The question requires calculating the amount of #CO_2# produced by each combustion, and then normalizing them to moles (or tonnes) of #CO_2#. Basically, it should be observable that the unit definition should mean that it will be the SAME for any source, because we are looking at one of the products, not the fuel.
Methane heat of combustion is -890 kJ/mol
#CH_4 + O_2 → CO_2 + 2H_2O# ONE mole of #CH_4# produces ONE mole of #CO_2#, so the energy release is the same: # -890 (kJ)/(mol)CO_2#
Propane heat of combustion is -2220 kJ/mol
#C_3H_8 + 5O_2 → 3CO_2 + 4H_2O# ONE mole of C_3H_8 produces THREE moles of #CO_2#, so the energy release is: #-2220/3 = 740 (kJ)/(mol) CO_2#
Let’s try a larger hydrocarbon:
#C_6H_12 + 9O_2 → 6CO_2 + 6H_2O DeltaH^o = -3924 (kJ)/(mol) C_6#
ONE mole of C_6H_12 produces SIX moles of #CO_2#, so the energy release is:
#-3924/6 = 654 (kJ)/(mol) CO_2#
How about alcohols?
#2 CH_3OH + 3 O_2 → 4 H_2O + 2CO_2 ∆H = -1452.8 (kJ)/(mol) C_1#
TWO moles of #CH_3OH# produces TWO moles of #CO_2#, so the energy release is the same:
#-1452.8/2 = 726.4 (kJ)/(mol) CO_2#
#C_2H_5OH + 3O_2 → 3H_2O + 2CO_2 ∆H = - 1009.8 (kJ)/(mol) C_2#
ONE mole of C_2H_5OH produces TWO moles of #CO_2#, so the energy release is: #-1009.8/2 = 504.9 (kJ)/(mol) CO_2#
#2C_3H_7OH + 9O_2→ 8H_2O + 6CO_2 ∆H = - 1606.2 (kJ)/(mol) C_3#
TWO moles of C_3H_8 produces SIX moles of #CO_2#, so the energy release is: #-1606.2 xx 2/6 = 535.4 (kJ)/(mol) CO_2#
Recapping:
#CH_4 → CO_2# at # -890 (kJ)/(mol)CO_2#
#C_3H_8 → CO_2# at # -740 (kJ)/(mol)CO_2#
#C_6H_12→ CO_2# at # -654 (kJ)/(mol)CO_2#
#CH_3OH→ CO_2# at # -726.4 (kJ)/(mol)CO_2#
#C_2H_5OH → CO_2# at # -504.9 (kJ)/(mol)CO_2#
#C_3H_7OH → CO_2# at # -535.4 (kJ)/(mol)CO_2#
So, yes, easily within the magnitude of any fuel, the amount of energy produced per tonne of #CO_2# emission is roughly the same.
The real differences are due to the ‘extra’ elements that need to be reacted. The higher the #C/H# ratio, the more energy is produced per mole (or tonne) of #CO_2# emission. It can be seen from an energy utilization perspective that 'clean burning'” alcohols are really less efficient at producing power for the amount of greenhouse gas emitted!
