1 CHAPTER THREE



CHEMICAL EQUATIONS & REACTION STOICHIOMETRY

Chapter Three Goals 2 1.

Chemical Equations 2.

Calculations Based on Chemical Equations 3.

Percent Yields from Chemical Reactions 4.

The Limiting Reactant Concept 5.

Concentrations of Solutions 6.

Dilution of solutions 7.

Using Solutions in Chemical Reactions

Chemical Equations

 1.

A chemical process is represented by a chemical equation Reaction of methane with O 2 : CH 4 + 2O 2 CO 2 + 2H 2 O

reactants products

3

2.

3.

4.

reactants on left side of reaction products on right side of equation relative amounts of each using stoichiometric coefficients

4 Chemical Equations

5 Chemical Equations

 Look at the information an equation provides: reactants yields 1 formula unit 3 molecules 1 mole 3 moles 159.7 g 84.0 g

2 Fe + 3 CO

2 2 atoms 2 moles 111.7 g products 3 molecules 3 moles 132 g

Chemical Equations 6



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.

Propane,C 3 H 8 , burns in oxygen to give carbon dioxide and water.



C

3

H

8 

5 O

2

3 CO

2 

4 H

2

O

7 Law of Conservation of Matter

 NH 3 burns in oxygen to form NO & water 

2 NH

3

+

5 2

O

2

2 NO + 3 H

2

O or correctly 4 NH

3

+ 5 O

2  

4 NO + 6 H

2

O

8 Law of Conservation of Matter

 C 7 H 16 burns in oxygen to form carbon dioxide and water.

You do it!

C 7 H 16 + 11 O 2 7 CO 2 + 8 H 2 O 

Balancing equations is a skill acquired only with lots of practice

– work many problems

9 Chemical Equations

 Look at the information an equation provides: reactants yields 1 formula unit 3 molecules 1 mole 3 moles 159.7 g 84.0 g

2 Fe + 3 CO

2 2 atoms 2 moles 111.7 g products 3 molecules 3 moles 132 g

Calculations Based on Chemical Equations 10

 How many CO molecules are required to react with 25 formula units of Fe 2 O 3 ?

25 Fe 2 O 3 + ? CO



Product 1Fe 2 O 3 needs 3 CO 25Fe 2 O 3

?

CO molecules

needs ? CO

= 25 formula units Fe 2 O 3  3 CO molecules 1 Fe 2 O 3 formula unit  75 molecules of CO

Calculations Based on Chemical Equations 11

 How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide?

Fe 2 O 3 + excess CO 1Fe 2 O 3



2Fe + 3CO gives 2Fe 2 2.5 X 10 5 Fe 2 O 3 gives ? Fe

?

Fe atoms = 2.50

 10 5 formula units Fe 2 O 3  2 Fe 1 formula atoms units Fe 2 O 3  5.00

 10 5 Fe atoms

Calculations Based on Chemical Equations 12

 What mass of CO is required to react with 146 g of iron (III) oxide?

Fe 2 O 3 + 3CO



Product MW(Fe 2 O 3 ) needs 3MW(CO) 146 g needs ?g CO

?

g CO = 146 g Fe 2 O 3  1 mol Fe 2 O 3 159 .

7 g Fe 2 O 3  3 mol CO 1 mol Fe 2 O 3  28.0

g CO 1 mol CO  76 .

8 g CO

Calculations Based on Chemical Equations 13

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

Fe 2 O 3 + excess CO



2Fe + 3CO 2 1mol Fe 2 O 3 gives 3 mol CO 2 0.540 mol Fe 2 O 3 gives ? mol CO 2



? mol CO 2 = ? g CO 2 /MW(g/mol) CO 2

? g CO 2  3  3 mol CO 2  g CO 1 mol CO 2 2 = 71.3 g CO 2

14 Calculations Based on Chemical Equations

 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!

Fe 2 O 3 + excess CO



2Fe + 3CO 2 ? g Fe 2 O 3 = 9.57 g Fe 2 O 3

Percent Yields from Reactions



Theoretical yield

is calculated by assuming that the reaction goes to completion.



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

15

from a reaction.

actual yield % yield = theoretical yield  100%

Percent Yields from Reactions

 A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5 . What is the percent yield?

CH 3 COOH + C 2 H 5 OH



CH 3 COOC 2 H 5 MW



MW + H 2 O 10.0 g



X (Theoretical Yield) 16

Percent Yields from Reactions 17

CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5  H 2 O 1.

Calculate the theoretic al yield ?

g CH 3 COOC 2 H 5 = 10.0

g C 2 H 5 OH  88.0

g CH 3 COOC 2 46 .

0 g C 2 H 5 OH  19 .

1 g CH 3 COOC 2 H 5 H 5 2 .

Calculate the percent yield.

% yield = 14.8

g CH 3 COOC 2 H 5 19.1

g CH 3 COOC 2 H 5  100 %  77 .

