Katy, as I figure it:
.delta.U = Q - W (right? and you know which signs heat and volume flow are, right?)
Q = -5.51083 * 10^5J
W = .delta.PV= 1 atm * (final volume - initial volume)
initial volume = that of 1.8 mol of oxygen (note: this is assumed to be an experimental value, hence #sig. figs = 2 only!) + that of the Fe to be consumed according to:
4 Fe + 3 O2 = 2 Fe2O3
assume mild steel density = 7.85 g cm^-3
then 1.8 mol O combusts 1.2 mol Fe (the question is a little ambiguous: does it mean 1.8 mol of oxygen as gas in the standard state, i.e. as O2? Or as O atoms? Will assume the latter, and check answer against literature values, which is what you should do if you ever consider a problem ambiguous!)
That's 0.9 mol O2 gas, PV = nRT = -0.9*8.31451J K^-1 mol^-1 * (273.15 + 28.7)K = -2258.76135915 J
Original steel volume (neglect the carbon content, why not) = 1.2 mol * 55.845 g mol^-1 / 7.85 g cm^-3
= 8.54 cm^3
Final rust volume = 0.6 mol Fe2O3 * 159.69 g mol^-1 / 5.242 g cm^3 = 18.28 cm^3
So solids expansion (Fe -> Fe2O3) adds V = (18.28-8.54) = 9.74 cm^3, equivalent to +0.99 J of PV work (you may have noticed this effect -- a substantial layer of rust on steel takes up quite a lot of space!)
So putting it all together we have
.delta.U = -5.51083*10^5J + 2258.76J - 0.99J = -5.4882523*10^5J. Notice: heat LOST is internal energy LOST by the system (there it went, out to the surround) (.delta.U is negative); volume LOST is internal energy GAINED by the system (if this seems hard to remember, think of volume GAINED is internal energy LOST, b/c the system had to PRODUCE work to push the piston out, and when you do work, you lose energy = become exhausted!).
Really, we should round to -5.5*10^5J -- why?
compare this with literature .delta.H formation Fe2O3 = -826kJ mol^-1 -> -4.96 * 10^5 J for 0.6 mol Fe2O3
generated. We're clearly in the right ballpark (i.e., assumption above was correct), although calorimetry *should* be more accurate than this!