Ammonia burns in oxygen

The burning of ammonia in oxygen to give nitrogen and steam is very exothermic, and thus the water formed is in gaseous state. The balanced equation should be:
4NH3­(g) + 3O2(g) → 6H2O(g) + 2N2(g)

Volume ratio NH3 : O2 = Mole ratio NH3 : O2 = 4 : 3
Volume of NH3 added = 0.5 L
Volume of O2 added = 0.2 L
Volume of NH3 needed to react O2 completely = 0.2 x (4/3) = 0.267 L
Hence, NH3 is in excess, and O2 is the limiting reactant (completely reacted).

Mole ratio O2 : H2O : N2 = 3 : 6 : 2

Volume of H2O formed = 0.2 x (6/3) = 0.4 L
Volume of N2 formed = 0.2 x (2/3) = 0.133 L

This is the Ostwald process:
http://en.wikipedia.org/wiki/Ostwald_process

The state of H2O should be gaseous, becoz the Q states that steam is produced, i.e.
4NH3(g) + 3O2(g) ---> 6H2O(g) + 2N2(g)