Question

The complete combustion of octane, a component of gasoline, proceeds as follows: 2 C8H18+25 O2 produces 16 CO2+18 H20 How many moles of O2 are needed to burn 1.50mol of C8H18? Octane has a density of 0.692g/ml at 20 degrees celsius . How many grams of O2 are required to burn 15.0gal of C8H18?

Answer 1

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

(Or)

C8H18+ (25/2) O2 --------------->  8 CO2 + 9 H2O

1) How many moles of O2 are needed to burn 1.50mol of C8H18?

   C8H18+ (25/2) O2 --------------->  8 CO2 + 9 H2O

1 mol     12.5 mol

1.5 mol     ? = 1.5 x 12.5 mol = 18.75 mol

Therefore,

18.75 moles of O2 are needed to burn 1.50mol of C8H18

2) How many grams of O2 are required to burn 15.0 gal of C8H18?

              1 gallon = 3785.41 mL

              15 gallon = 15 x 3785.4 mL = 56781 mL

       density of Octane = 0.692 g/ml

Hence, mass of 15 gallon octane = density x volume = 0.692 g/ml x 56781 mL = 39292.4 g

C8H18        + (25/2) O2 --------------->  8 CO2 + 9 H2O

1 mol           12.5 mol

114 g            12.5 x32 = 400 g

39292.4 g      ?

            ? = (39292.4 g/ 114 g) x 400 g O2

              = 137868 g O2

Therefore, 137868 grams of O2 are required to burn 15.0 gal of C8H18.