For the nitrogen fixation reaction 3H2(g) + N2(g) = 2NH3(g), Kc = 6.0 x 10-2 at 500°C. If 0.253 M H2 and 0.044 M NH3 are
present at equilibrium, what is the equilibrium concentration
of N2?

The equilibrium concentration  of N₂ : 1.992

Given

Kc = 6.0 x 10⁻² at 500°C

0.253 M H₂ and 0.044 M NH₃

Reaction

3H₂(g) + N₂(g) = 2NH₃(g)

Required

The equilibrium concentration  of N₂

Solution

Kc for the reaction :

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.