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

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