The complete combustion of octane, C₈H₁₈, a component of gasoline, proceeds as follows: 2 C₈H₁₈ (l) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O (g) (c) Octane has a density of 0.692 g/mL at 20 °C. How many grams of O₂ are required to burn 15.0 gal of C₈H₁₈ (the capacity of an average fuel tank)?

Video transcript

Hello everyone. Today we have the following problem. The complete combustion of octane, a component of gasoline precedes as follows. We have problems seen that says octane has a density of 0.692 g per milliliter at 20 °C. How many grams of oxygen are required to burn 22 gallons of octane? So first, we need to calculate the mass of octane that we have. So we have the mass of octane in grams is equal to first, we have our 22 gallons and we need to multiply by the conversion factor that one gallon is equivalent to 3.7854 leaders. We then can multiply to get from liters to milliliters with the following factor conversion factor that 1 mL is equal to 10 to the negative third liters. And then lastly, we can multiply by our density which states that 1 mL is equal to 0.692 g. Our units will cancel that and we will arrive at 5 57,628 0.93 g of our octane. We then can perform a multiple ratio. So we have our 57,628 0.93 g of our octane. We first multiply by its molar mass. So we have one mole of octane is equal to eight carbons plus 18 hydrogens for 1 14.23 g. And then we multiply from moles of octane. We have one mole of octane in our chemical equation. And we see that it's like 25 miles. We have two moles, excuse me. And then we have 25 moles of our oxygen gas. One mole of our oxygen gas is equal to its smaller mass, which is 32 g of it. When we cancel out all our units and sulfur the mass of oxygen, we get 4.04 times 10 to the fifth grams. And that aligns with answer choice D and with that, we have solved the problem overall, I hope this helped. And until next time.