I wanted to calculate the theoretical capacity of NiMnO3, which will be use as anode material for LIB. It is very hard to get it through literature. So

I have calculated it like this:

NiMnO3 + 6Li + 6e = Ni + Mn + 3Li2O 

n x f / M.W x 3600

= 6 x 96485 / 161.63 x 3600 =994.9 mAh/g

MW of NiMnO3 = 161.63 g/mol

Electron involved = 6 e-

Please suggest.

Thanks

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