Limiting Reactant Help!?

Molar mass of Zn = 65.4 g/mol
Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol
Molar mass of ZnO = (65.4 + 16.0) g/mol = 81.4 g/mol

Initial moles of Zn = (35.8 g) × (1 mol / 65.4 g) = 0.5474 mol
Initial moles of O₂ = (12.9 g) × (1 mol / 32.0 g) = 0.4031 mol

2 Zn(s) + O₂(g) → 2 ZnO(s)
Mole ratio Zn : O₂ = 2 : 1

If Zn is completely reacted,
O₂ needed = (0.5474 mol) × (1/2) = 0.2737 mol < 0.4031
Hence, O₂ is in excess.
The limiting reactant is Zn.

The answer: A. Zn

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According to the above equation, mole ratio Zn : ZnO = 2 : 2
Moles of ZnO produced = (0.5474 mol) × (2/2) = 0.5474 mol
Mass of ZnO produced = (0.5474 mol) × (81.4 g/mol) = 44.6 g

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Mass of O₂ remains after the reaction
= (Total initial mass of the reactants) - (Mass of ZnO produced)
= [(35.8 + 12.9) - 44.6] g
= 4.1 g

OR:
According to the above calculation:
Moles of O₂ reacted = 0.2737 mol
Mass of O₂ reacted = (0.2737 mol) × (32.0 g/mol) = 8.8 g
Mass of O₂ remains after the reaction = (12.9 - 8.8) g = 4.1 g