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 2.05 mol Mg and 0.891 mol N₂, find the following.

(a) the limiting reactant (b) the excess reactant (c) the number of moles of magnesium nitride produced

Answer 1

3 Mg(s) + N₂(g) → Mg₃N₂(s)

from the equation;

3 moles of Mg metal will react with 1 mole of N2 will produce 1 mole of Mg3N2.

No. of moles of each Mg and N2 present in the reaction are 2.05 mol and 0.891 mol respectively.

Mg needed to react = (1/3) x 2.05 = 0.683 moles

no. of moles of N2 present (0.891 mol) greater than no. of moles of Mg.

One which is present in the lower no. of moles will be the limiting reagent.

a)

So, limiting reagent is Mg

b)

Excess reagent (N2) present is = 0.891 -(1/3 x 2.05) = 0.891 - 0.683 = .2076 moles

weight of N2 present = 0.2076 x 28 = 5.815 grams

c)

No. of moles of Mg3N2 formed = 0.683 mole