Hess's Law problem, Please help

Question

Given the following data:

Fe2O3 (s) + 3 CO (g) → 2 Fe (s) + 3 CO2 (g) ΔHO = -23.4 kJ

3 Fe2O3 (s) + CO (g) → 2 Fe3O4 (s) + CO2 (g) ΔHO = -48.3 kJ

Fe3O4 (s) + CO (g) → 3 FeO (s) + CO2 (g) ΔHO = +21.8 kJ

Use Hess’s law to calculate ΔHO for the reaction:

FeO (s) + CO (g) → Fe (s) + CO2 (g)

Answer 1

The easiest way to begin is to switch the equations (and change the enthalpy sign!) to match the reaction.

So, we can tell that the third reaction: Fe₃O_{4 (s)} + CO → 3FeO_(s) + CO_{2 (g)} ΔH= 21.8 needs to flipped to this.

3 FeO_(s)+ CO₂ → Fe₃O_{4 (s)} + CO ΔH= -21.8

Looking down the list we can see that second reaction needs to be flipped to cancel out the Fe₃O₄ from the third equation. 
2 Fe₃O₄ + CO₂ → 3 Fe₂O₃ + CO ΔH= 48.3

Your full reaction synthesis will now look like this in:
3 FeO + CO₂ → Fe₃O₄ + CO

2 Fe₃O₄ + CO₂ → 3 Fe₂O₃ + CO

Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂
Now comes hard part. My preference is to tally how much of each species is on the reactant and product side.

Reactant side: 6 FeO, 2 CO₂, 2 Fe₃O₄, 1 Fe₂O₃, 3 CO

Product side: 3 CO₂, 1 Fe₃O₄, 3 Fe₂O₃, 2 Fe ,2 CO

So it looks like to cancel out the Fe₂O₃ and Fe₃O₄, we'll need to find the least common denominator, which is 2 Fe₃O₄ and 3 Fe₂O₃. So you multiply the first reaction by 2 and the third reaction by 3. 

The overall reaction will look like this when put together:

6 FeO + 3 CO₂ + 9 CO + 2 Fe₃O₄ + 3 Fe₂O₃ → 6 Fe + 3 CO + 9 CO2 + 2 Fe₃O₄ + 3 Fe₂O₃

However, when simplified, it becomes:

 

6 FeO + 6 CO → 6 Fe + 6 CO₂

 

Which is the overall net reaction, by a factor of six.

Now, we can calculate the change in enthalpy (finally!), which is just adding up the ΔH values (keeping in mind the switching of signs and the multiplication) and dividing by six:

ΔH= ((2*-21.8)+48.3 + (3*-23.4))/6= -10.9 kJ

So the final answer would ΔH= -10.9 kJ