols SO₂ (÷8->0.16)

For H₂S (molar mas 34.1 g/mol): 83.0 g x 1 mol / 34.1 g = 2.43 mols H₂S (÷16->0.15)

Since 0.15 is less than 0.16, H₂S is the limiting reactant, and the moles of H₂S (2.43 mols) will determine the maximum amount of product (S₈) that can be produced.

Max. mass of S₈ (molar mass 256.5 g/mol):

2.43 mol H₂S x 3 mol S₈ / 16 mol H₂S x 256.5 g S₈ / mol S₈ = 117 g S₈

(be sure to check all of the math)