Question

A gas mixture of 20 moles nitrogen and 70 moles hydrogen is fed to a reactor in which ammonia is produced. Assuming that the reaction proceeds to completion, determine the composition of the exit gas stream both in mole and mass %. If 75% of nitrogen gets converted, what will be the corresponding fractions?

Answer 1

Complete reaction

mole %: 0% N₂; 20% H₂; 80% NH₃

Gram %: 0% N₂; 2,9% H₂; 97,1% NH₃

75% reaction:

mole%: 8,3% N₂; 41,7% H₂; 50% NH₃

Gram: 20% N₂; 7,2% H₂; 72,8% NH₃

Step-by-step explanation:

The global reaction in the reactor is:

N₂(g) + 3 H₂ (g) → 2 NH₃ (g)

For a complete reaction of 20 moles of N₂(g) you need:

20 moles N₂ ₓ
`(3 mol H_2)/(1 mol N_2)` = 60 moles of H₂(g)

So, 10 moles of H₂(g) will stay in the end.

The moles produced of NH₃ are:

20 moles N₂ ₓ
`(2 mol NH_3)/(1 mol N_2)` = 40 moles of NH₃(g)

Thus, the final composition in moles is:

0 moles N₂; 10 moles H₂; 40 moles of NH₃

The mole % is:

0% N₂; ¹⁰/₅₀ ₓ 100 = 20% H₂; ⁴⁰/₅₀ ₓ100 = 80% NH₃

In grams you have:

10 moles H₂
`(2,02 g)/(1 mol)` = 20,2 g of H₂

40 moles NH₃
`(17,03 g)/(1 mol)` = 681,2 g of NH₃

0% N₂;
`(20,2 g)/(701,4)` ₓ 100 = 2,9% H₂;
`(681,2 g)/(701,4)` ₓ100 = 97,1% NH₃

With a 75% of conversion you have that the moles produced of NH₃ are:

40 moles of NH₃(g) × 75% = 30 moles of NH₃

The necessary moles to produce these moles of NH₃ are:

30 moles NH₃ ₓ
`(3 mol H_2)/(2 mol NH_3)` = 45 moles of H₂(g)

So, 25 moles of H₂(g) will stay in the end.

30 moles NH₃ ₓ
`(1 mol N_2)/(2 mol NH_3)` = 15 moles of N₂(g)

So, 5 moles of H₂(g) will stay in the end.

Thus, the final composition in moles is:

5 moles N₂; 25 moles H₂; 30 moles of NH₃

The mole % is:

⁵/₆₀ ₓ 100 =8,3% N₂; ²⁵/₆₀ ₓ 100 = 41,7% H₂; ³⁰/₆₀ ₓ100 = 50% NH₃

In grams you have:

5 moles N₂
`(28 g)/(1 mol)` = 140 g of N₂

25 moles H₂
`(2,02 g)/(1 mol)` = 50,5 g of H₂

30 moles NH₃
`(17,03 g)/(1 mol)` = 510,9 g of NH₃

`(140 g)/(701,4)` ₓ 100 =20% N₂;
`( 50,5g)/(701,4)` ₓ 100 = 7,2% H₂;
`(510,9 g)/(701,4)` ₓ100 = 72,8% NH₃

I hope it helps!