How many grams of oxygen are required to react with 200 grams of octane (C_{8}*H_{18}) in the combustion of octane in gasoline?

Answer 1

The combustion of octane is as follows:

2C₈H₁₈ + 25O₂ ==> 16CO₂ + 18H₂O ... balanced equation

How many moles of octane are present?

200 g C₈H₁₈ x 1 mol C₈H₁₈ / 114.2 g = 1.75 mols octane

According to the balanced equation, how many moles of O₂ are needed?

1.75 mols C₈H₁₈ x 25 mols O₂ / 2 mols C₈H₁₈ = 21.88 mols O₂ needed

Convert mols of O₂ to grams of O₂:

21.88 mols O₂ x 32 g O₂ / mol = 700 g O₂ needed (rounded to 1 sig. fig based on 200 g octane)