Determine :delta:H, q, w, and :delta:E at 298 K and 1 atm for the complete reaction of 8.610 g of N2O4.

N2O4(g) + 2N2H4(l) 3N2(g) + 4H2O(l).

The heat of formation of liquid hydrazine (N2H4) is 50.63 kJ/mol. Use tabulated data for other heats of formation.

The tolerance on each question is only 0.03 kJ, so express all answers to 0.01 kJ.

Wow I have been stuck on this problem for an hour now. I can't get the :delta:H right for some reason. I looked up the values and N2O4's heat of formation is 9.66, N2H4 as stated in problem is 50.63, N2 is 0 and H20 is -285.83.

Now to get :delta:H I did products heat of formation minus the reactants and I still can't get it.

Heres what it looks like:

(3(0) + 4(-285.83)) - (9.66 + 2(50.63)) = -1254.24 kJ

It is wrong. What am I doing wrong?