Energetics Question Help

Any ideas how to attack this monster..?
5. Nitric acid is produced industrially from ammonia, air and water using the following sequence of reactions:

(1) 4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = –909 kJ mol−1
(2) 2NO(g) + O2(g) → 2NO2(g) ∆H = –115 kJ mol−1
(3) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) ∆H = –117 kJ mol−1
Which is the enthalpy change (in kJ mol−1) for the following reaction?
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
A −679
B −794
C −1024
D −1139

thanks

using data from equation 1 and 2 enthalpy change can be calculated Enthalpy change= -909 + 2(-115) = -1139 KJ/mol

Any ideas how to attack this monster..?
5. Nitric acid is produced industrially from ammonia, air and water using the following sequence of reactions:

(1) 4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = –909 kJ mol−1
(2) 2NO(g) + O2(g) → 2NO2(g) ∆H = –115 kJ mol−1
(3) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) ∆H = –117 kJ mol−1
Which is the enthalpy change (in kJ mol−1) for the following reaction?
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
A −679
B −794
C −1024
D −1139

thanks

Firstly, you don't need equation 3- that is a red herring

Multiply equation 2 by 2 to get 4NO +2O2 --> 4NO2 delta H for this is 2 x -115Kj = -230KJ
Then combine this with equation 1, the 4NO's appear on both sides and cancel out to give:

4NH3 + 7O2 --> 4NO2 + 6H20 delta H is -230 + (-909) = -1139

See attachment for details

love your method, but how is it possible that we can cancel both sides, wouldn't the NO2 of both sides still change the entahlpy? is it because you don't need no enthaply to change no2 to no2?