How many grams of Li3N are produced when 25 grams of N2 react with excess Li?

Question

Balanced form: 6 Li + N2 = 2 Li3N

Answer 1

Basic stoichiometry

25g N₂/1 x [ 1 mol N₂ /28g N₂ ] x [2 mol Li₃N / 1 mol N_2] x [34.82g Li₃N / 1 mol Li₃N] = 62 g Li₃N

Answer 2

This is a straightforward stoichiometry problem. Starting from 25 grams N₂, we will covert thusly:

25 grams N₂ --(molar mass)-> moles N₂ --(mole ratio)-> moles Li --(molar mass)-> grams Li

With the conversion factors, and rounding to 2 significant figures, we end up with:

25 g N₂ * (1 mol N₂/28.014 g N₂) * (6 mol Li/1 mol N₂) * (6.941 g Li/1 mol Li) = 37 g Li

I hope this helped!