Question

Ammonia (NH3) can be produced by the reaction of hydrogen gas with nitrogen gas:

3H2 + N2 = 2NH3

A chemist reacts 2.00 mol H2 with excess N2. The reaction yields 0.54 mol NH3. What is the percent yield of the reaction?

a) 25%
b) 40%
c) 60%
d) 80%

Answer 1

Answer: The correct option is b.

Explanation: To calculate the percentage yield, we use the formula:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100` ....(1)

For a given reaction:

`3H_2+N_2\rightarrow 2NH_3`

• Experimental yield calculations:

0.54 moles of ammonia is formed.

So, amount of ammonia formed will be calculated using the formula:

`Moles=\frac{\text{given mass}}{\text{Molar mass}}` ....(2)

Molar mass of
`NH_3` = 17.031g/mol

`0.54=\frac{\text{given mass}}{17.031}`

Amount of
`NH_3` = 9.197g

Experimental yield : 9.197 g

• Theoretical yield calculations:

By Stoichiometry of the reaction,

Here, limiting reagent is hydrogen gas because it limits the formation of product.

3 moles of hydrogen gas is producing 2 moles of ammonia

So, 2 moles of hydrogen gas will produce =
`(2)/(3)* 2=1.33` moles of ammonia.

Amount of ammonia is calculated by using equation 2, we get:

`1.33moles=\frac{\text{Given mass}}{17.031g/mol}`

Theoretical yield of ammonia =22.65 grams

Now, putting values in equation 1, we get:

`\%\text{ yield}=(9.197)/(22.65)* 100`

% yield=40.60 %

Hence, the correct option is b.

Answer 2

b) 40%
The balanced equation indicates that for every 3 moles of H2 used, 2 moles of NH3 will be produced. So the reaction if it had 100% yield would produce (2.00 / 3) * 2 = 1.333333333 moles of NH3. But only 0.54 moles were produced. So the percent yield is 0.54 / 1.3333 = 0.405 = 40.5%. This is a close enough match to option "b" to be considered correct.