Question

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

Answer 1

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