Question is below in the photo ?

Molar mass of KClO₃ = (39.1 + 35.5 + 16.0×3) g/mol = 122.6 g/mol
Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol

2KClO₃ → 2KCl + 3O₂
Mole ratio  KClO₃ : O₂ = 2 : 3

Moles of KClO₃ reacted = (73.9 g) / (122.6 g) = 0.6028 mol
Moles of O₂ produced = (0.6028 mol) × (3/2) = 0.9042 mol
Nass of O₂ produced = (0.9042 mol) × (32.0 g/mol) = 28.9 g

====
OR:

(73.9 g KClO₃) × (1 mol KClO₃ / 122.6 g KClO₃) × (3 mol O₂ / 2 mol KClO₃) × (32.0 g O₂ / 1 mol O₂)
= 28.9 g O₂

Try moving this to Chemistry...