I thought I'd look around for more on this reaction, and I see both products given. Here's a page that gives MgO and NH3 as the products: http://dl.clackamas.cc.or.us/ch104-04/fourth.htm. Here's one that says Mg(OH)2 and NH3 are the products: http://en.wikipedia.org/wiki/Ammonia. Possibly MgO forms during the course of the reaction if both protons of one water molecule are transferred to the nitride ion. Leaving off the Mg2+ ion, we would have: H2O + N3- 
NH2- + OH-
OH- + NH2- 
NH2- + O2-
This gives oxide ion, O2-, in the form of MgO, plus amide ion, NH2-, also with Mg2+ as its counterion. Then another water molecule could react with the amide ion to form ammonia:
H2O + NH2- 
NH3 + OH-
But I doubt if the reaction could be controlled to give the oxide, MgO, as the only product. Even if MgO forms, it would react with another water molecule, if available, to give Mg(OH)2. Possibly MgO results as long as excess water is not added. But for sure, in the actual experiment, after you add the water (5 to 10 drops, which is in excess with respect to the very small amount of Mg3N2 present), you have to reheat in order to decompose the Mg(OH)2. Interesting question!
>> I think they need to attend your class <<
Hey thanks... Bring 'em on! 
Steve
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