Question

Ammonia (NH3) reacts with oxygen to form nitric oxide (NO) and water vapor: 4NH3 + 502 4NO + 6H2O b) When 20.0 g NH3 and 50.0 g O2 are allowed to react, which is the limiting reagent? Which reactant is left over and how much left over at the end of reaction? Ans: 2.94 g c) What is the theoretical yield of NO if 20.0 g NH3 and 50.0 g O2 are allowed to react? Ans: 35.29 g d) What is the theoretical yield of H20 if 20.0 g NH3 and 50.0 g O2 are allowed to react? Ans: 31.74 g

Answer 1

Answer: a) . Ammonia is the limiting reagent

b. Oxygen is left over and 0.1375 g of oxygen is left over.

c. The theoretical yield of $NO$ is 35.29 g.

d. The theoretical yield of $H_2O$ is 31.74 g.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

For $NH_3$

Given mass of ammonia = 20.0 g

Molar mass of ammonia = 17.031 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{20.0g}{17.031g/mol}=1.17mol$

For $O_2$

Given mass of oxygen gas = 50.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{50.0g}{32g/mol}=1.6mol$

The chemical equation for the reaction is

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

By Stoichiometry of the reaction:

4 moles of ammonia reacts with = 5 moles of oxygen

So 1.17 moles of ammonia will react with = $\frac{5}{4}\times 1.17=1.4625mol$ of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent and (1.6-1.4625)= 0.1375 g of oxygen is left unreacted.

Thus ammonia is considered as a limiting reagent because it limits the formation of product.

1. By Stoichiometry of the reaction:

4 moles of ammonia produces = 4 moles of $NO$

1.17 moles of ammonia will produce = $\frac{4}{4}\times 1.17=1.17moles$ of $NO$

Mass of $NO=moles\times {\text{Molar Mass}}=1.17\times 30=35.29g$

Thus Theoretical yield of $NO$ is 35.29 grams.

2. By Stoichiometry of the reaction:

4 moles of ammonia produces = 6 moles of $H_2O$

1.2 moles of ammonia will produce = $\frac{6}{4}\times 1.2=1.8moles$ of $H_2O$

Mass of $H_2O=moles\times {\text{Molar Mass}}=1.8\times 18.015=31.74g$  $H_2O$

Thus Theoretical yield of $H_2O$ is 31.74 grams.