what mass of O2 is needed to react with 5.95 g of NH3?

The products of the reaction between O₂ and NH₃ depend on reaction conditions.

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Case I : When NH₃ is burned in pure O₂
4NH₃(g) + 3O₂(g) → 2N₂(g) + 6H₂O(g)
OR: Mole ratio NH₃ : O₂ = 4 : 3

Molar mass of NH₃ = (14.0 + 1.0×3) g/mol = 17.0 g/mol
No. of moles of NH₃ reacted = (5.95 g) / (17.0 g/mol) = 0.350 mol
No. of moles of O₂ needed = (0.35 mol) × (3/4) = 0.2625 mol

Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
Mass of O₂ needed = (0.2625 mol) × (32.0 g/mol) = 8.40 g

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Case II : When NH₃ is heated in air in the presence of platinum catalyst
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
OR: Mole ratio NH₃ : O₂ = 4 : 5

Molar mass of NH₃ = (14.0 + 1.0×3) g/mol = 17.0 g/mol
No. of moles of NH₃ reacted = (5.95 g) / (17.0 g/mol) = 0.350 mol
No. of moles of O₂ needed = (0.35 mol) × (5/4) = 0.4375 mol

Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
Mass of O₂ needed = (0.4375 mol) × (32.0 g/mol) = 14.0 g