Ammonia, NH3 (Delta. Hf = –46.2 kJ), reacts with oxygen to produce water (Delta. Hf = –241.8 kJ) and nitric oxide, NO (Delta. Hf = 91.3 kJ), in the following reaction:

4N H 3 (g) + 5 O2 (g) ---> 6 H 2O (g) + 4O (g).

What is the enthalpy change for this reaction?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
–900.8 kJ
–104.6 kJ
104.6 kJ
900.8 kJ