Question

1. In an experiment, suppose 10.0 g of potassium iodide was used with excess lead (II) nitrate & 9.50 g of precipitate was collected. Calculate the percentage yield of the reaction.(I already know how to do this but just for reference for second question)

2.In the previous example, 10.0 g of potassium iodide was reacted with lead (II) nitrate. What if the potassium iodide was only 85.0% pure? How much precipitate would we expect then?

Answer 1

Answer:

Percent yield represents the ratio between what is experimentally obtained and what is theoretically calculated, multiplied by 100%.

% yield=actual yieldtheoretical yield⋅100%

So, let's say you want to do an experiment in the lab. You want to measure how much water is produced when 12.0 g of glucose (C6H12O6) is burned with enough oxygen.

C6H12O6+6O2→6CO2+6H2O

Since you have a 1:6 mole ratio between glucose and water, you can determine how much water you would get by

12.0 g glucose⋅1 mole glucose180.0 g⋅6 moles of water1 mole glucose⋅18.0 g1 mole water=7.20g

This represents your theoretical yield. If the percent yield is 100%, the actual yield will be equal to the theoretical yield. However, after you do the experiment you discover that only 6.50 g of water were produced.

Since less than what was calculated was actually produced, it means that the reaction's percent yield must be smaller than 100%. This is confirmed by

% yield=6.50 g7.20 g⋅100%=90.3%

You can backtrack from here and find out how much glucose reacted

65.0 g of water⋅1 mole18.0 g⋅1 mole glucose6 moles water⋅180.0 g1 mole glucose=10.8g

So not all the glucose reacted, which means that oxygen was not sufficient for the reaction - it acted as a limiting reagent.

As a conclusion, percent yield problems always have one reactant act as a limiting reagent, thus causing a difference between what is calculated and what is actually obtained. A percent yield that exceeds 100% is never possible, under any circumstances, and means that errors were made in the calculations.

Explanation:

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