Question

What is the enthalpy change for the first reaction?

Fe2O3(s) → 2Fe(s) + 3/2O2(g) ΔH =

4Fe(s) + 3O2(g) → 2Fe2O3 (s) ΔH = -1,652 kJ

I think I'm supposed to rearrange the equation and double everything but enthalpy isn't my strong suit

Answer 1

You can do this two ways.
The first is to take equation 1 and use
dHrxn = (n*dHf products) - (n*dHf reactants). When you look these up you will find dH Fe is zero and dH O2 is zero so it is a simple calculation and you don't have anything else to do.
OR
you can recognize that equation 1 is the reverse of the heat of formation which is written as equation 2 EXCEPT that heat of formation actually is
2Fe + 3/2O2 ==> Fe2O3 so what is written as equation 2 is twice heat formation. Take half of that to get -851 kJ and change the sign to make it +851 kJ for the reverse and that will be kJ for equation 1.

Answer 2

Consider the reaction:

2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)

The ΔHf for Fe2O3(s) = -824.3 kJ/mole. The ΔHf for Al2O3(s) = -1675.7 kJ/mole.

Finish the equation.

ΔHrxn = [(1)( kJ/mole) + (2)( kJ/mole)] - [(1)( kJ/mole) + (2) ( kJ/mole)]

Answer 3

It's actually more simple than that.
Divide the entire equation by two in order to make the Fe2O3 be equal to one mol of the substance.
"(2Fe(s) + 3/2 O2(g) → Fe2O3 (s) ΔH = -826 kJ)"

This will also divide the enthalpy by 2 (-1652/2 = -826). Then, use the molar mass of Fe, times the mol-to-mol ratio of Fe and Fe2O3, times conversion factor provided by the enthalpy of the reaction (-826 KJ/1 mol)

11.2g Fe (1 mol/55.85)(1 Fe2O3 mol/ 2 Fe mol)(-826 KJ/1 mol)
11.2g (1/55.85)(1/2)(-826/1) = -82.8KJ

Hope it helps!