PROBLEM 96P

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?

Solution 96P

Given:-

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