Consider the following reaction: 4NH₃(g) + 7O₂(g) $\rightarrow$ 4NO₂(g) + 6H₂O(l)
Consider an experiment in which you react ammonia and oxygen. At the end of the experiment, you find that you produced 27.0 g of water, and 8.52 g of ammonia is left over. Calculate the initial mass of ammonia. Assume the reaction went to completion.

Question

Consider the following reaction: 4NH₃(g) + 7O₂(g)

\rightarrow

4NO₂(g) + 6H₂O(l)
Consider an experiment in which you react ammonia and oxygen. At the end of the experiment, you find that you produced 27.0 g of water, and 8.52 g of ammonia is left over. Calculate the initial mass of ammonia. Assume the reaction went to completion.

Quizplus · Accepted Answer

The balanced chemical equation for the reaction between ammonia and oxygen is:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

From the balanced equation, we can see that 4 moles of ammonia react with 5 moles of oxygen to produce 6 moles of water. Therefore, the theoretical yield of water from the given 8.52 g of ammonia is:

(6/4) x (8.52 g / 17.03 g/mol) x ( 18.02 g/mol) = 27.0 g

where 17.03 g/mol and 18.02 g/mol are the molar masses of NH3 and H2O respectively.

This means that all the ammonia has reacted to give the observed 27.0 g of water. Therefore, the initial mass of ammonia which reacted with oxygen is:

8.52 g

However, the question asks for the initial mass of ammonia before the reaction. To get that, we need to add the mass of ammonia which did not react:

Initial mass of ammonia = 8.52 g + mass of unreacted NH3

The mass of unreacted NH3 can be calculated by subtracting the mass of NH3 which reacted from the initial mass of NH3:

Initial mass of ammonia = 8.52 g + (mass of NH3 which did not react)

Initial mass of ammonia = 8.52 g + [(initial mass of NH3 - final mass of NH3)]

We don't know the final mass of NH3, but we know that all of the ammonia did not react. Therefore:

final mass of NH3 = 0 g

Substituting this value in the equation above, we get:

Initial mass of ammonia = 8.52 g + [(initial mass of NH3 - 0 g)]

Initial mass of ammonia = 8.52 g + initial mass of NH3

Initial mass of ammonia = 8.52 g + x (say)

Now, we know that the theoretical yield of water from the given mass of ammonia is the same as the observed yield of water. Therefore:

(6/4) x (8.52 g / 17.03 g/mol) x ( 18.02 g/mol) = x + 8.52 g

Solving for x, we get:

x = 25.5 g

Therefore, the initial mass of ammonia was 25.5 g.

Consider the Following Reaction: 4NH3(g) + 7O2(g) →\rightarrow→ 4NO2(g) + 6H2O(l) Consider an Experiment in Which You React

Consider the Following Reaction: 4NH₃(g) + 7O₂(g) $\rightarrow$ 4NO₂(g) + 6H₂O(l)
Consider an Experiment in Which You React