Question

Octane (C8H18) is a component of gasoline. Complete combustion of octane yields H2O and CO2. Incomplete combustion produces H2O and CO, which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gallon (gal) of octane is burned in an engine. The total mass of CO, CO2, and H2O produced is 11.53 kg. Calculate the efficiency of the process; that is, calculate the fraction of octane converted to CO2. The density of octane is 2.650 kg/gal.

Answer 1

$86.5\; \%$ of octane had been converted to carbon dioxide CO₂.

Explanation

Octane has a molar mass of

$12.01 \times 8 + 1.008 \times 18 = 114.22 \; \text{g} \cdot \text{mol}^{-1}$

1.000 gallon of this fuel would have a mass of 2.650 kilograms or $2.65 \times 10^{3} \; \text{g}$, which corresponds to $2.65 \times 10^{3} / 114.22 = 23.2\; \text{mol}$ of octane.

Octane undergoes complete combustion to produce carbon dioxide and water by the following equation:

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

An incomplete combustion of octane that gives rise to carbon monoxide and water but no carbon dioxide would consume not as much oxygen:

$2\; \text{C}_8\text{H}_{18} + 17 \; \text{O}_2 \to 16 \; \text{CO} + 18 \; \text{H}_2\text{O}$

The mass of the product mixture is $11.53 - 2.65 = 8.88 \; \text{kg}$ heavier than that of the octane supplied. Thus $8.88 \; \text{kg} = 8.88 \times 10^{3} \; \text{g}$ of oxygen were consumed in the combustion. There are $277.5 \; \text{mol}$ of oxygen molecules in $8.88 \times 10^{3} \; \text{g}$ of oxygen.

Let the number of moles of octane that had undergone complete combustion as seen in the first equation be $x$ ($0 \le x \le 23.2$). The number of moles of octane that had undergone incomplete combustion through the second equation would thus equal $23.2 - x$.

25 moles of oxygen gas is consumed for every two moles of octane that had undergone complete combustion and 17 moles if the combustion is incomplete.

$n(\text{O}_2, \; \text{Complete Combustion}) + n(\text{O}_2, \; \text{Incomplete Combustion} ) = n(\text{O}_2, \; \text{Consumed})$

$\frac{25}{2} \; x + \frac{17}{2} \; (23.2 - x) = 277.5\\4 \; x = 277.5 - \frac{17}{2} \times 23.2\\x = 20.1$

Therefore $20.1 \; \text{mol}$ out of the 23.2 moles of octane had undergone complete combustion to produce carbon dioxide.

$\%n(\text{Complete Combustion}) = 20.1 / 23.2 \times 100 \; \%= 86.5\; \%$