when 8gram hydrogen reacts with 8g oxygen how much water is produced?

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 = 18.0 g/mol

2H₂ + O₂ → 2H₂O
2 moles of H₂ completely reacts with 1 mole of O₂ to produce 2 moles of H₂O.

Initial number of moles of H₂ = (8 g) / (2.0 g/mol) = 4 mol
Initial number of moles of O₂ = (8 g) / (32.0 g/mol) = 0.25 mol

When 0.25 mol of O₂ completely reacts :
Number of moles of H₂ needed = (0.25 mol) × 2 = 0.50 mol < 4 mol
Hence, H₂ is in excess, and O₂ is the limiting reactant (completely reacts).

Number of moles of O₂ reacts = 0.25 mol
Number of moles of H₂O produced = (0.25 mol) × 2 = 0.50 mol
Mass of H₂O produced = (18.0 g/mol) × (0.50 mol) = 9 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 = 18.0 g/mol
2H₂ + O₂ → 2H₂O 2 moles of H₂ completely reacts with
1 mole of O₂ to produce 2 moles of H₂O.
Initial number of moles of H₂ = (8 g) / (2.0 g/mol)
= 4 mol Initial number of moles of O₂
= (8 g) / (32.0 g/mol)
= 0.25 mol
When 0.25 mol of O₂ completely reacts :
Number of moles of H₂ needed = (0.25 mol) × 2 = 0.50 mol < 4 mol
Hence, H₂ is in excess, and O₂ is the limiting reactant (completely reacts).
Number of moles of O₂ reacts = 0.25 mol
Number of moles of H₂O produced = (0.25 mol) × 2 = 0.50 mol
Mass of H₂O produced = (18.0 g/mol) × (0.50 mol) = 9 g