Question

Air is 23.19% o2 and 75.46% n2 by weight) would be needed to burn a pound of gasoline by a reaction whereby c8h18 reacts with o2 to form co2 and h2o?

Answer 1

Approximately $15.10$ pounds.

Start by balancing the equation for the complete combustion of octane in oxygen:

$\text{C}_8\text{H}_{18} + 25/2 \;\text{O}_2 \to 8 \; \text{CO}_2 + 9 \; \text{H}_2\text{O}$

$2 \; \text{C}_8\text{H}_{18} + 25 \;\text{O}_2 \to 16 \; \text{CO}_2 + 18 \; \text{H}_2\text{O}$

Thus it would take $25$ molecules of oxygen to completely react with $2$ molecules of octane. Oxygen and octane thus react at a ratio of

$25 \; \text{mol} \; \text{O}_2$ to $2 \; \text{mol} \; \text{C}_8 \text{H}_{18}$

Given the molar mass of the species:

• $M(\text{O}_2) = 2 \times 16.00 = 32.00 \; \text{g} \cdot \text{mol}^{-1}$
• $M(\text{C}_8\text{H}_{18}) = 8 \times 12.01 + 18 \times 1.008 =  114.22 \; \text{g} \cdot \text{mol}^{-1}$

The mass ratio between the two species when the combustion proceeds to completion would thus equal:

$25 \; \text{mol} \times 32.00 \; \text{g} \cdot \text{mol}^{-1} = 800.00 \; \text{g} \;$ to

$2 \; \text{mol} \times 114.22 \; \text{g} \cdot \text{mol}^{-1} = 228.45 \; \text{g}$

It would thus take

$\begin{array}{lll}m(\text{O}_2)& = & 1 \; \text{lb} \; \text{C}_8\text{H}_{18} \times (800.00 \; \text{g}\; \text{O}_2) / ( 228.45 \; \text{g}\; \text{C}_8 \text{H}_{18})\\ &=& 3.5019 \; \text{lb} \; \text{O}_2 \end{array}$

to combust completely one pound of octane.

Apply the mass ratio stated in the question:

$\begin{array}{lll}m(\text{Air}) & = & m(\text{O}_2) / \frac{m(\text{O}_2)}{m(\text{Air})}\\ & = & 3.5019 \; \text{lb} / 0.2319 \\ & = & 15.10 \; \text{lb}\end{array}$