Solve This...
Percent Yield
The difference between what you should’ve gotten and
what you actually got
Solved:
5.00 g (KClO3)
122.55 g/mol (KClO3)
x
3 mol (O2)
2 mol (KClO3)
When 5.00 g of potassium chlorate is heated it decomposes
according to the equation:
2KClO3(s) ! 2KCl(l) + 3O2(g)
a) Calculate the theoretical yield of oxygen (this is just basic
stoichiometry)
Solve This (part 2)...
x
32.00 g/mol (O2)
= 1.96 g (O2)
When 5.00 g of KClO3 is heated it decomposes according to the
equation:
2KClO3 ! 2KCl + 3O2
Therefore, the theoretical yield of oxygen is
1.96 grams when 5.00 grams of potassium
chlorate is decomposed.
a) Theoretical yield of oxygen is 1.96 grams.
b) Give the % yield if 1.78 g of O2 is produced.
That’s it? Try this:
Solved:
% yield =
actual yield
theoretical yield
% yield =
1.78 g
1.96 g
=
90.8%
x 100
What is the % yield of water if 58.0 grams of water are
produced by combining 60.0 grams of oxygen and 7.0
grams of hydrogen?
x 100
Solution:
Solution:
2H2 + O2 ! 2H2O
2H2 + O2 ! 2H2O
Find your limiting reagent first
Determine your % yield using the amount of
product produced by the limiting reagent
H2O from O2
60.0 g
32.00 g/mol
x
2 mol (H2O)
1 mol (O2)
x
2 mol (H2O)
2 mol (H2)
x
18.02 g/mol (H2O)
= 67.6 g (H2O)
% yield =
H2O from H2
7.0 g
2.02 g/mol
x
18.02 g/mol (H2O)
your limiting reagent
= 62 g (H2O)
% yield =
=
actual yield
theoretical yield
58.0 g
62 g
94%
x 100
x 100