Chapter 12 - Stoichiometry

“SUPER

DIMENSIONAL

ANALYSIS”

Review of moles…..

• 1 mole = 6.022 x 1023 particles

• Molar mass = X grams

1 moleX = molar mass of substance( )

Use atomic mass

Calculate molar mass of CaBr2

Ca: 40.1 x 1 = 40.1 g/mol

Br: 79.9 x 2 = 159.8 g/mol

= 199.9 g/mol

What is mass of 0.250 mol CO2?

C: 12 x 1 = 12 g/mol

O: 16 x 2 = 32 g/mol

= 44 g/mol

0.250 mol CO2

1 mol CO2

44 g CO2 = 11 g CO2

Stoichiometry

• Defn – study of mass relationships between reactants and products in a chemical reaction

• What does this mean?– How much of something (products) can be

made?– How much of starting materials were there?

Stoichiometry

You must have a _____________________

to do stoichiometry calculations.

balanced chemical reaction

To make a basic cheeseburger:

2 buns + 1 meat patty + 1 cheese

1 cheeseburger

1 c-burger

2 buns

1 c-burger

1 meat patty

1 c-burger

1 cheese

1 meat patty

2 buns

1 cheese

1 meat patty

2 buns

1 cheese

Mole Ratio

• Defn – ratio btwn # of moles of any two substances in a balanced chem rxn

Ex reaction

2 Al + 3 Br2 2 AlBr3

• give all the mole ratios

2 mol Al

3 mol Br2

2 mol Al

2 mol AlBr3

2 mol Al

3 mol Br2

3 mol Br2

2 mol AlBr3

2 mol Al

2 mol AlBr3

3 mol Br2

2 mol AlBr3

4 steps to a basic stoichiometry problem

1) Balance the equation

2) Identify the given, convert to moles

3) Identify unknown, do a mole to mole ratio between given and unknown (KEY STEP)

4) Convert unknown to unit specified in problem

Stoichiometry Flow Chart

grams A grams Bmoles A moles B

MolarMass A

MolarMass B

MoleRatio

BA

examples

• How many moles of H2 are formed when 65 g HCl is used?

• If 3.7 mol KBr reacts with calcium, how many moles of CaBr2 are formed?

A

A

B

B

Mole A to Mole B Conversions

grams Bmoles A moles B

MolarMass A

MolarMass B

MoleRatio

BA

grams A

__ C3H8 + __ O2 __ CO2 + __ H2O

• How many moles of CO2 are produced when 10.0 moles of O2 is used?

1 45 3

10.0 mol ? mol

10.0 mol O2

MoleRatio

BA

5 mol O2

3 mol CO2

A B

= 6 mol CO2

Mole A to Mole B Conversions

Mole A to Mass B Conversion

grams Bmoles A moles B

MolarMass A

MolarMass B

MoleRatio

BA

grams A

___ Na + ___ Cl2 ___ NaCl

• How many grams of sodium chloride is formed when 1.25 moles of sodium react w/ chlorine gas?

Mole A to Mass B Conversion

2 21

1.25 mol ? g

1.25 mol Na 2 mol NaCl

2 mol Na 1 mol NaCl

58.5 g NaCl

AB

MoleRatio

BA

MolarMass B

= 73.1 g NaCl

Mass A to Mole B Conversion

grams Bmoles A moles B

MolarMass A

MolarMass B

MoleRatio

BA

grams A

___ Na + ___ Cl2 ___ NaCl

• How many moles of chlorine gas is needed to make 50 g NaCl?

