Question

The reaction of magnesium with nitrogen produces magnesium nitride, as follows.

3 Mg(s) + N₂(g) → Mg₃N₂(s)

If the reaction is started with 2.70 mol Mg and 0.805 mol N₂, find the following.

What is the the limiting reactant and the number of moles of magnesium nitride produced?

Answer 1

3 Mg(s) + N2(g) → Mg3N2(s)

moles of Mg in th abov balanced reaction = 3

moles of N in th abov balanced reaction = 1

Mole ratio Mg:N = 3:1

It means 1 mole of N Reacts with 3 moles of Mg

Moles of Mg and N given = 2.70 and 0.805 respectively

Mole ratio = 2.70:0.805 = 3.348:1

((HINT: Multiply both 2.7 and 0.805 by 1.24. So that these may be convrted to simple integers)

According to the stiochiometery of the reaction MOUNT OF N IS LESS. Hence N is the LIMITTING REAGENT.

(b) One mole of N gives one moles of Mg₃N₂(s).

Hence 0.805 mols of N will give 1 x 0.805 = 0.805 moles of Mg₃N₂

(HINT:moles of Mg3N2 formed = Moles of Mg3N2 formed by one mole of N X Moles of N)