How many grams of oxygen are required to react with 19.0 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

Balanced equation for the reaction:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
Mole ratio C₈H₁₈ : O₂ = 2 : 25

No. of moles of C₈H₁₈ reacted = (19.0 g) / (114.0 g/mol) = 0.1667 mol
No. of moles of O₂ required = (0.1667 mol) × (25/2) = 2.084 mol
Mass of O₂ required = (2.084 mol) × (32.0 g/mol) = 66.7 g

Sep 7, 2015 - Balance the following chemical equation, then answer the following question. C8H18(g) + O2(g) →CO2(g)+H2O(g) How many grams of oxygen are required to react with 10.0 grams of octane (C8H18) in the combustion of octane in gasoline?

First, balance the equation:

2C8H18 + 25O2 --> 16CO2 + 18H2O

Now, find moles of octane - divide 19.0 by the molecular weight of octane. (It's about 210 g/mole but you are going to want it to three significant figures). Now convert to moles of O2 (multiply by 25/2) and convert to grams of O2 (multiply by 32 g/mole but again you need three significant figures.)