Question

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

Answer 1

The answer would be 7

Answer 2

Answer:

15.64 moles

81.8% (3 s.f.)

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.

$\boxed{percentage \: yield = \frac{actual \: yield}{theoretical \: yield} \times 100\%}$

∴ Percentage yield of NH₃

$= \frac{12.8}{15.64} \times 100\%$

= 81.8% (3 s.f.)