[tomariemd]

*delete me* I am studying for the ACS exam and can't figure this out...

Combustion of ammonia is in the below equation:
4NH3 + 5O2>4NO +6H20  delta Hrxn -904.8kj

What is the enthalpy of formation of NH3??

Given entalhpy formation
NO 90.4
H20 -241.8

I know its production-reactants but no H for NH3???

[sdekivit]

you have the formation enthalpy for NO and H2O and the overall reaction enthalpy change.

So in order to form NO and H2O you need to break the bondings in the NH3 to convert NH3 into its elemnts: N2 and H2. The enthalpy change of this process is the negative value of the enthalpy of formation of NH3.

So your problem is actually very easy:

-(4 * enthalpy of formation of NH3) + (90.4 * 4) + (6 * -241.8 ) = -904.8