Question

The complete combustion of octane, C8H18, the main component of gasoline, proceeds as follows: 2C8H18 (l) 25 O2(g) --> 16 CO2(g) 18 H2O (g)
a. How many moles of 02 are needed to burn 1.50 mol of C8H18?
b. How many grams of 02 are needed to burn 10.0 g of C8H18?
c. Octane has a density of 0.692 g/ml at 20 degrees Celsius. How many grams of 02 are required to burn 15.0 gal of C8H18 (the capacity of an average file tank)?

Answer 1

Answer:

a. 18.75 mol of O₂

b. 35.008 g   of O₂

c. 137868 g   of O₂

Explanation:

a.

Based on the balanced equation the molar ratio of C₈H₁₈ : O₂(g) = 2:25  Therefore, 1.50 mol of C₈H₁₈ = 25/2 x 1.5 = 18.75 mol of O₂

b.

Given, mass of C₈H₁₈ = 10.0 g  

Molar mass of C₈H₁₈ = (12 x 8) + (18 x 1) = 114 g/mol

Moles of C₈H₁₈ = 10.0 g/ 114 g/mol = 0.08772

Based on the balanced equation the molar ratio of C₈H₁₈ : O₂(g) = 2:25  

Therefore, 0.08772 mol of C₈H₁₈ = 25/2 x 0.08772 = 1.0965 mol of O₂

Molar mass of O₂ = 32 g/mol

Mass of O₂ = 1.0965 mol x 32 g/mol = 35.008 g  

c.

Given, density of C₈H₁₈ = 0.692 g/ml

Volume of C₈H₁₈ = 15.0 gal

Volume of C₈H₁₈ = 15.0 gal x 3785.41 ml/ 1 gal = 56781.15 ml

Mass of C₈H₁₈ = density of C₈H₁₈ x Volume of C₈H₁₈

                         = 0.692 g/ml x 56781.15 ml = 39292.5 g  

Molar mass of C₈H₁₈ = 114 g/mol

Moles of C₈H₁₈ = 39292.5 g / 114 g/mol = 344.67 mol

Based on the balanced equation the molar ratio of C₈H₁₈ : O₂(g) = 2:25  

Therefore, 344.67 mol of C₈H₁₈ = 25/2 x 344.67 = 4308.375 mol of O₂

Molar mass of O₂ = 32 g/mol

Mass of O₂ = 4308.375 mol x 32 g/mol = 137868 g  

Answer 2

The ratio of mole number of the reactants and products is equal to the coefficients. So the answer is a. 18.75 mol. b. 35.1 g. c. 1.38 * 10^5 g.