Question

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?

Answer 1

Answer: The percentage yield of the given reaction is 77.33%.

Explanation:

Moles is calculated by using the formula:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

• Moles of Ammonia:

Given mass of ammonia = 150g

Molar mass of ammonia = 17 g/mol

Putting values in above equation, we get:

$\text{Moles of ammonia}=\frac{150g}{17g/mol}=8.82moles$

• Moles of Oxygen

Given mass of oxygen = 150g

Molar mass of oxygen = 32 g/mol

Putting values in above equation, we get:

$\text{Moles of oxygen}=\frac{150g}{32g/mol}=4.6875moles$

For the given chemical equation:

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

By Stoichiometry,

5 moles of oxygen reacts with 4 moles of ammonia.

So, 4.6875 moles of oxygen will react with = $\frac{4}{5}\times 4.6875=3.75moles$ 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 = $\frac{4}{5}\times 4.6875=3.75moles$ of nitric oxide

Now, to calculate the theoretical amount of nitric oxide, we use equation 1:

Molar mass of nitric oxide = 30 g/mol

$3.75mol=\frac{\text{Given mass}}{30g/mol}$

Given mass of nitric oxide = 112.5 g

Now, to calculate the percentage yield, we use the formula:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 87 g

Theoretical yield = 112.5 g

Putting values in above equation, we get:

$\%\text{ yield}=\frac{87}{112.5}\times 100=77.33\%$

Hence, the percentage yield of the given reaction is 77.33%.