For the reaction 6 li + n2 → 2 li3n , what is the maximum amount of li3n (34.8297 g/mol) which could be formed from 14.18 mol (6.941 g/mol) of li and 16.37 mol of n2 (28.0134 g/mol)? answer in units of mol.

Moles Li = 3.50 g / 6.941 g/mol= 0.504
the ratio between Li and N2 is 6 : 1
moles N2 required = 0.504 /6=0.0840
we have 3.50 g / 28.0134 g/mol=0.125 moles of N2 so N2 is in excess
the ratio between Li and Li3N is 6 : 2
moles Li3N = 0.504 x 2 /6=0.168
mass Li3N = 0.168 mol x 34.8297 g/mol=5.85 g

Outermost (valence) electrons

The teacher is mostly talking about the distillation of a mixture