Octane has a density of 0.692 g/mL at 20°C. How many grams of O2 are required to burn 8.40 gal. of C8H18?
2007-06-07 15:26:11 · 2 answers · asked by Ham Wallet 1 in Science & Mathematics ➔ Chemistry
First you need a balanced equation
2C8H18 + 25O2 == 16CO2 + 18H2O
You need to convert the gallons to liters (3.89 liters / gal) and then to mL. Now use the density to get grams of octane. Now divide ny the molecular mass to get moles.
The stoichiometry is straight from this point. Use the right molar ration and get moles of O2. Multiply by the molecular mass and you will have grams
2007-06-07 15:36:10 · answer #1 · answered by reb1240 7 · 0⤊ 0⤋
That wouldn't be too daunting given a conversion value between gallons and liters. Put your gallons into liters, then
Start with a balanced equation for the combustion of octane.
(Value you found) liters C8H18 (1,000 mL/1 L)(.692 g/ 1 mL)(1 mole C8H18/molar mass of C8H18)(coefficient moles O2/coefficient moles C8H18)(32 grams/1 mole O2)
2007-06-07 15:36:22 · answer #2 · answered by Molly H 1 · 1⤊ 0⤋