Stoichiometry

1
Chapter 12

• Stoichiometry

2
Stoichiometry

• The study of the quantitative, or measurable,
  relationships that exist in chemical formulas and
  chemical reactions.

3

• Interpreting Balanced Chemical Equations

2 H2 O2 ? 2 H2O

• Based on the mole ratio

4
Mole Ratio

• indicated by coefficients in a balanced equation

2 H2 O2 ? 2 H2O
Molar ratio of H2 to H2O is 22 (Simplify
11) Molar ratio of O2 to H2O is 12
5
In terms of moles
2 H2 O2 ? 2 H2O

• 2 moles of hydrogen react with 1 mole of oxygen
  to produce 2 moles of water.

6
In terms of mass
2 H2 O2 ? 2 H2O

• 4 grams of hydrogen react with 32 grams of
  oxygen to produce 36 grams of water.

7
In terms of molecules
2 H2 O2 ? 2 H2O

• 2 hydrogen molecules react with 1 oxygen
  molecule to produce 2 water molecules.
• Notice that the number of molecules is NOT the
  same on each side of the arrow.

8
In terms of atoms
2 H2 O2 ? 2 H2O

• 4 atoms of hydrogen react with 2 atoms of oxygen
  to produce 2 water molecules, which are 4 atoms
  of hydrogen and 2 atoms of oxygen.

9
In terms of volumes
2 H2 O2 ? 2 H2O

• 44.8 L of hydrogen gas react with 22.4 L of
  oxygen gas to produce 44.8 L of water vapor.
• Notice that the number of liters of gas is NOT
  the same on each side of the arrow.

10
Law of Conservation of Matter Mass
VERIFIED!
2 H2 O2 ? 2 H2O
2 moles H2 react with 1 mole of O2 to form 2
moles of H2O.


4 g
32 g
36 g
11
Practice Problem 1
Lead will react with hydrochloric acid to produce
lead (II) chloride and hydrogen. How many moles
of hydrochloric acid are needed to completely
react with 0.36 moles of lead?
Oh, no where do I start?
12
Write an equation and balance it.
Lead will react with hydrochloric acid to produce
lead chloride and hydrogen.
Pb HCl ? PbCl2 H2
2
13
Determine Mole Ratio
How many moles of hydrochloric acid are needed to
completely react with 0.36 moles of lead?
Pb HCl ? PbCl2 H2
2
hydrochloric acid
lead
given
wanted
Coefficient 1
Coefficient 2
14
Set up mole ratio wanted to given
How many moles of hydrochloric acid are needed to
completely react with 0.36 moles of lead?
Pb HCl ? PbCl2 H2
2
X moles of HCl
2

1
0.36 moles of lead
15
Solve for wanted
How many moles of hydrochloric acid are needed to
completely react with 0.36 moles of lead?
Pb HCl ? PbCl2 H2
2
X 2 (0.36 moles)
X 0.72 moles HCl
16
Umm It would require 0.72 moles of hydrochloric
acid. Is that right?
Thats Correct!
17
Hem, hem you have 9 more problems ready, go.
18
How to solve, given an amount of one substance to
another substance.

• Stoich Problems

19
Stoichiometry Steps

• 1. Write a balanced equation.
• 2. Identify wanted given.
• 3. Convert given information to moles.
• 4. Determine Mole Ratio.
• (Moles of wanted Moles of given)
• 5. Calculate Moles of wanted.
• 6. Convert to required units.

• Determine Mole ratio
• (moles of wanted to moles of given)

KEY step in all stoichiometry problems!
20
Mass-Mass Problems

• What mass of bromine is produced when fluorine
  reacts with 1.72 g of potassium bromide?

Help I really need to know this right now!
21
1. Write a balanced equation.

• What mass of bromine is produced when fluorine
  reacts with 1.72 g of potassium bromide?

F2 2KBr ? 2KF Br2
Help I really need to know this right now!
22
2. Identify wanted and given

• What mass of bromine is produced when fluorine
  reacts with 1.72 g of potassium bromide?

F2 2KBr ? 2KF Br2
Help I really need to know this right now!
given
wanted
23
3. Convert given information to moles

• 1.72 g of potassium bromide

Molar Mass of KBr 39 80 119 g/mol
Help I really need to know this right now!
1 mol
1.72 g

0.01445 mol KBr
119 g
24
4. Determine Mole Ratio
F2 2KBr ? 2KF Br2
wanted
given
Help I really need to know this right now!
Coefficients tell Mole Ratio wanted to given
1
2

25
5. Calculate Moles of wanted
F2 2KBr ? 2KF Br2
Moles Br2
1

Moles KBr
Help I really need to know this right now!
2
X
1

2
0.01445 mol KBr
2 X 0.01445 mol
X 0.007225 mol
26
6. Convert to required units
X 0.007225 mol Br2
Molar Mass Br2
80 80 160 g/mol
0.007225 mol
160 g

