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What mass of lithium is required to react completely with 59.9 mL of N2 gas at STP?
Lithium reacts with nitrogen gas according to the following reaction:
6Li(s)+N2(g)→2Li3N(s)
What mass of lithium is required to react completely with 59.9 mL of N2 gas at STP?
1 Answer
- anonymousLv 74 years ago
Use the fact that 1 mole of any gas at STP occupies 22.4 L = 22,400 mL
The first two factors below give you the number of moles of N₂ and then it's using the balanced equation followed by a molar mass conversion.
(59.9 mL N₂) * (1 mole N₂ / 22,400 mL) * (6 moles Li / 1 mole N₂) *
(6.94 g Li / 1 mole Li)
= 0.111 g Li