posted by James .
What is the maximum mass of S8 that can be produced by combining 85.0 g of each reactant?
8SO2 + 16H2S yields 3S8 +16H2O
Convert each to mols S8 produced.
mols SO2 = 85/64 = about 1.3 estimated. You need to do it exactly.
mols H2S = 85/34 = about 2.5
Use the coefficients in the balanced equation to convert mols of each to mols S8.
1.3 x (3 mols S8/8 mols SO2) = about 5 mols S8 produced if we had 1.3 mols SO2 and all of the H2S we needed.
2.5 x (3 mol S8/16 mol H2S) = aboaut 4.5 mol S8 produced if we had 2.5 mol H2S and all of the SO2 we needed.
You have two different values for S; you know one of them must be wrong. The correct value in limiting regent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
mass S8 = mols S8 x molar mass S8 = ?grams.