the molecularity of both Mo and N in the starting materials MoCl5 and NH3 are 1:1. So you just mix 1 Mole of NH3 which is equal to 17g , and 273g MoCl5  the molar ratio will be 1:1.

Hi Angela

Agreed, the literature around the formation of MoN from MoCl5 by thermal ammonolysis is rather vague. Since Mo is reduced on going from MoCl5 to MoN, I think the following equation is probably valid.

2MoCl5 + 4NH3 --> 2MoN + 10HCl + H2 + N2

A number of papers report the formation of HCl as a byproduct of MoN synthesis. At lower reaction temperatures, NH4Cl may form as a byproduct (see attached paper), in which case the following formula is probably valid.

2MoCl5 + 14NH3 --> 2MoN + 10NH4Cl + H2 + N2

Best of luck for your ongoing work

Geoff