Enthalpy Change using Enthalpies of Formation

Don't know how to tackle this problem, has been puzzling me for about 2 hours and I just don't understand!!!

I'm not interested in the final answer, I'm hoping to find out HOW to solve the equation.

Use the enthalpies of formation below to calculate the enthalpy change for the following reaction:

3 Fe(s) + 4 H2O(g) → 4 H2(g) + Fe3O4(s)

ΔH f: H2O(g) -242; Fe3O4(s) -1117 kJ mol-1

All help is highly appreciated!!!!!

∆Hrxn = ∆Hformation all products - ∆Hformation all reactants
the thing is this: elements don't have ∆Hformation values since they are elements
plus, using the balanced equation, you need to take kJ/mole and multiply by the number of moles of the compounds, for Fe3O4 = 1 and H2O = 4

∆Hrxn = ∆Hf Fe3O4 - ∆Hf H2O
∆Hrxn = -1117kJ - (4)(-242kJ) = -149kJ

it is the delta h of the products - the delta h of the reactants. you have to multiply the enthalpy of formation of each compound by their stoichiometric coefficients.

Here are some solved example problems using standard enthalpies of formation:

http://www.chemteam.info/Thermochem/Hess…