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
