Calculation Question!!! (1 Viewer)

[x-ray1018]

One way of minimizing the greenhouse effect is to minimize the amount of carbon dioxide formed per kilojoule of energy produced. Arrange the following fuels in order of increasing amount of carbon dioxide per kilojoule of energy released. For each substance the amount of the energy released in kJ per mole during combustion is given in brackets: coal (455), methanol (726), octane (5460), ethanol (1360), 1-propanol (2021), methane (1360).
thx

 

[LostAuzzie]

From my Calculations (assuming complete combustion):

Octane - 17.93mL per kJ

2C8H18 + 17O2 -> 18H2O + 8CO2
4mol CO2 for every 1mol Octane (5460kJ energy)
Thus for every 5460kJ, 4 x 24.47 = 97.88L CO2 released
For every 1kJ 97.88/5460 x 1000 = 17.93 mL released

Methane - 17.99mL per kJ

CH4 + 3O2 -> 2H2O + CO2
1mol CO2 for every 1mol Methane (1360kJ energy)
Thus for every 1360kJ, 24.47L CO2 released
For every 1kJ 24.47/1360 x 1000 = 17.99 mL released

Methanol - 33.7mL per kJ

2C3H3OH + 3O2 -> 4H2O + 2CO2
1mol CO2 for every 1mol Methanol (726kJ energy)
Thus for every 726kJ, 24.47L CO2 released
For every 1kJ 24.47/726 x 1000 = 33.70 mL released

Ethanol - 35.99mL per kJ

C2H5OH + 3O2 -> 3H2O + 2CO2
2mol CO2 for every 1mol Ethanol (1360kJ energy)
Thus for every 1360kJ, 2 x 24.47 = 48.94L CO2 released
For every 1kJ 48.94/1360 x 1000 = 35.99mL released

Propanol - 36.32mL per kJ

2C3H7OH + 9O2 -> 8H2O + 6CO2
3mol CO2 for every 1mol Propanol (2021kJ energy)
Thus for every 2021kJ, 3 x 24.47 = 73.41L CO2 released
For every 1kJ 73.41/2021 x 1000 = 36.32 mL released

Correct me if theres anything wrong there