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
