Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.

Question

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.

N2(g)+3H2(g)⟶2NH3(g)

Assume 0.220 mol N2

and 0.705 mol H2

are present initially.

After complete reaction, how many moles of ammonia are produced?

NH3:____?____mol

How many moles of H2

 remain?

H2:___?____mol

How many moles of N2

 remain?

N2:____?____mol

What is the limiting reactant?

A.hydrogen

B.nitrogen

Answer 1

0.220 moles N₂ x 2 moles NH₃ = 0.44 moles NH₃

1mole N₂

0.705 moles H₂ x 2 mole NH₃ = 0.47 moles NH₃

3 moles H₂

The lowest amount tells you the maximum amount (theoretical yield) of the product that can be produced.

Therefore the amount of NH₃ that can be produced in this reaction is 0.44 moles.

Since N₂ yielded the least amount of product, then N₂ is the limiting reactant. Therefore, H₂ is the excess reactant

To find the amount of excess H₂. Use N₂ to find how much of the H₂ was used in the reaction.

0.220 moles N₂ x 3 moles H₂ = 0.66 moles NH₃

1mole N₂

Excess amount of H₂ = 0.705 moles - 0.66 moles

The excess amount of the reactant (H₂) is 0.465 moles.

Since N₂ is the limiting reactant, it was used up first in the reaction. Therefore 0 moles of N₂ remain.