1 mol
27

• Stoichiometry

28
Limiting Reactants

• Available Ingredients
• 4 slices of bread
• 1 jar of peanut butter
• 1/2 jar of jelly

• Limiting Reactant
• bread

• Excess Reactants
• peanut butter and jelly

29
Limiting Reactants

• Available Ingredients
• Copper Wire
• 0.5 g AgNO3

• Limiting Reactant
• 0.5 grams AgNO3

• Excess Reactants
• Copper Wire

30
Limiting Reactant

• The reactant that limits the amount of product
  that can be formed.

31

• When quantities of reactants are available in the
  exact ratio described by the balanced equation,
  they are said to be in Stoichiometric proportions.

32
Limiting Reactants

• Limiting Reactant
• used up in a reaction
• determines the amount of all products formed
• Excess Reactant
• added to ensure that the other reactant is
  completely used up
• usually cheaper easier to recycle

33
Solving Problems Limiting Reactants

• 1. Write a balanced equation.
• 2. For each reactant, calculate the amount
  of product formed.
• 3. The reactant that produces the smaller amount
  of product is the limiting reactant.

Very similar to mass-mass problems!
34
Step 1 Write a balanced equation.

• Identify the limiting reactant when 1.22 g of
  oxygen reacts with 1.05 g of hydrogen to produce
  water.

O2 2H2 ? 2 H2O
35
Step 2
For each reactant, calculate the amount of
product formed.

• Identify the limiting reactant when 1.22 g of
  oxygen reacts with 1.05 g of hydrogen to produce
  water.

O2 2H2 ? 2 H2O
36
Step 2

• 1.22 g oxygen

1 mole

0.038 mol O2
32 g
O2 2H2 ? 2 H2O
given
wanted
2
wanted
X

1
given
0.038 mol
X

0.076 mol H2O
37
Step 2

• 1.05 g H2

1 mole

0.525 mol H2
2 g
O2 2H2 ? 2 H2O
wanted
given
2
wanted
X

2
given
0.525 mol
X

0.525 mol H2O
38
Step 3
The one that produces the smallest amount is your
limiting reactant.

• 1.22 g of O2 would produce 0.0763 mol H2O

Oxygen is your limiting reactant!
1.05 g of H2 would produce .525 mol H2O
39
Limiting Reactants

• Identify the limiting reactant when 1.7 g of
  sodium reacts with 2.6 L of chlorine gas at STP
  to produce sodium chloride.

40
Step 1 Write a balanced equation.

• Identify the limiting reactant when 1.7 g of
  sodium reacts with 2.6 L of chlorine gas at STP
  to produce sodium chloride.

2Na Cl2 ? 2NaCl
41
Step 2
For each reactant, calculate the amount of
product formed.

• Identify the limiting reactant when 1.7 g of
  sodium reacts with 2.6 L of chlorine gas at STP
  to produce sodium chloride.

2Na Cl2 ? 2NaCl
42
Step 2

• 1.7 g Na

1 mole

0.0739 mol Na
23 g
2Na Cl2 ? 2NaCl
wanted
given
2
wanted
X

2
given
0.0739 mol
X

0.0739 mol NaCl
43
Step 2

• 2.6 L Cl2

1 mole

0.116 mol Cl2
22.4 L
2Na Cl2 ? 2NaCl
wanted
given
2
wanted
X

1
given
0.116 mol
X

0.232 mol NaCl
44
Step 3
The one that produces the smallest amount is your
limiting reactant.

• 1.7 g Na would produce 0.0739 mol NaCl

Sodium is your limiting reactant!
2.6 L Cl2 would produce 0.232 mol NaCl
Click on the real player file called
Sodium_Chlorine_2 to see a demo of this reaction
45
Percent Yield
46
Percent Yield

• When 45.8 g of K2CO3 react with excess HCl,
  46.3 g of KCl are formed. Calculate the
  theoretical yield and yield of KCl.

K2CO3 2HCl ? 2KCl H2O CO2
45.8 g
? g
actual 46.3 g
47
Percent Yield
K2CO3 2HCl ? 2KCl H2O CO2
45.8 g
? g
actual 46.3 g

• Theoretical Yield

45.8 g K2CO3
1 mol K2CO3 138 g K2CO3
2 mol KCl 1 mol K2CO3
74 g KCl 1 mol KCl
49.1 g KCl
48
Percent Yield
K2CO3 2HCl ? 2KCl H2O CO2
45.8 g
49.1 g
actual 46.3 g

• Theoretical Yield 49.1 g KCl

? 100
94.3
Yield