how many grams of oxygen are required to react with 14.0 grams of octane in the combustion of octane in gasoline

Question

chemistry

Answer 1

First, we must write the correctly balanced equation for this reaction:

Octane = C₈H₁₈

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

Next, find mols of octane present:

molar mass C₈H₁₈ = 114 g/mol

14.0 g x 1 mol / 114 g = 0.123 mol

Finally, find mols O₂ needed and then convert to grams O₂ needed:

0.123 mol C₈H₁₈ x 25 mols O2 / 2 mol C₈H₁₈ = 1.54 mols O₂

1.54 mols O₂ x 32 g O₂ / mol = 49.3 g O₂