[ChmNerd111]

Here is the question:
If the heat of combustion of CH4 is -802 kJ/mol, use the following information (and nothing else, don't just look up data from tables) to determine the enthalpy of reaction (kJ/mol) for the partial oxidation of methane by oxygen gas to give carbon monoxide and hydrogen gas.

CH4 (g) + H2O (g) --> CO + 3 H2 (g) delta H= 206kJ
CH4 (g) + CO2 (g) --> 2 CO + 2 H2 (g) delta H= 246 kJ

What I did:
I flipped the second equation and added the two together and got:

H2O + CO  --> CO2 + 1 H2 delta H= -40kJ

I know this is wrong, because I want to get methane and oxygen on the right side and carbon monoxide and hydrogen on the right side, but I don't know where to go from here. Also, oxygen gas isn't in either of these equations, so how can I form that? Any help would be greatly appreciated. (Note: This is not a graded hw problem. I have an exam coming up and this is a suggested review problem, but its aneven numbered problem so the answer is not in the back of the book. Again, thanks in advance for any help.)

[GollumsLunch]

I would start by writing the equation for the reaction for which you want the enthalpy.  You then have 3 known reactions to manipulate (not just the two). In addition to the two you are explicitly given you also know the enthalpy for the combustion of methane.

The reactions you know are thus:
CH4 (g) + H2O (g) --> CO + 3 H2 (g) delta H= 206kJ
CH4 (g) + CO2 (g) --> 2 CO + 2 H2 (g) delta H= 246 kJ
CH4 (g) + 2O2 (g) --->  CO2 (g)  + 2H2O (g) delta H = -802 kJ