Back with another question similar to the one I asked before, but this one's a bit more confusing

Question

4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H= –909 kJ/mol

2NO(g) + O2(g) → 2NO2(g) ∆H= –115 kJ/mol

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

Thank you so much for dealing with this question :D

Answer 1

eqn 1 + 2*eqn 2 = ? See if doing this doesn't give you the correct equation you want.

Ignore eqn 3. You don't need it.
A favorite trick when profs want you to think REALLY hard. Eqn 3 is a red herring.