Given:

*Fe2O3(s)+3CO(s)-->2Fe(s)+3CO2(g)(delta h is -23kJ)
*3Fe2O3(s)+CO(g)-->2Fe3O4(s)+CO2(g)(delta h is -39kJ)
*Fe3O4(s)+CO(g)-->3FeO(s)+CO2 (delta h is 18kJ)
Desired:FeO(s)+CO(g)-->Fe(s)+CO2(g)
find delta h for the desired reaction.

which reactions do i reverse and then how do i cancel the out the ones i don't need?

ok this is what i have gotten:

3Fe2O3(s)+9CO(s)-->6Fe(s)+9CO2(g)
2Fe3O4(s)+CO2(g)-->3Fe2O3(s)+CO(g)
6FeO(s)+2CO2(g)-->2Fe3O4(s)+2CO(g)
but the desired reaction is in a one to one ratio, and also the CO(g) is on the left side in the desired reaction. does that mean i have to switch over again?

Start with the end equation.

FeO(s) + CO(g) ==> Fe(s) + CO2(g)

I want an Fe(s) on the right and CO2(g) on the right. The first equation has that so write that one as is (at least for the time being).
I would reverse equations 2 and 3
and multiply equation 1 by 3, equation 2 by 1, and equation 3 by 2. Then add everything on the left side and everything on the right side, then cancel those molecules/atoms that are on both sides. If I didn't goof I obtained the equation you want by doing that.

I don't think so. Let's see how it plays out. I'll go through some of it and let you finish.

3Fe2O3 at top left cancels with 3Fe2O3 on right in equation 2.
2Fe3O4 on left in equation 2 cancels with 2Fe3O4 on right of equation 3. Then 9CO in equation 1(left side) cancels with 1 CO on right of equation 2 and 1 CO on right of equation 3 to leave 6 CO on left side. For CO2, we have 2CO2 on left of equation 3 and 1 CO2 on left of equation 2 which subtracts from 9 CO2 on right of equation 1 leaving 6 CO2 on right side. If I've done all that right (you should confirm all of this), that leaves 6FeO + 6CO ==> 6Fe + 6 CO2. So whatever delta H adds up, just divide by 6 to get delta H for the equation you want (1:1:1:1).

You might want to check your work. Since you multiplied equation 1 by 3, did you multiply delta H by 3? And you multiplied equation 3 by 2, did you multiply delta H by 2? If those are delta Hs for the reaction as shown, then the multiplications are in order and that will change the final answer.

To find the delta H for the desired reaction, you can use a combination of the given reactions and apply the Hess's Law. The strategy is to reverse the necessary reactions and manipulate them to cancel out the undesired species.

1. Reverse the reactions:
i) Reverse the first reaction to give: 2Fe(s) + 3CO2(g) -> Fe2O3(s) + 3CO(s)
ii) Reverse the second reaction to give: 2Fe3O4(s) + CO2(g) -> 3Fe2O3(s) + CO(g)
iii) Reverse the third reaction to give: 3FeO(s) + CO2(g) -> Fe3O4(s) + CO(g)

2. Manipulate the reactions to cancel out the undesired species:
i) Multiply the first reaction by 3: 3(2Fe(s) + 3CO2(g) -> Fe2O3(s) + 3CO(s)) gives 6Fe(s) + 9CO2(g) -> 3Fe2O3(s) + 9CO(s)
ii) Multiply the second reaction by 2: 2(2Fe3O4(s) + CO2(g) -> 3Fe2O3(s) + CO(g)) gives 4Fe3O4(s) + 2CO2(g) -> 6Fe2O3(s) + 2CO(g)

3. Add the manipulated reactions together to eliminate the common species on both sides:
6Fe(s) + 9CO2(g) + 4Fe3O4(s) + 2CO2(g) -> 3Fe2O3(s) + 9CO(s) + 6Fe2O3(s) + 2CO(g)
Simplify it as: 6Fe(s) + 11CO2(g) + 4Fe3O4(s) -> 9CO(s) + 9CO(g)

4. Simplify the equation by canceling out the common species that occur on both sides:
Since CO2 and CO(g) appear on both sides, they can be canceled out.
Resulting in: 6Fe(s) + 11CO2(g) + 4Fe3O4(s) -> 9CO(s)

Now, you have the equation: 6Fe(s) + 4Fe3O4(s) + 11CO2(g) -> 9CO(s)

To find the delta H for the desired reaction, add up the delta H values of the individual reactions:

Delta H for the desired reaction = Delta H1 + Delta H2 + Delta H3

Delta H1 = -23 kJ (from the equation Fe2O3(s) + 3CO(s) -> 2Fe(s) + 3CO2(g))
Delta H2 = -39 kJ (from the equation 2Fe3O4(s) + CO2(g) -> 3Fe2O3(s) + CO(g))
Delta H3 = -18 kJ (from the equation 3FeO(s) + CO2(g) -> Fe3O4(s) + CO(g))

Delta H for the desired reaction = -23 kJ + (-39 kJ) + (-18 kJ)
Delta H for the desired reaction = -80 kJ

Therefore, the delta H for the desired reaction FeO(s) + CO(g) -> Fe(s) + CO2(g) is -80 kJ.

i got -1/3 for delta H. thank you very much, you have been a big help.