Hess's Law Problem Help

[Littledude]

I find these problem fun and relatively easy, but this one is stumping me and I'd like some help. This is for some extra credit chemistry homework.

Target equation:  FeO(s) + CO(g) -> Fe(s) + CO2(g)

3Fe2O3(s) + CO(g) -> 2Fe3O4(s) + CO2(g)       ΔH=-47.0 kJ
Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)           ΔH=-25.0 kJ
Fe3O4(s) + CO(g) -> 3FeO(s) + CO2(g)            ΔH=10.0 kJ

If you don't know what Hess's Law is but think you could help, what you have to do is arrange the 3 given equations, by multiplying them or flipping them or whatever algebra rules you want to use, so that when, combined the equations will cancel out to result in the target equation. Then you add up the ΔH values to find the ΔH of the target equation.

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[raegirl1]

Sorry, can't help ya. Looks like a bunch of soup to me.

[ChappersTeddy]

Sorry, can't help ya. Looks like a bunch of soup to me.

forget

[Deathwish]

Wow you people do Hess' law weirdly, don't you do the whole cycle stuff?

[Littledude]

Wow you people do Hess' law weirdly, don't you do the whole cycle stuff?

Um, no?

[Nom|Aplem|Nom]

I find these problem fun and relatively easy, but this one is stumping me and I'd like some help. This is for some extra credit chemistry homework.

Target equation:  FeO(s) + CO(g) -> 2Fe3O4(s) + CO2(g)

That's all I can help you with.
Do I get a sticker?

[Oasis]

I now cannot wait for Chemistry.

Those look beautiful.

[pocketrocket300]

You really arranged your mechanism weird.  The target reactants and the target products are both on the products side of the equation.  Are the reactions reversible?

[King Leo]

Your main reaction makes no sense at all. Are you sure you didn't mean
FeO(s) + CO(g) -> Fe(s) + CO2(g)   ?

[Littledude]

Your main reaction makes no sense at all. Are you sure you didn't mean
FeO(s) + CO(g) -> Fe(s) + CO2(g)   ?

woops I must've started reading the next line down. You're right.

[King Leo]

Without checking CO/CO2 balance, I would suggest
-2*rx3
-1*rx1
+3*rx2

Jupp, thats right. So your reaction enthalpy is then:
(-2*10)+(-1*-47)+(3*-25)=-48 kJ/6mol

So for 1 mol, just divide by 6,
Δ_rxH=-8 kJ/mol

[DrenDran]

Sorry, can't help ya. Looks like a bunch of soup to me.

Then don't post?

[Littledude]

Without checking CO/CO2 balance, I would suggest
-2*rx3
-1*rx1
+3*rx2

Jupp, thats right. So your reaction enthalpy is then:
(-2*10)+(-1*-47)+(3*-25)=-48 kJ/6mol

So for 1 mol, just divide by 6,
Δ_rxH=-8 kJ/mol

why thank you good sir