# What is an example of a percent yield practice problem?

Jan 17, 2015

Here's a nice example of a percent yield problem. Consider the following chemical equation:

$2 {\text{FePO"_4 + 3"Na"_2"SO"_4 -> "Fe"_2("SO"_4)_3 + 2"Na"_3"PO}}_{4}$

Let's say you have $\text{34.4 g}$ of ${\text{FePO}}_{4}$ and excess ${\text{Na"_2"SO}}_{4}$ for this reaction, how many grams of "Fe"_2("SO"_4)_3 can you produce with 100% yield?

Since we know that ${\text{Na"_2"SO}}_{4}$ will not act as a limiting reagent, and that we have a $\text{2:1}$ mole ratio between ${\text{FePO}}_{4}$ and "Fe"_2("SO"_4)_3, we can say that

"34.4 g" * ("1 mole FePO"_4)/("150.8 g") * ("1 mole Fe"_2("SO"_4)_3)/("2 moles Fe""PO"_4) * ("399.8 g")/("1 mole Fe"_3"PO"_4) = "45.6 g Fe"_2("SO"_4)_3

This value represents our theoretical yield - what is produced for a 100% yield reaction.

What if, instead of 45.6 g, you actually produce 28.9 g, what can you say about ${\text{Na"_2"SO}}_{4}$ and about the reaction's percent yield?

Well, since we've obtained less than expected, we must conclude that ${\text{Na"_2"SO}}_{4}$ acts as a limiting reagent this time. So, what is produced this time will be the actual yield, which will determine the reaction's percent yield at

$\text{% yield" = ("actual yield")/("theoretical yield") * "100%}$

$\text{% yield" = ("28.9 g")/("45.6 g") * "100%" = "63.4%}$

We can work backwards and determine how much of the ${\text{FePO}}_{4}$ actually reacted.

${\text{28.9 g" * ("1 mole Fe"_2("SO"_4)_3)/("399.8 g") * ("2 moles FePO"_4)/("1 mole Fe"_2("SO"_4)_3) * ("150.8 g")/("1 mole FePO"_4) = "21.8 g FePO}}_{4}$

You can double-check your calculations by using the reaction's percent yield. Since we know that we started with 34.4 g of ${\text{FePO}}_{4}$, we can determine how much actually reacted by

$\text{actual" = "%yield" * "theoretical" = 63.4/100 * 34.4 = "21.8 g}$.

As you can see, percent yield links how much of a reactant actually reacted with how much of a product was actually produced.