I want to get an stequiometry 1:1 (MoN) using NH3 gas and MoCl5 in powder form, so i want to know, how to get or calculate the relation of this two precursors to form my compound. I will really appreciate your contribution to this questions. Thank you.
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
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