Question

2. If 7.82 mol of nitrogen, N2, are reacted with excess hydrogen, what is the theoretical

yield of ammonia, NH3, in moles?
What is the percent yield of NH3 if the actual yield is 12.8 mol?

Answer 1

The answer would be 7

Answer 2

15.64 moles

81.8% (3 s.f.)

Step-by-step explanation:

Let's start by writing a balanced equation.

N₂ +H₂ → NH₃

To balance the equation, ensure that the number of atoms of each element is the same on both side of the arrow. On the left, we have 2 N atoms and only 1 N on the right. Thus, write '2' in front of NH₃ to balance the N.

N₂ +H₂ → 2NH₃

Now, balance the number of H atoms. Currently, there are 2 Hs on the left and 6 Hs on the right. To balance the equation, write a 3 in front of H₂.

N₂ +3H₂ → 2NH₃

The equation is now balanced.

`\boxed{N₂ +3H₂ → 2NH₃}`

Given that hydrogen is in excess, the number of moles of NH₃ is dependent on the number of moles of N₂, which is the limiting reactant.

The mole ratio of N₂ to NH₃ produced is 1: 2.

Thus with 7.82 mol of N₂,

number of moles of NH₃

= 2(7.82)

= 15.64 moles

This is the theoretical yield since the calculations were based from the chemical equation.

However, in reality, the percentage yield may not be 100% as some products are lost in the process.

`\boxed{percentage \: yield = (actual \: yield)/(theoretical \: yield) * 100\%}`

∴ Percentage yield of NH₃

`= (12.8)/(15.64) * 100\%`

= 81.8% (3 s.f.)