Chapter 9

Stoichiometry

• The study of quantitative, or

measurable, relationships that

exist in chemical formulas and

chemical reactions.

Mole – Mole Relationship

• Use a balanced chemical equation to

determine the mole – mole

relationship of any substance.

• Mole – Mole relationship can be

applied to any substance in the

balanced equation.

• N2H

4 + 2 H

2O

2 N

2 + 4 H

2O

Practice Mole-Mole

• Determine the number mole-

mole relationship for Ag in the

following equation:

• 3 AgNO3 + Al Al(NO

3)3 + 3 Ag

Mole – Mole Problems

• How many moles of HCl are

needed to react with 2.3 moles

of Zn?

• 2 HCl + Zn ZnCl2 + H

2

Practice Problems

• 1) How many moles of Na are

needed to react with 55.5 moles

of Cl2,

in the production of salt?

Na + Cl2 NaCl

• 2) How many moles of Al(OH)3 are

produced when 2.53 moles of H2O

react.

Al2O

3 + H

2O Al(OH)

3

Answers

1) 2 Na + Cl2 2NaCl

2) Al2O

3 + 3 H

2O 2 Al(OH)

3

9-2 Ideal Stoichiometric

Calculations

• Using all other conversions from

chapter 7, along with the mole-

mole to determine unknown

amounts of each substance.

• 1 mol = 6.02 x 1023

particles

• 1 mol = 22.4L at STP

• 1 mol = formula mass(g)

Example: Mass - Mass

• Determine the mass of sodium

hydroxide produced when .25g of

sodium reacts with water.

2 Na + 2H2O 2NaOH + H

2

.25g x g

Practice Problem #1

• What mass of bromine, is

produced when fluorine reacts

with 3.55 moles of potassium

bromide?

F2 + 2 KBr 2 KF + Br

2

3.55 mol x g

Mass – Volume :

Example

• Find the mass of aluminum

required to produce 1.32 L of

hydrogen gas at STP.

2 Al + 3 H2SO

4 Al

2(SO

4)3 + 3 H

2

x g 1.32 L

Volume – Volume:

Relationship

• The volume relationship of 2

gases at STP are directly

proportional to the mole-mole

relationship of those 2 gases.

• Why?

• All gases at STP occupy the

same volume, 22.4L / mol.

Example:

Volume - Volume

• What volume of hydrogen gas is

necessary to produce 16.0L of

ammonia (NH3)?

N2 + 3 H

2 2 NH

3

x L 16.0 L

Chapter 9–3: Limiting

and Excess Reactants

• Limiting reactant

– The reactant that limits the amount

of product obtained.

– Completely consumed in the

reaction. No trace remains in the

product.

• Excess reactant

– The reactant that isn’t completely

used in the reaction and remains

after the reaction is over.

• Limiting Excess comparison

Example Problem:

• Determine the mass of NaCl produced

when 1.2 mol of Na react with .85 mol

Cl2.

2 Na + Cl2 2 NaCl

1.2mol .85mol ?mol

1st determine the limiting reactant.

Convert moles of one given to moles of

another.

Determining Limiting

and Excess

• Compare: Have vs. Need

• Have = Given amount

• Need = Calculated amount

• If Have > Need, excess reactant

• If Need > Have, limiting reactant

Comparing

• Determine the mass of NaCl

produced when 1.2 mol of Na react

with .85 mol Cl2.

2 Na + Cl2 2 NaCl

1.2mol .85mol ?g

22 Cl .6mol

Na 2mol

Cl 1mol

1

molNa2.1x

You have .85mol Cl2 you need .6mol

Cl2.

Chlorine is the excess, which means

Sodium is the limiting.

Finishing the Problem

• Determine the mass of NaCl produced

when 1.2 mol of Na react with .85 mol

Cl2.

2 Na + Cl2 2 NaCl

1.2mol .85mol ?g

You must begin with the limiting given

to determine the unknown amount.

Practice Problem

3ZnO + 2Al(NO3)3 Al

2O

3 + 3Zn(NO

3)2

Considering 25g of ZnO react with

.555moles of Al(NO3)3. How many

grams of Al2O

3 will be produced?

a) Determine limiting/excess reactant.

b) Determine excess that will remain.

c) Determine the unknown.

• A)

.

• B)

• C)

• Limiting and Excess Demo

Percent Yield

• Yield = Product

• Actual Yield

– Amount of product given in the

problem.

• Theoretical Yield (Expected)

– Amount of product determined

by a stoichiometry calculation.

• Percent Yield

100 x yield ltheoretica

yield actual YieldPercent

Sample Problem

• Determine the percent yield

when 2.80g Al(NO3)3 react to

produce .966g of Al(OH)3.

Al(NO3)3 + 3NaOH Al(OH)

3 + 3NaNO

3

2.80g .966g

Identifying what to do!

• 1) Regular stoichiometry:

• Given one value.

• 2) Limiting/Excess:

• Given 2 values both are reactants (left

side of equation)

• 3) %yield:

• Given 2 values 1 reactant and 1 product

• All problems involve a mole-mole

conversion.

Practice Problem

• Determine the percent yield when

1.50g Al(OH)3 decomposes to produce

.25g of H2O.

2Al(OH)3 Al

2O

3 + 3H

2O

1.50g .25g