5 %

Percent Yields from Reactions

 Salicylic acid reacts with acetic anhydride to form aspirin, acetylsalicylic acid. If the percent yield in this reaction is 78.5%, what mass of salicylic acid is required to produce 150. g aspirin?

2 C 7 H 6 O 3 + C 4 H 6 O 3



salicylic acid acetic anhydride

MW = 138 g/mol

2 C 9 H 8 O 4 aspirin + H 2

MW = 180 g/mol

O

29 billion tablets are consumed by Americans each year

18

Percent Yields from Reactions 78.5 = actual Yield (150 g) Theoretical Yield (g) x 100 2 C 7 H 6 O 3 + C 4 H 6 O 3



2 C 9 H 8 O 4 salicylic acid acetic anhydride aspirin 2MW X + H 2 O

 

2MW 191.08

(Theoretical Yield)

Answer:

X = 146.5 g 19

Limiting Reactant Concept

1.

2.

3.

In a given reaction, there is not enough of one reagent to use up the other reagents completely.

The reagent in short supply LIMITS the quantity of the product that can be formed.

How many bikes can be made from 10 frames and 16 wheels?

1 frame + 2 wheels

excess limiting

1 bike

20

21 Limiting Reactant Concept

Limiting Reactant Concept When 100.0 g mercury is reacted with 100.0 g bromine to form mercuric bromide, which is the limiting reagent?

Hg + Br 2 HgBr 2 22 Thus the limiting reagent is

Limiting Reactant Concept 23



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?

 2

1 mol 3 mol 1 mol 2 mol 76.2 g 3(32.0 g) 44.0 g 2(64.1 g)

Limiting Reactant Concept 24

CS 2  3 O 2  CO 2  2 SO 2 ?

mol SO 2  95 .

6 g CS 2  1 mol CS 2 76.2

g  2 mol SO 2 1 mol CS 2  64 .

1 g SO 2 1 mol SO 2  161 g SO 2 ?

mol SO 2  110 g O 2  1 mol O 2 32.0

g O 2  2 mol SO 2 3 mol O 2  64 .

1 g SO 2 1 mol SO 2  147 g SO 2 

Which is limiting reactant?

 

Limiting reactant is O 2 .

What is maximum mass of sulfur dioxide?



Maximum mass is 147 g.

Limiting Reactant Concept 3PbO 2 + Cr 2 (SO 4 ) 3 + K 2 SO 4 + H 2 O



3PbSO 4 + K 2 Cr 2 O 7 + H 2 SO 4

If 25.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant?

(a) PbO 2 (b) H 2 O (c) K 2 SO 4 (d) PbSO 4 (e) Cr 2 (SO 4 ) 3 2MnO 2 + 4KOH + O 2 + Cl 2



2KMnO 4 + 2KCl + 2H 2 O

If 20.0 g of each reactant were used in performing the

25

following reaction, which would be the limiting reactant?

(a) MnO 2 (b) KOH (c) O 2 (d) Cl 2 (e) KMnO 4

Concentration of Solutions 26



Definition of Solution: a homogeneous mixture of two or more substances dissolved in another.



A solution is composed of two parts: (1) Solute : dissolved substance (or substance in the lesser amount).

(2) Solvent : dissolving substance (or substance in the greater amount).

–

In aqueous solutions, the solvent is water.

Example: Solution of NaCl in water, H 2 O: NaCl: solute, H 2 O: solvent

Concentration of Solutions 27 Amount of solute Concentration = Mass or Volume of solution

Relative terms:

Dilute solution:

small amount of solute in large amount of solvent.

Concentrated solution:

large amount of solute in smaller amount of solvent (e.g. the amount of sugar in sweet tea can be defined by its concentration).

We will discuss 2 concentration units: 1-

Percent by mass

(do not confuse with % by mass of element in compound) 2-

Molarity

Concentration of Solutions 28 1- Percent by mass

: % by mass of solute mass of = mass of solute solution  100 % mass of solution = mass of solute + mass of solvent % by mass of solute has the symbol % w/w

Note:

if the question says the solution is aqueous oe does not Specify the solvent, the solvent is water, H 2 O.

Concentration of Solutions 29

  Calculate the mass of potassium nitrate, KNO 3 required to prepare 250.0 g of solution that is 20.0 % KNO 3 by mass.

What is the mass of water in the solution?

(a) % by mass = 20.0 % = g KNO 3 g solution g KNO 3 250.0 g x 100 x 100 20.0 % x 250.0 g g KNO 3 = = 50.0 g 100

30 Concentration of Solutions

(b) mass of solution = mass of KNO 3 + mass H 2 O mass H 2 O = mass of solution - mass of KNO 3 mass H 2 O = 250.0 g - 50.0 g mass H 2 O = 200.0 g

31 Concentration of Solutions

 Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.

