What is the maximum mass of S8 that can be produced by combining 78.0g of each reactant? 8SO2+16H2S⟶3S8+16H2O

Answer 1

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

Since we are given the mass of BOTH reactants, we must first find out which reactant is limiting. To do this, we can calculate the amount of S8 formed from each reactant separately, and the least amount will represent the limiting reactant, and will also give us our answer.

For SO_2:

78.0 g SO₂ x 1 mol SO₂/64.1 g x 3 mol S₈/8 mol SO₂ x 256 g S₈/mol = 117 g S₈

For H₂S:

78.0 g H₂S x 1 mol H₂S/34.1 g x 3 mol S₈/16 mol H₂S x 256 g S₈/mol = 110 g S₈

So, H₂S is limiting and the answer is 110 g (3 sig. figs.) S₈ is the maximum mass that can be formed.