Question

Octane(C8H18) is a component of gasoline. Complete combustion of octane yields only CO2 and H2O in the combustion products. If insufficient oxygen is supplied during the combustion process, the combustion products will include CO and possibly unreacted carbon particles. These other products reduce the efficiency of the engine since they are capable of undergoing further combustion and producing more energy from the initial charge of hydrocarbon. In a test run, 1.000 gallon of octane is burned in an engine and the total mass of CO, CO2 and H2O produced is 11.53 kg. If the density of octane is

Answer 1

2.650 kg/gallon, calculate the efficiency of the engine by calculating the percent of octane converted to carbon dioxide.

The answer to this question is around 87% however when I do my calculations I get 70%. Can someon please verify the answer. I will post my solution to the problem tomorrow to search for mistakes on. Thanks.

Answer 2

Balancing both sides of the equation, we deduce that the mol ratio of octane:water is 1:9
\[C _{8}H _{18} + O _{2}\rightarrow (?)CO + (?)CO _{2} + 9H _{2}O\]
We also deduce that the sum of moles of CO and CO2 = 8.
Let the number of moles of CO be n.
\[C _{8}H _{18} + O _{2}\rightarrow nCO + (8-n)CO _{2} + 9H _{2}O\]
Mr of octane = 114
Amount in mol of octane = 23.24 mol (deduced from question)
mol ratio of octane:water = 1:9
Amount in mol of water = 209.2105 mol
Mass of water = 3765.789 g
Mass of CO and CO2 = 7764.211 g
Mr of CO = 30
Mr of CO2 = 44
mol ratio of octane:CO:CO2 = 1:1:1
Hence,
\[(23.24)(n)(30) + (23.24)(8-n)(44)=7764.211g\]
Divide through by 23.24 on both sides yields
\[(n)(30) + (8-n)(44)=334.08\]
n=1.28
8-n = 6.72
Amt in mol of CO = 1.28(23.24)
Amt in mol of CO2 = 6.72(23.24)
I'm not exactly sure what your question is asking but these are the values I calculated