Chapter 9 Chemical Equations & Reaction Stoichiometry

Chapter 9Chemical Equations &

Reaction Stoichiometry

2

Chemical Equations

• Symbolic representation of a chemical reaction that shows:

1. reactants on left side of reaction2. products on right side of equation3. relative amounts of each using stoichiometric

coefficients

3

Chemical Equations

• Attempt to show on paper what is happening at the laboratory and molecular levels.

4

Chemical Equations

• Look at the information an equation provides:

Fe O + 3 CO 2 Fe + 3 CO2 3 2

5

Chemical Equations


reactants yields productsFe O + 3 CO 2 Fe + 3 CO2 3 2

6

Chemical Equations


reactants yields products

1 formula unit 3 molecules 2 atoms 3 molecules

Fe O + 3 CO 2 Fe + 3 CO2 3 2

7

Chemical Equations




1 mole 3 moles 2 moles 3 moles

Fe O + 3 CO 2 Fe + 3 CO2 3 2

8

Chemical Equations




1 mole 3 moles 2 moles 3 moles159.7 g 84.0 g 111.7 g 132g

Fe O + 3 CO 2 Fe + 3 CO2 3 2

9

Chemical Equations

• Law of Conservation of Matter – There is no detectable change in quantity of matter in an

ordinary chemical reaction.– Balanced chemical equations must always include the

same number of each kind of atom on both sides of the equation.

– This law was determined by Antoine Lavoisier.

• Propane,C3H8, burns in oxygen to give carbon dioxide and water.

OH 4 CO 3 O 5 HC 22283

10

Calculations Based on Chemical Equations

• Can work in moles, formula units, etc.• Frequently, we work in mass or weight (grams

or kg or pounds or tons).

Fe O + 3 CO 2 Fe + 3 CO2 3 2

11


• Example 1: How many CO molecules are required to react with 25 formula units of Fe2O3?

CO of molecules 75

unit formula OFe 1

molecules CO 3OFe units formula 25 = molecules CO ?

3232

12


• Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

325 OFe units formula 102.50=atoms Fe ?

13


• Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

OFe units formula 1

atoms Fe 2

OFe units formula 102.50=atoms Fe ?

32

325

14


• Example 2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

atoms Fe 105.00 OFe units formula 1

atoms Fe 2

OFe units formula 102.50=atoms Fe ?

5

32

325

15


• Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

32

3232 OFe g 7.159

OFe mol 1OFe g 146 = CO g ?

16


• Example 3: What mass of CO is required to react with 146 g of iron (III) oxide?

3232

3232 OFe mol 1

CO mol 3

OFe g 7.159


17


• Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

CO g 8.76CO mol 1

CO g 28.0

OFe mol 1

CO mol 3

OFe g 7.159


3232

3232

18


• Example 4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

OFe mol 1

CO mol 3OFe mol 540.0CO g ?

32

2322

19


• Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

2

2

32

2322 CO mol 1

CO g 0.44

OFe mol 1

CO mol 3OFe mol 540.0CO g ?

20


• Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

? g CO mol Fe O3 mol CO

1 mol Fe O

g CO

mol CO

= 71.3 g CO

2 2 32

2 3

2

2

2

0 54044 0

1.

.

21


• Example 5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

You do it!You do it!

22


3232

32

2

32

2

2232

O Feg 5.10O Femol 1

O Feg 7.159

CO mol 3

O Femol1

CO g 44.0

molCO 1CO g 8.65O Feg ?

• Example 5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

23

Limiting Reactant Concept

• Kitchen example of limiting reactant concept.1 packet of muffin mix + 2 eggs + 1 cup of milk

12 muffins

• How many muffins can we make with the following amounts of mix, eggs, and milk?

24

Limiting Reactant Concept• Mix Packets Eggs Milk

1 1 dozen 1 gallonlimiting reactant is the muffin mix

2 1 dozen 1 gallon3 1 dozen 1 gallon4 1 dozen 1 gallon5 1 dozen 1 gallon6 1 dozen 1 gallon7 1 dozen 1 gallon

limiting reactant is the dozen eggs

25


• Example 8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

CS2 3 O2 CO2 2 SO2

26



CS2 3 O2 CO2 2 SO2

1 mol 3 mol 1 mol 2 mol

27



CS O CO 2 SO

1 mol 3 mol 1 mol 2 mol

76.2 g 3(32.0 g) 44.0 g 2(64.1 g)

2 2 2 2 3

28



g 76.2

CS mol 1CS g 6.95 SO mol ?

SO 2 CO O 3 CS

222

2222

29


22

2

2

2

222

2222

SO g 161SO mol 1

SO g 1.64

CS mol 1

SO mol 2

g 76.2

CS mol 1CS g 6.95 SO g ?

SO 2 CO O 3 CS

What do we do next?You do it!You do it!

30


22

2

2

2

2

222

22

2

2

2222

2222

SO g 147SO mol 1

SO g 1.64

O mol 3

SO mol 2

O g 32.0

O mol 1O g 110SO g ?

SO g 161SO mol 1

SO g 1.64

CS mol 1

SO mol 2

g 76.2

CS mol 1CS g 6.95 SO g ?

SO 2 CO O 3 CS

• Which is limiting reactant?• Limiting reactant is O2.• What is maximum mass of sulfur dioxide?• Maximum mass is 147 g.

31

Percent Yields from Reactions• Theoretical yield is calculated by assuming that the

reaction goes to completion.– Determined from the limiting reactant calculation.

• Actual yield is the amount of a specified pure product made in a given reaction.– In the laboratory, this is the amount of product that is

formed in your beaker, after it is purified and dried.• Percent yield indicates how much of the product is

obtained from a reaction.

% yield = actual yield

theoretical yield100%

32

Percent Yields from Reactions

• Example 9: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What is the percent yield?

33


523

52

52352523

2523523

HCOOCCH g 1.19

OHHC g 0.46

HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?

yield al theoretic theCalculate 1.

OH HCOOCCH OHHC + COOHCH

34


523

52

52352523

2523523

HCOOCCH g 1.19

OHHC g 0.46




35


yield.percent theCalculate .2

HCOOCCH g 1.19

OHHC g 0.46




523

52

52352523

2523523

36


%5.77%100HCOOCCH g 19.1

HCOOCCH g 14.8= yield %

yield.percent theCalculate .2

HCOOCCH g 1.19

OHHC g 0.46




523

523

523

52

52352523

2523523

Documents

Chapter 9 Chemical Equations & Reaction Stoichiometry