?

g solution = 32.0

g NaOH  100.0

g solution 8.00

g NaOH  400 .

g sol' n

32 Concentration of Solutions

 Calculate the mass of NaOH in 300.0 mL of an 8.00% w/w NaOH solution. Density is 1.09 g/mL.

You do it!

?

g NaOH = 300.0

mL sol' n  1.09

g sol' 1 mL sol' n n  8.00

g NaOH 100 g sol' n  26 .

2 g NaOH

33 Concentrations of Solutions

 What volume of 12.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL.

You do it!

?

mL solution = 40.0

g KOH  100.0

g solution 12.0

g KOH  300 .

mL solution  1 mL solution 1.11

g solution

34 Concentrations of Solutions

2 Second common unit of concentration : Molarity

35 Concentrations of Solutions

2 Second common unit of concentration : Molarity molarity  number of moles of solute number of liters of solution

M

 moles L

M

 mmol mL

Concentrations of Solutions 36

 Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.

You do it!

?

mol H 2 SO 4 L sol' n  12.5

g H 2 SO 4 1 .

75 L sol' n  1 mol H 2 SO 4 98.1

g H 2 SO 4  0 .

0728 mol H 2 SO L  0 .

0728

M

H 2 SO 4 4

37 Concentrations of Solutions

 Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO 3 ) 2 .

You do it!

? g Ca(NO 3 ) 2  .

L  0.800 mol Ca(NO 3 ) 2 L 164 g Ca(NO 3 ) 2 1 mol Ca(NO 3 ) 2  459 g Ca(NO 3 ) 2 

38 Concentrations of Solutions

 One of the reasons that molarity is commonly used is because:

M x L = moles solute and M x mL = mmol solute

Concentrations of Solutions 39

 The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?

specific gravity = 1.185

tells us density = 1.185

g/mL or 1185g/L ?

mol HCl/L = 1185 g solution L solution  36 .

31 g HCl 100 g sol' n  1 mol HCl 36.46

g HCl  11.80

M

HCl

Dilution of Solutions



To dilute a solution, add solvent to a concentrated solution.

–

One method to make tea “less sweet.”



The number of moles of solute (before and after dilution) in the two solutions remains constant.

40



The relationship M 1 V 1 = M 2 V 2 is appropriate for dilutions, but not for chemical reactions.

41 Dilution of Solutions

 Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb.

Dilution of Solutions 42 Take-Home Calculations

 

(M x V) A = (M x V) B M x V = moles of solute



M x V = W/MW



(M x V) A = (W/MW) A OR

 

(M x V) A = (W/MW) B W = M x V x MW

Dilution of Solutions 43

 If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mL of solution, what is the concentration of the solution?

12 .

0

M M

1 V 1  10 .

0 mL

M

2  

M M

2 2 V 2  100.0

mL  12 .

0

M

 10 .

0 100.0

mL  1 .

20

M

mL

Dilution of Solutions 44

 What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?

You do it!

M

1 V 1 V 1 V 1  

M M

2 V 2 2  V 2 

M

1 2.50

L  2.40

M

18.0

M

 0.333

L or 333 mL

45 Using Solutions in Chemical Reactions

 Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution.

Using Solutions in Chemical Reactions 46

 What volume of 0.500 M BaCl 2 is required to completely react with 4.32 g of Na 2 SO 4 ?

Na 2 SO 4 + BaCl 2  BaSO 4 + 2 NaCl ?

L BaCl 2  4.32

gNa 2 SO 4  1 mol Na 2 SO 142 g Na 2 SO 4 4  1 mol BaCl 2 1 mol Na 2 SO 4  1 L BaCl 2 0.500

mol BaCl 2  0.0608

L

47 Using Solutions in Chemical Reactions

 (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate, Al(NO 3 ) 3 ?

Al  NO 3  3  3 NaOH  Al   3  3 NaNO 3

You do it!

Using Solutions in Chemical Reactions 48

 (a) What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate?

Al  NO 3  3  3 NaOH  Al(OH) 3  3 NaNO 3 ?

mL NaOH = 50.0

mL Al(NO 3 ) 3 sol' n  1 L 1000 mL 0.200

mol Al(NO 3 ) 3 sol' n 1 L Al(NO 3 ) 3 sol' n  1 3 mol mol NaOH Al(NO 3 ) 3  1 L NaOH 0.200

mol NaOH  0.150

L or 150 mL NaOH sol' n

49 Using Solutions in Chemical Reactions

 (b) What mass of Al(OH) 3 precipitates in (a)?

?

g Al(OH) 3  50.0

mL Al(NO 3 ) 3 sol' n  1 L 1000 mL 0.200

mol Al(NO 3 ) 3 1 L Al(NO 3 ) 3 sol' n   1 mol 1 mol Al(OH) Al(NO 3 0 .

780 g Al(OH) 3 3 ) 3  78 .

0 g Al(OH) 3 1 mol Al(OH) 3

50

Homework Assignment One-line Web Learning (OWL): Chapter 3 Exercises and Tutors – Optional

51 End of Chapter 3