Percent yield - GRCI

[Pages:2]Percent Yield

The difference between what you should've gotten and what you actually got

Solve This...

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)

Solved:

1.96 g x x = 5.00 g (KClO3)

122.55 g/mol (KClO3)

3 mol (O2) 2 mol (KClO3)

32.00 g/mol (O2)

(O2)

Therefore, the theoretical yield of oxygen is 1.96 grams when 5.00 grams of potassium

chlorate is decomposed.

Solve This (part 2)...

When 5.00 g of KClO3 is heated it decomposes according to the equation:

2KClO3 ! 2KCl + 3O2 a) Theoretical yield of oxygen is 1.96 grams. b) Give the % yield if 1.78 g of O2 is produced.

Solved:

actual yield

% yield =

x 100

theoretical yield

% yield =

1.78 g 1.96 g x 100

= 90.8%

That's it? Try this:

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?

Solution:

2H2 + O2 ! 2H2O

Find your limiting reagent first

H2O from O2

60.0 g

x

32.00 g/mol

H2O from H2

7.0 g

x

2.02 g/mol

2 mol (H2O) 1 mol (O2)

2 mol (H2O) 2 mol (H2)

67.6 g x 18.02 g/mol (H2O) =

(H2O)

62 g x 18.02 g/mol (H2O) =

(H2O)

your limiting reagent

Solution:

2H2 + O2 ! 2H2O

Determine your % yield using the amount of product produced by the limiting reagent

actual yield

% yield =

x 100

theoretical yield

% yield =

58.0 g 62 g x 100

= 94%

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