Please help i am horrible at Stoichiometry?

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

The balanced equation for the reaction:
2 C₈H₁₈(g) + 25 O₂(g) → 8 CO₂(g) + 9 H₂O(g)
Mole ratio C₈H₁₈ : O₂ = 2 : 25

No. of moles of C₈H₁₈ = (18.0 g) / (114.0 g/mol) = 0.1579 mol
No. of moles of O₂ = (0.1579 mol) × (25/2) = 1.974 mol
Mass of O₂ needed = (1.974 mol) × (32.0 g/mol) = 63.2 g

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OR:

(18.0 g C₈H₁₈) × (1 mol C₈H₁₈ / 114.0 g C₈H₁₈) × (25 mol O₂ / 2 mol C₈H₁₈) × (32.0 g O₂ / 1 mol O₂)
= 63.2 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


How many grams of oxygen are required to react with 18.0 grams of octane in the combustion of octane in gasoline?


ratio is 800/225  grams
800/225 = x/18
x = 64 g