The balanced equation for the reaction is as follows; 2H₂S + SO₂ —> 2H₂O + 3S Stoichiometry of H₂S to SO₂ is 2:1 Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present. Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂ According to molar ratio of 2:1 If we assume H₂S to be the limiting reactant 2 mol of H₂S reacts with 1 mol of SO₂ Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂ But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant. H₂S is the limiting reactant Amount of S produced depends on amount of H₂S present Stoichiometry of H₂S to S is 2:3 2 mol of H₂S forms 3 mol of S Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
