Question

Consider the following reaction in a closed container

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

If initially 68 g ammonia and 192 g oxygen is present, calculate the approximate mass % NO(g) formed at 80% completion of the reaction.


28%


45%


25%


37%

Answer 1

25

Answer:

25

Explanation:

Answer 2

The correct option is 37%

GIVEN

Initially 68 g ammonia and 192 g oxygen is present.

TO FIND

Approximate mass % NO(g) formed at 80% completion of the reaction.

SOLUTION

We can simply solve the above problem as follows;

We can observe from the reaction;

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Given mass of Ammonia = 68 grams

Moles of Ammonia = 68/17 =4 moles.

Given mass of O₂ = 192 grams

Moles of Oxygen given = 192/32 = 64 moles.

4 moles of NH₃ gives 4 moles of NO

5 moles of O₂ Will give 4 moles

64 moles of O₂ will give, (4/5)× 64 = 51.2 grams of NH₃

So, Ammonia is the limiting reagent.

Now,

Moles of H₂O formed from 4 moles of NH₃ = 6 moles

It is given that reactions is only 80% completion

= 80% of 68 grams of NH₃ = 54.4 grams of NH₃ will react in the reaction.

Number of moles of NH₃ = 3.2 moles

Therefore,

Number of Moles of NO formed from 3.2 moles of NH₃ = 3.2 moles = 96 grams Of NO

Number of H₂O formed from 3.2 moles of NH₃ = 4.8 moles = 86.4 moles of H₂O

Mass% of NO = (96/260)× 100 = 36.9% ≈ 37%

Hence, The correct option is 37%

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