What is the maximum mass of s8 that can be produced by combining 87.0 g of each reactant? 8so2+16h2s=3s8+16h20?

The balanced chemical reaction is expressed as:

8SO2 + 16H2S = 3S8 + 16H2O

We are given the initial amount of the reactants. From there, we determine the limiting reactant. We do as follows:

87.0 g SO2 (1 mol / 64.07 g) = 1.36 mol SO2 (16 mol H2S / 8 mol SO2) = 2.72 mol H2S

87.0 g H2S (1 mol / 34.08 g) = 2.55 mol H2S (8 mol SO2 / 16 mol H2S) = 1.28 mol SO2

Therefore, the limiting reactant would be H2S. We calculate the maximum amount of S8 that can be produced from the amount of H2S.

2.55 mol H2S (3 mol S8 / 16 mol H2S) (256.48 g / 1 mol) = 122.63 g S8