Question

if the reaction occurs in the laboratory and produces 2.8 moles of ammonia, calculate the percent yield for the experiment. Use the quantity of the product that is produced by the limiting reagent as the theoretical yield.

Answer 1

The question is incomplete, here is the complete question:

The Haber process can be used to produce ammonia,  $NH_3$  and it is based on the following reaction.

$N_2+3H_2\rightarrow 2NH_3$

If the reaction occurs in the laboratory and 5 moles of each hydrogen and nitrogen gas reacts and produces 2.8 moles of ammonia, calculate the percent yield for the experiment. Use the quantity of the product that is produced by the limiting reagent as the theoretical yield.

Answer: The percent yield of the reaction is 84.08 %.

Explanation:

We are given:

Moles of nitrogen gas = 5 mole

Moles of hydrogen gas = 5 mole

For the given chemical equation:

$N_2+3H_2\rightarrow 2NH_3$

By Stoichiometry of the reaction:

3 moles of hydrogen gas reacts with 1 mole of nitrogen gas

So, 5 mole of hydrogen gas will react with = $\frac{1}{3}\times 5=1.67mol$ of nitrogen gas

As, given amount of nitrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of hydrogen gas produces 2 moles of ammonia

So, 5 moles of hydrogen gas will produce = $\frac{2}{3}\times 5=3.33mol$ of ammonia

To calculate the percentage yield of ammonia, we use the equation:

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

Experimental yield of ammonia = 2.8 moles

Theoretical yield of ammonia = 3.33 moles

Putting values in above equation, we get:

$\%\text{ yield of ammonia}=\frac{2.8mol}{3.33mol}\times 100\\\\\% \text{yield of ammonia}=84.08\%$

Hence, the percent yield of the reaction is 84.08 %.