If 50.0 g of H2 and 100.0 g of O2 react how many moles of H2O can be produced in the reaction below? 2H2(g) + O2(g) -> 2H2O(g)?

Molar mass of H₂ = 1.0×2 = 2.0 g/mol
Molar mass of O₂ = 16.0×2 = 32.0 g/mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol

Initial moles of H₂ = (50.0 g) / (2.0 g/mol) = 25.0 mol
Initial moles of O₂ = (100.0 g) / (32.0 g/mol) = 3.125 mol

2H₂ + O₂ → 2H₂O
Mole ratio H₂ : O₂ : H₂O = 2 : 1 : 2

If 3.125 mol O₂ completely reacts, moles of H₂ needed = (3.125 mol) × (2/1) = 6.25 mol < 25.0 mol
Hence, H₂ is in excess, and O₂ completely reacts.

Moles of O₂ reacted = 3.125 mol
Moles of H₂O produced = (3.125 mol) × 2 = 6.25 mol