Lithium reacts with nitrogen gas according to the reaction:
6 Li(s) + N2(g) → 2 Li3N(s)
How many grams of lithium are required to completely react with 58.5 mL of N2 gas measured at STP?
Volume of N2 is 58.5 ml
We have to calculate the amount of Li requires to complete the reaction.
6Li + N2 ------> 2 Li3N
1mole N2 gas=28g
1mole N2 gas occupies 22.4L at STP
therefore 28g N2= 22.4 L or 22400 ml
Thus 58.5 ml of N2 is,
58.5ml N2 28g/22400 ml = 0.073 g N2
The atomic weight of Li = 6.94g/m
Hence the amount of Li requires to complete the reaction can be calculated as,
0.073gN2/28g/m6Li/N26.94g/m Li = 0.109g Li
Thus 0.109g Li requires to complete the reaction