Question

If the reaction of 150. G of ammonia with 150. G of oxygen gas yields 87. G of nitric oxide (no), what is the percent yield of this reaction?.

Answer 1

The percentage yield of the given reaction is 77.33%.

What is percentage yield?
Reactants frequently produce fewer product quantities than predicted by the chemical reaction's formula. The percentage of a theoretical yield that's been produced in a reaction is calculated using the percent yield formula. Working through a stoichiometry problem yields the theoretical yield, which is the ideal quantity of the final product. The actual yield is determined by calculating the volume of the product formed. We can determine the percentage yield by dividing the actual yield by the theoretical yield.

Moles is calculated by using the formula:
Moles of Ammonia:
Given mass of ammonia = 150g
Molar mass of ammonia = 17 g/mol
Putting values in above equation, we get:

Moles of Oxygen
Given mass of oxygen = 150g
Molar mass of oxygen = 32 g/mol
Putting values in above equation, we get:
For the given chemical equation:

By Stoichiometry,
5 moles of oxygen reacts with 4 moles of ammonia.
So, 4.6875 moles of oxygen will react with =  of ammonia
As, moles of ammonia required is less than the calculated moles. Hence, ammonia is present in excess and is termed as excess reagent.
Therefore, oxygen is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the given reaction:
5 moles of oxygen gas produces 4 moles of nitric oxide
So, 4.6875 moles of oxygen gas will produce =  of nitric oxide

Now, to calculate the theoretical amount of nitric oxide, we use equation 1: Molar mass of nitric oxide = 30 g/mol
 Given mass of nitric oxide = 112.5 g
Now, to calculate the percentage yield, we use the formula:
Experimental yield = 87 g
Theoretical yield = 112.5 g
Putting values in above equation, we get:
Hence, the percentage yield of the given reaction is 77.33%.

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