# Help me find the percentage of the volume of nitrogen gas N2 (g) and hydrogen gas H2(g)?

The reaction for the Haber process, the industrial production of ammonia, is:

N2(g) + 3H2(g) 🡪 2NH3 (g) When the equilibrium arrived, the NH3(g) has a 25% volume of the Haber. Find the percentage of the volume of nitrogen gas N2 (g) and hydrogen gas H2(g), respectively.

### 2 Answers

- Trevor HLv 76 months agoFavorite Answer
The balanced equation:

N2(g) + 3H2(g) → 2NH3(g)

1 mol N2 reacts with 3 mol H2 to produce 2 mol NH3

For gases , moles = volumes

1 volume N2 reacts with 3 volumes H2 to produce 2 volume NH3

Now you ask for the % volumes when equilibrium has been achieved:

Let us say that you have Haber reactor that has a volume of 100 L

You load the reactor with 25 L of N2 and 75 L of H2

In going to completion :

25 L N2 reacts with 75 L H2 to produce 50 L NH3

The reaction proceeds until an equilibrium is achieved , when the NH3 is 25% of the volume of the reactor.

This means that 25 L of NH3 has been produced.

.

We know that 25 L N2 produces 50 L NH3 at completion

Therefore 12.5 L N2 produced 25 L NH3 at equilibrium

and 37.5 L H2 produce the 25 L NH3

Therefore , remaining unreacted :

25L - 12.5 L N2 = 12.5 L N2

75 L - 37.5 L H2 = 37.5 L H2

% composition of the gases in the reactor at equilibrium:

NH3 = 25 L = 25%

N2 = 12.5 L = 12.5%

H2 = 37.5 L = 37.5%

- oeman50Lv 76 months ago
At a constant final pressure in the reaction vessel, the volume % is equal to the mole %. Since the NH3 is occupying 25%, the N2 and the H2 are occupying 75%, with 1 mole of N2 and 3 moles of H2 for a total of 4 moles.

For the N2:

1/4 X 75% = 18.75%

For the H2:

3/4 X 75% = 56.25%