Here's a nice example of a percent yield problem. Consider the following chemical equation:
#2"FePO"_4 + 3"Na"_2"SO"_4 -> "Fe"_2("SO"_4)_3 + 2"Na"_3"PO"_4#
Let's say you have #"34.4 g"# of #"FePO"_4# and excess #"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 #"Na"_2"SO"_4# will not act as a limiting reagent, and that we have a #"2:1"# mole ratio between #"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 #"Na"_2"SO"_4# and about the reaction's percent yield?
Well, since we've obtained less than expected, we must conclude that #"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
#"% yield" = ("actual yield")/("theoretical yield") * "100%"#
#"% yield" = ("28.9 g")/("45.6 g") * "100%" = "63.4%"#
We can work backwards and determine how much of the #"FePO"_4# actually reacted.
#"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 #"FePO"_4#, we can determine how much actually reacted by
#"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.
