Gas stoichiometry: what volume of oxygen at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53 grams of propane?

Answer is: volume of oxygen is 14.7 liters.
Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
m(C₃H₈-propane) = 5.53 g.
n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).
n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.
n(C₃H₈) = 0.125 mol.
From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.
n(O₂) = 0.625 mol.
T = 25° = 298.15K.
p = 1.04 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
V(O₂) = n·R·T / p.
V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.
V(O₂) = 14.7 L.