What is the maximum mass of S8 that can be produced by combining 86.0 g of each reactant? 8SO2 + 16H2S —-> 3S8 + 16H2O

Answer 1

8SO2 + 16H2S —-> 3S8 + 16H2O

Molar mass SO2 = 64.07 g/mol

Molar mass H2S = 34.08 g/mol

Molar mass S8 = 256 g/mol

moles SO2 = 86 g x 1 mol/64.07 g = 1.34 mol

mols H2S = 86 g x 1 mol/34.08 g = 2.52 mols

According to the balanced equation it takes twice as much H2S as SO2 so H2Swill be limiting.

mass S8 produced = 2.52 mols H2S x 3 mols S8 / 16 mols H2S x 256 g / mol = 121 g S8