How many grams of oxygen are required to react with 10.0 grams of octane (C8H18) in the combustion of octane in gasoline?  ?

Molar mass of C₈H₁₈ = (12.0×8 + 1.0×18) g/mol = 114.0 g/mol

Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol

According to the equation, mole ratio C₈H₁₈ : O₂ = 2 : 25

Moles of C₈H₁₈ reacted = (10.0 g) / (114.0 g/mol) = 0.08772 mol

Moles of O₂ required = (0.08772 mol) × (25/2) = 1.097 mol

Mass of O₂ required = (1.097 mol) × (32.0 g/mol) = 35.1 g

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OR:

(10.0 g C₈H₁₈) × (1 mol C₈H₁₈ / 114.0 g C₈H₁₈) × (25 mol O₂ / 2 mol C₈H₁₈) × (32.0 g O₂ / 1 mol O₂)

= 35.1 g O₂

Octane + Oxygen ➜ Carbon Dioxide + water

2C₈H₁₈ + 25O₂ ➜ 16CO₂ + 18H₂O

molecular weights

C = 12

H = 1

O = 16

2C₈H₁₈ = 2(8•12+18) = 228

25O₂ = 25•16•2 = 800

16CO₂ = 16(12+32) = 704

18H₂O = 18•18 = 324

check 228+800 = 704+324 = 1028 ok

2 mole of C₈H₁₈ + 25 moles of O₂ ➜ 16 mole of CO₂ + 18 moles of H₂O

228 grams of C₈H₁₈ + 800 grams of O₂ ➜ 704 grams of CO₂ + 324 grams of H₂O

ratio by grams of  O₂ to C₈H₁₈ is 800/228

x/10 = 800/228

x = 35.1 g

While it is possible to calculate an answer to your question as a general stoichiometry exercise, you should know that there is hardly any n-octane in gasoline.  Less than 0.5% of gasoline is "octane."  Gasoline is a mixture of over 100 hydrocarbons, and the composition varies with brand and the time of year (winter blend vs summer blend).  But just because gasoline has an "octane rating" doesn't mean that octane is a major component of gasoline.  You would need over 2000 grams of gasoline to have 10g of octane.

The octane rating of a sample of gasoline is the percentage of iso-octane (2,2,4-trimethylpentane) in a mixture of it and n-heptane which performs like the test gasoline.

And now, on with the "show."  (But keep in mind that this "show" is a work of fiction.)

2C8H18(g) + 25O2(g) --> 16CO2(g) + 18H2O(g)10.0g................?g

10.0g C8H18 x (1 mol C8H18 / 114.0g C8H18) x (25 mol O2 / 2 mol C8H18) x (32.0g O2 / 1 mol O2) = 35.1g O2