Chapter 9

Limiting Reactants

and Percent Yield

A. Limiting Reactant1. If you are given one dozen loaves of

bread, a gallon of mustard, and three pieces of ham, how many ham sandwiches can you make?

2. The limiting reactant is the reactant you run out of first.

3. The excess reactant is the one you have left over.

4. The limiting reactant determines how much product you can make

5. What is the limiting reactant of your sandwich supplies?

ham

6. What is the excess reactant of your sandwich supplies?

gallon of mustard

B. How do you find out?

1. Do two stoichiometry problems.

2. The one that makes the least product is the limiting reagent.

3. For example

Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

• If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?

• 2Cu + S Cu2S

10.6 g Cu 63.55g Cu 1 mol Cu

2 mol Cu 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 13.3 g Cu2S

3.83 g S 32.06g S 1 mol S

1 mol S 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 19.0 g Cu2S

= 13.3 g Cu2S

Cu is Limiting

Reactant

C. Clarify1. Limiting Reactant1. Limiting Reactant

–used up in a reaction–determines the amount of product

2. Excess Reactant2. Excess Reactant–added to ensure that the other

reactant is completely used up–cheaper & easier to recycle

3. Example

a. Available Ingredientsa. Available Ingredients–4 slices of bread–1 jar of peanut butter–1/2 jar of jelly

b. Limiting Reactantb. Limiting Reactant–bread

c. Excess Reactantsc. Excess Reactants–peanut butter and jelly

E. Calculating Limiting Reactants

1. Write a balanced equation.

2. For each reactant, calculate the

amount of product formed.

3. Smaller answer indicates:

–limiting reactant

4. Do Stoichiometry!

7. Still another example• If 10.3 g of aluminum are reacted with

51.7 g of CuSO4 how much copper will be produced?

• How much excess reagent will remain?

• Al + CuSO4 -> Al2(SO4)3 + Cu

• Balanced?

• 2 Al + 3 CuSO4 -> Al2(SO4)3 + 3Cu

2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu

1st do Al

Now CuSO4

26.98g Al

1mol Al 3mol Cu

2mol Al 1 mol Cu 63.55g Cu

= 36.4g Cu

159.56g CuSO4

1mol CuSO4

3 mol CuSO4

3mol Cu 63.55g Cu

1 mol Cu

=20.6g Cu

51.7g CuSO4

10.3g Al

So, limiting reactant?

Al: 36.4g Cu

CuSO4: 20.6g Cu

Limiting reactant: CuSO4

Excess: Al

A. Percent Yield

1. The amount of product made in a chemical reaction.

2. There are three types:

a. Actual yield- what you get in the lab when the chemicals are mixed

b. Theoretical yield- what the balanced equation tells what should be made

c. Percent yieldc. Percent yield =

Actual Theoretical

3. Details

• Percent yield tells us how “efficient” a reaction is.

• Percent yield can not be bigger than 100 %.

X 100%

100yield ltheoretica

yield actualyield %

calculated on paper

measured in lab

B. Example 1

• 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.

• 2Al + 3 CuSO4 Al2(SO4)3 + 3Cu

• What is the actual yield?

• What is the theoretical yield?

• What is the percent yield?

2Al + 3 CuSO4 Al2(SO4)3 + 3Cu

Actual yield? 6.78g

Why? “cause it told me already!!!!

Theoretical yield?

Stoichiometry!!!!

3.92g Al1mol Al26.98g Al 2mol Al

3 mol Cu

1 mol Cu

63.55g Cu

= 13.85g Cu

So, Theoretical yield is 13.85g Cu

What is percent yield?

Actual yield

Theoretical yield

6.87g Cu

13.85g Cu

X 100%

X 100% = 48.95%

Example 2• When 45.8 g of K2CO3 react with excess

HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.

K2CO3 + 2HCl 2KCl + H2O + CO2

45.8 g ? g

actual: 46.3 g

45.8 gK2CO3

1 molK2CO3

138.21 gK2CO3

= 49.4g KCl

2 molKCl

1 molK2CO3

74.55g KCl

1 molKCl

K2CO3 + 2HCl 2KCl + H2O + CO2

45.8 g ? g

actual: 46.3 g

Theoretical Yield:

Theoretical Yield = 49.4 g KCl

% Yield =46.3 g

49.4 g 100

=93.7%

K2CO3 + 2HCl 2KCl + H2O + CO2

45.8 g 49.4 g

actual: 46.3 g

Yet another example

If 36.8g of C6H6 react with excess of Cl2And the actual yield of C6H5Cl is 38.8g,

what is the percent yield?

1st find theoretical yield of C6H5Cl

78.06g C6H6

1 mol C6H6

1mol C6H6

1mol C6H5Cl36.8g

C6H6 1mol C6H5Cl

112.5g C6H5Cl

= 53.03g C6H5Cl

Percent yield

38.8g

53.03gX 100% = 73.17%

One more time!

If 1.85g of Al react with excess CuSO4 to produce 3.70g of Cu, what is the percent yield?

Al + CuSO4 -> Al2(SO4)3 + Cu

Balance

2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu

Now find theoretical yield

2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu

1.85g Al26.98g Al

1mol Al

2mol Al3mol Cu 63.55g Cu

1mol Cu

= 6.54g CuNow percent yield

3.70g

6.45gX 100% = 57.36%