Ammonia Reaction with O2 I asked a similar question earlier that lead me to ask this question.
NH3 and O2 react in the following way $$\ce{4NH3 + 3O2 -&gt; 2N2 + 6H2O}$$  From my understanding Ammonia will explode(react quickly/release a lot of heat quickly is how I understand the term explode) when there is 15%-25% ammonia vapor in Air.

If I injected ammonia vapor into the air (above autoignition temperature of 651°C so that it's fuel to air ratio was more like 30% ammonia to air.  Would it follow this reaction slowly until it reached the 25% mark then explode/react quickly until it reached the 15% ammonia balance air, and then react slowly from 15% ammonia to lower?

Here is the link to my similar question. Meaning of Flammability Limits and reaction in Air of Ammonia
No.  Because remember, the oxygen is being consumed too.  If we assume it's reacting via that reaction (realistically, you're going to get a complex mix of products) then we start with something like: 30% NH3 56% N2 14% O2 (the mix is 70% air, which is an 80/20 mix of N2 and O2).  For every 4% reduction in Ammonia, you're getting a 3% decrease in oxygen.  So O2/NH3 reactions will be rarer (slower), and slow down heat generation.  Also, the amount of inert gas in the vessel will be increasing, which also need to be heated up, so less of the enthalpy of reaction will go into increasing the temperature.

That said, above the autoignition temperature, it's going to be a pretty fast reaction anyway...  What we can be sure of is it will stop reacting at about 10% NH3 when all the oxygen is gone.