Efficient Combustion of CH4: Enthalpy, Energy Output & CO2 Emission Calculations

[mojo4king]

for eg. combustion of CH4:
CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l)

enthalpy of combustion: (-393.51) + 2(-241.82) - (-74.81) = -802.34 kj/mol

energy output per unit mass:
802.34 x 1000/16(molar mass) = -50146.25 kj/kg

energy output per unit volume:
-50146.25 x 0.645(mass density) = -32344.33 kj/m3

However I have no clue on how to calculate the carbon dioxide output per unit mass and volume, any help would be kindly appreciated.
Many thanks.

 

[mbeychok]

mojo4king:

I am not sure that I completely understand your questions, but here goes:

(1) The molecular weight of carbon dioxide is 44 and the molecular weight weight of methane is 16. Since the combustion equation shows that 1 mole of methane (when combusted) yields 1 mole of carbon dioxide, that means that 44 mass units of carbon dioxide are formed when 16 mass units of methane are combusted.

(2) Since 1 mole of any gas has the same volume as 1 mole of any other gas, and since 1 mole of carbon dioxide is produced from combusting 1 mole of methane, that means that 1 volume unit of of carbon dioxide is produced for every volume of methane combusted.

Milt Beychok

 

[Adrian4English]

I could go on for hours about combustion and will write a fuller reply. Essentially Beychock is correct, you must remember the stoicheometry.
One mole of methane requires two moles of oxygen to produce one mole of carbon dioxide and two moles of water so equating moles to volumes (which is not quite true) and assuming everything starts and ends at the same pressure and temperature then one volume of methane will produce one volume of carbon dioxide.
Incidentally, the combustion figure you quote of 802.34 kJ/kgmol is known as the "Inferior Heating Value" (LHV). Here the water vapour produced remains as water vapor. The "Superior Heating Value" (SHV) is where the water vapour is condensed to liquid water and the latent heat of liquifaction is recovered.
Given this, the overall volumes are the same, you start with one volume of methane and two volumes of oxygen and end up with one volume of carbon dioxide and two volumes of water vapour.
If you use the SHV then the volume of, now liquid, water is negligable and the original three volumes of methane and oxygen combust to one volume of carbon dioxide and a negligable volume of water.
I don't know where you are, so I've used the term volume to mean either cubic meters or cubic feet, also I don't know what level of chemistry you are studying and I don't want to confuse you with lots of technical terms. I don't mean to be patronising but...