The first step in the ostwald process for producing nitric acid is 4nh3(g) + 5o2(g) -> 4no(g) + 6h2o(g). 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?

The percentage yield of the given reaction is 77.33%.

Explanation:

Moles is calculated by using the formula:

Given mass of ammonia = 150g

Molar mass of ammonia = 17 g/mol

Putting values in above equation, we get:

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%.