Chemistry help!!?

Molar mass of C₂H₅OH (ethanol) = (12.0 + 1.0×6 + 16.0) g = 46.0 g/mol
Molar mass of O₂ (oxygen gas) = 16.0 × 2 g/mol = 32.0 g/mol
Molar mass of H₂O (water) = (1.0×2 + 16.0) g/mol = 18.0 g/mol

Initial number of moles of C₂H₅OH = (3.8 ml) × (0.789 g/ml) / (46.0 g/mol) = 0.06518 mol
Initial number of moles of O₂ = (12.5 g)/ (32.0 g/mol) = 0.3906 mol

Balanced equation for the reaction :
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
Mole ratio C₂H₅OH : O₂ : H₂O = 1 : 3 : 3

If 0.06518 mol C₂H₅OH completely reacts, O₂ needed = (0.06518 mol) × 3 = 0.1955 mol < 0.3906 mol
Hence, O₂ is in excess, and thus C₂H₅OH is the limiting reactant.

Number of moles of C₂H₅OH reacted = 0.06518 mol
Maximum number of moles of H₂O formed = (0.06518 mol) × 3 = 0.1955 mol
Theoretical yield of H₂O = (0.1955 mol) × (18.0 g/mol) = 3.52 g

Actual yield of H₂O = (3.10 ml) × (1 g/ml) = 3.10 g
Percent yield of H₂O = (3.10/3.52) × 100(%) = 88.1%

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