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.

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Chemistry

Eliezer

Today, 17:15

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.

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Andrew Bowman · Answer 1

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