Mass A to Mole B Conversion

2 21

50 g? mol

50 g NaCl

58.5 g NaCl

1 mol NaCl

2 mol NaCl

1 mol Cl2

AB

MoleRatio

BA

MolarMass A

= 0.43 mol Cl2

Mass A to Mass B Conversion

grams Bmoles A moles B

MolarMass A

MolarMass B

MoleRatio

BA

grams A

___NH4NO3 ___ N2O + ___ H2O

• Determine the mass of water formed from decomposition of 25.0 g NH4NO3

Mass A to Mass B Conversion

1 1 2

25 g ? g

25 g NH4NO3

80 g NH4NO3

1 mol NH4NO3

1 mol NH4NO3

2 mol H2O

1 mol H2O

18 g H2O

= 11.25 g H2O

A B

Limiting Reactant

• In basic stoichiometry problems, you are provided with one given quantity. In LR problems, you are given both reactants. Before you can solve the problem, you have to determine which of the two given quantities to use as your given.

Limiting Reactant

You need to choose the one that will run out first, known as the __________________. It controls how much product you can make. The other reactant is known as the _________________ because there will be some left over.

limiting reactant

excess reactant

1 cheese + 2 bread slices 1 cheese sandwich

Given: 10 pieces of cheese, 50 bread slices

How many cheese sandwiches can you make?

What is the limiting reactant?What is the excess reactant?

How many pieces of excess reactant used?How many pieces of excess reactant left over?

10

20

30

cheesebread

How to find limiting reactant

1) Convert both amounts of reactants to moles

2) Divide the mole amount of each reactant by its coefficient in the balanced equation

3) Compare two numbers. The one that is smaller is the limiting reactant. Other one is excess reactant.

____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3

Find the limiting reactant if 6.9 g Al and 0.35 mol CuCl2are available.

Reactant #1: Reactant #2:

6.9 g Al 1 mol Al

27 g Al

= 0.256 mol Al = 0.35 mol CuCl2

1) Convert both amounts of reactants to moles.

____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3

Reactant #1: Reactant #2:

0.256 mol Al 0.35 mol CuCl2

2 3

2) Divide each mole amount by its coefficient inthe balanced equation

= 0.128 = 0.117

CuCl2 is the limiting reactant

____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3

Find the limiting reactant if 6.2 g Al and 48.5 g CuCl2are available.

Reactant #1: Reactant #2:

6.2 g Al 1 mol Al

27 g Al

= 0.230 mol Al = 0.360 mol CuCl2

1) Convert both amounts of reactants to moles.

48.5 g CuCl2 1 mol CuCl2134.5 g CuCl2

____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3

Reactant #1: Reactant #2:

0.230 mol Al 0.360 mol CuCl2

2 3

2) Divide each mole amount by its coefficient inthe balanced equation

= 0.115 = 0.120

Al is the LRCuCl2 is excess reactant

Based on the LR in #4, how many grams of copper will be produced?

1 mol Al

2 mol Al

3 mol Cu

1 mol Cu

63.5 g Cu

____ Al + ___ CuCl2 ___ Cu + ___ AlCl32 23 3

= 21.9 g Cu

27 g Al

6.2 g Al

Percent Yield

• You buy a 500 g ketchup bottle, do you ever use all 500 g?

No. Some is still left on the sides youcannot retrieve. Not all 100% is used.

Percent Yield

• Theoretical Yield – max amount of product that can be produced (what you expect to get)

• Actual Yield – amt of product you actually produced (always less than theoretical)

• Percent Yield =actual

theoreticalx 100

Ex problem #1

• Joe does experiment to form carbon dioxide. The maximum he can obtain is 34.5 grams. He performs the experiment and only obtains 18.6 grams. What is his percent yield?

AT

=18.6 g34.5 g

X 100 = 53.9%

Ex problem #2

• What is the theoretical yield of Ag2CrO4 if 0.500 g AgNO3 reacts with excess K2CrO4?

__K2CrO4 + __ AgNO3 __KNO3 + __Ag2CrO42 21 1

0.500 g ? g

0.5 g K2CrO4

170 g K2CrO4

1 mol K2CrO4

1 mol K2CrO4

1 mol Ag2CrO4

1 mol Ag2CrO4

331.7 g

= 0.49 g Ag2CrO4

Ex problem #2

• If 0.455 g of Ag2CrO4 is produced, what is the percent yield?

AT

=0.455 g0.49 g

X 100 = 93.2%