How many grams of oxygen are required to react with 17.0 grams of octane (c8h18) in the combustion of octane in gasoline?

The balanced equation for the combustion of octane is as follows
2C₈H₁₈ + 25O₂ ---> 16CO₂  + 18H₂O
stoichiometry of C₈H₁₈  to O₂ is 2:25
number of octane moles reacted - 17.0 g / 114.2 g/mol = 0.149 mol 
according to molar ratio 
if 2 mol of octane reacts with 25 mol of O₂
then 0.149 mol of octane reacts with - 25 /2 x 0.149 mol = 1.86 mol of O₂
 mass of O₂  - 1.86 mol x 32 g/mol = 59.5 g
59.5 g of O₂ is required to react with 

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