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

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.

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!

To determine the enthalpy change for the first reaction (Fe2O3(s) → 2Fe(s) + 3/2O2(g)), we can use the concept of Hess's Law and manipulate the given equations to obtain the desired reaction.

The given equations are:
1. 4Fe(s) + 3O2(g) → 2Fe2O3 (s) ΔH = -1,652 kJ (Equation 1)
2. Fe2O3(s) → 2Fe(s) + 3/2O2(g) ΔH = ? (Equation 2)

To obtain Equation 2, we need to reverse and double Equation 1. Here's how you can do it step-by-step:

1. Reverse Equation 1: Multiply the equation by -1, which changes the sign of ΔH.
-4Fe(s) - 3O2(g) → -2Fe2O3 (s) ΔH = 1,652 kJ

2. Double Equation 1: Multiply both the reactants and products by 2.
-8Fe(s) - 6O2(g) → -4Fe2O3 (s) ΔH = 3,304 kJ

Now, Equation 2 is the desired reaction, but we need to find the enthalpy change for this reaction. Since we manipulated Equation 1 to obtain Equation 2, the enthalpy change for Equation 2 will be the negative of the enthalpy change for Equation 1.

Therefore, the enthalpy change for the first reaction (Fe2O3(s) → 2Fe(s) + 3/2O2(g)) is +3,304 kJ.

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)]