Question

Q2- octane is a gasoline. Complete combustion of octane yields component of CO2 combustion produces co and Hzo. In a certain run, and H Incomplete 2.345 gal of octane is burned in an engine. The total mass of co, com, and H20 produced is 14.053 kg. Calculate the efficiency of the process, that is, calculate the fraction of octane converted to co2. The density of octane is 2.65 kg/gal

Answer 1

Given data:

Mass of octane (C8H18) undergone combustion = 2.345 gal., i.e. 2.345*2.65 Kg = 6.214 Kg

No. of moles of octane undergone combustion = 6214/114, i.e. 54.51.

Total mass of CO, CO2 and H2O produced = 14.053 Kg

Therefore, the total mass of O2 required for combustion of C8H18 = 14.053 - 6.214, i.e. 7.839 Kg

No. of moles of O2 required for combustion of C8H18 = 7839/32, i.e. 244.97.

C8H18 : O2 = 54.51 : 244.97, i.e. 1 : 4.49 ~ 1: 9/2

The balanced equation according to the moles ratio of C8H18 and O2 can be written as

C8H18 + 9/2 O2 8C + 9H2O

Here, the fraction of octane converted to CO2 = 0.

Note: Check the given mass of octane and total masses of CO, CO2 and H2O once again. If you mistyped any value, recheck the calculations accordingly and if you get, for e.g., 5CO2 like in the below equation

C8H18 + 10 O2 5CO2 + 9H2O + 1CO

Here, 5 moles of carbon is converted to CO2, i.e. the fraction of octane converted = 5*12/114, i.e. 52.6%