H2S(g) + O2(g) → SO2(g) + H2O(g) Determine the maximum number of moles of SO2 produced from 6.00 moles of H2S and 2.00 moles of O2.?

Balanced equation :
2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g)
Mole ratio in reaction H₂S : O₂ : SO₂ = 2 : 3 : 2

No. of moles of H₂S before reaction = 6.00 mol
No. of moles of O₂ before reaction = 2.00 mol

For complete reaction of H₂S, No. of moles of O₂ needed = 6.00 × (3/2) = 9.00 mol
Obviously, H₂S is in excess, and O2 is the limiting reactant (completely reacted).

No. of moles of O₂ reacted = 2.00 mol
Max. no. of moles of SO₂ produced = 2.00 × (2/3) = 1.33 mol ...... (Ans)