[80z_chick]

Hi Again.  Sorry to post another question again, but I'm hoping that someone can explain so I can "get" it:)
"Write the balanced thermochemical equation for the molar heat of combustion for propane, c₃H_{8 (g)} using the table of molar heats of combustion." 
So, the table says that the heat of combustion of propane is -2220 kJ/mol.  I know that the equation has to be balanced somehow, but can't figure out how to do it. I guess this is an alkane, which reacts with oxygen to produce CO₂ and water.  So the answer might go something like:
C₃H ₈ + 0₂------->C0₂ + H₂0 ΔH_c = -2220 kJ  I suspect that I have to add other numbers in there somewhere?

This is the example I'm using that is confusing me even more:
"Write the balanced thermochemical equation for the molar heat of combustion of ethane, C₂H_{6 (g)}.  They give the answer as:
C₂H_6(g) + 7/2 0_{2 (g)}------->2CO_2(g) + 3H₂O ΔH_c=-1560 kJ

Please help me understand how they did this.  I get that you have to balance these with fractions, but how did they get the 7/2 of oxygen, the 2 of CO₂ and the 3 H₂0?

[Borek]

They use fractions to make sure there is one mole of the hydrocarbon on the left hand side.

I am not sure what is your problem. Properly balanced reaction equation for ethane is

2C₂H₆ + 7O₂ 4CO₂ + 6H₂O

For one mole of ethane you have to divide all coefficients by 2, that will yield the equation you listed.

Note - by IUPAC defined rules state symbols should be written using plain text, so it is C₂H₆(g).