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Liono4ka [1.6K]
3 years ago
8

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.
Chemistry
1 answer:
Yuki888 [10]3 years ago
4 0
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
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2)  0.0300

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