Suppose 1.25 g of sodium peroxide is added to a large excess of water. What mass of oxygen gas will be produced?

Molar mass of Na₂O₂ = (23.0×2 + 16.0×2) g/mol = 78.0 g/mol
Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol

The balanced equation for the reaction :
2Na₂O₂(s) + 2H₂O(l) → 4NaOH(aq) + O₂(g)
Mole ratio Na₂O₂ : O₂ = 2 : 1

No. of moles of Na₂O₂ reacted = (1.25 g) / (78.0 g/mol) = 1.25/78.0 mol
No. of moles of O₂ produced = (1.25/78.0 mol) × (1/2) = 1.25/156 mol
Mass of O₂ produced = (1.25/156 mol) × (32.0 g/mol) = 0.256 g