Grams of oxygen required to burn 1.00 gallons of octane (C8H18)? (Octane has density of .692 g/mL at 20 oC)?

Question

Grams of oxygen required to burn 1.00 gallons of octane (C8H18)? (Octane has density of .692 g/mL at 20 oC)?

Answer 1

Balanced Reaction:

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

1 gallon = 3785.41178 ml

(3785.41178 ml)(0.692 g/mL) = 2619.50495176 g C8H18

To find the mass of O2 required to burn 2619.50495176 g of C8H18 first compute the moles of octane and relate the moles of octane to the moles of oxygen per mole of octane, and use the molar mass of oxygen to find the mass of oxygen.

(2619.50495176 g C8H18)(1 mol C8H18/ 114 g C8H18)(25 mol O2/2 mol C8H18)(32 g/mol) = 9 191.24544 g O2

Answer 2

1 gal (US) = 3.7853 liters

3785 mL x 0.692 g/mL =  2619 g

2619 g / 114.2 g/mol = 22.9 mol

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

22.9 mol x (25/2) = 286.3 mol

286.3 x 32 g/mol = 9160 g