Question

Question 2 of 10 >

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia,
8H8
N2(g) + 3 H2(g) — 2NH,(8)
Assume 0.170 mol N, and 0.561 mol H, are present initially,
After complete reaction, how many moles of ammonia are produced?
NH,
How many moles of H, remain?

Answer 1

Moles of H₂ left = 0.051 mol

Moles of ammonia formed = 0.34 mol

Step-by-step explanation:

Given data:

Moles of N₂ = 0.170 mol

Moles of H₂ = 0.561 mol

Moles of ammonia formed = ?

Moles of H₂ left = ?

Solution:

Balance chemical equation:

N₂ + 3H₂ → 2NH₃

Now we will compare the moles of ammonia with nitrogen because nitrogen is limiting reactant and limit the yield of ammonia.

N₂ : NH₃

1 : 2

0.170 : 2×0.170 = 0.34 mol

Moles of ammonia formed = 0.34 mol

Moles of H₂ reacted:

N₂ : H₂

1 : 3

0.170 : 3/1×0.170 = 0.51 mol

0.51 moles of hydrogen react with 0.170 moles of nitrogen.

Moles of H₂ left:

Moles of H₂ left = Total - moles reacted

Moles of H₂ left = 0.561 mol - 0.51 mol

Moles of H₂ left = 0.051 mol