chemical equations & 3 reaction stoichiometry

3 Chemical Equations &

Reaction Stoichiometry

2

Chapter Three Goals 1. Chemical Equations 2. Calculations Based on Chemical Equations 3. The Limiting Reactant Concept 4. Percent Yields from Chemical Reactions 5. Sequential Reactions 6. Concentrations of Solutions 7. Dilution of solutions 8. Using Solutions in Chemical Reactions 9. Synthesis Question

3

Chemical Equations

• Symbolic representation of a chemical reaction that shows:

1. reactants on left side of reaction 2. products on right side of equation 3. relative amounts of each using

stoichiometric coefficients

4

Chemical Equations

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

5

Chemical Equations

• Look at the information an equation provides:

Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →

6

Chemical Equations


reactants yields products

Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →

7

Chemical Equations



1 formula unit 3 molecules 2 atoms 3 molecules

Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →

8

Chemical Equations



1 formula unit 3 molecules 2 atoms 3 molecules

1 mole 3 moles 2 moles 3 moles

Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →

9

Chemical Equations


reactants yields products 1 formula unit 3 molecules 2 atoms

3 molecules 1 mole 3 moles 2 moles

3 moles 159.7 g 84.0 g 111.7 g 132g

Fe O + 3 CO 2 Fe + 3 CO2 3 2∆ →

10

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 +→+ ∆

11

Law of Conservation of Matter

• NH3 burns in oxygen to form NO & water You do it!

12


• NH3 burns in oxygen to form NO & water

OH 6 + NO 4 O 5 + NH 4correctlyor

OH 3 + NO 2 O + NH 2

223

2225

3

→

→

∆

∆

13


• C7H16 burns in oxygen to form carbon dioxide and water.

You do it!

14



OH 8 + CO 7 O 11 + HC 222167 →∆

15



• Balancing equations is a skill acquired

only with lots of practice – work many problems

OH 8 + CO 7 O 11 + HC 222167 →∆

16

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∆ →

17

Calculations Based on Chemical Equations • Example 3-1: How many CO molecules

are required to react with 25 formula units of Fe2O3?

CO of molecules 75unit formula OFe 1

molecules CO 3OFe units formula 25 = molecules CO ?32

32

=

×

18

Calculations Based on Chemical Equations • 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 ? ×

19



=×

×

OFe units formula 1atoms Fe 2

OFe units formula 102.50=atoms Fe ?

32

325

20



atoms Fe 105.00 OFe units formula 1

atoms Fe 2OFe units formula 102.50=atoms Fe ?

5

32

325

×=×

×

21

Calculations Based on Chemical Equations • 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 ? ×

22



3232

3232 OFe mol 1

CO mol 3OFe g 7.159

OFe mol 1OFe g 146 = CO g ? ××

23



CO g 8.76CO mol 1CO g 28.0

OFe mol 1CO mol 3

OFe g 7.159OFe mol 1OFe g 146 = CO g ?

3232

3232

=×

××

24

Calculations Based on Chemical Equations • 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?

OFe mol 1CO mol 3OFe mol 540.0CO g ?

32

2322 ×=

25



2

2

32

2322 CO mol 1

CO g 0.44 OFe mol 1

CO mol 3OFe mol 540.0CO g ? ××=

26



? 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 01

..

27

Calculations Based on Chemical Equations • Example 3-5: 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!

28

Calculations Based on Chemical Equations

3232

32

2

32

2

2232

O Feg 5.10O Femol 1O Feg 7.159

CO mol 3 O Femol1

CO g 44.0 molCO 1CO g 8.65O Feg ?

=

×××=

• Example 3-5: 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?

29

Calculations Based on Chemical Equations • Example 3-6: How many pounds of carbon

monoxide would react with 125 pounds of iron (III) oxide?

You do it!

30

Calculations Based on Chemical Equations

CO lb 7.65CO g 454

CO lb 1CO mol 1CO g 28

OFe mol 1CO mol 3

OFe g 7.159OFe mol 1

OFe lb 1OFe g 454OFe lb 125 = CO lb ?

3232

32

32

3232

=×

×××

×

YOU MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!!

31

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?

32

Limiting Reactant Concept • Mix Packets Eggs Milk 1 1 dozen 1 gallon

limiting reactant is the muffin mix 2 1 dozen 1 gallon 3 1 dozen 1 gallon 4 1 dozen 1 gallon 5 1 dozen 1 gallon 6 1 dozen 1 gallon 7 1 dozen 1 gallon

limiting reactant is the dozen eggs

33


• Example 3-7: Suppose a box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box?

( )( )( )

numbersmallest by the determinedsets 55 is make can number we maximum the

sets 99nut 1set 1nuts 99

sets 55 washers2set 1 washers110

sets 87bolt 1set 1bolts 87

=

=

=

34


• Look at a chemical limiting reactant situation.

Zn + 2 HCl→ ZnCl2 + H2

35


• Example 3-8: 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

36



CS2 + 3 O2 → CO2 + 2 SO2

1 mol 3 mol 1 mol 2 mol

37



CS O CO 2 SO1 mol 3 mol 1 mol 2 mol76.2 g 3(32.0 g) 44.0 g 2(64.1 g)

2 2 2 2+ → +3

38



g 76.2CS mol 1CS g 6.95 SO mol ?

SO 2 CO O 3 CS

222

2222

×=

+→+

39


22

2

2

2

222

2222

SO g 161SO mol 1

SO g 1.64CS mol 1

SO mol 2g 76.2

CS mol 1CS g 6.95 SO g ?

SO 2 CO O 3 CS

=××

×=

+→+

What do we do next? You do it!

40


22

2

2

2

2

222

22

2

2

2222

2222

SO g 147SO mol 1

SO g 1.64O mol 3

SO mol 2O g 32.0O mol 1O g 110SO g ?

SO g 161SO mol 1

SO g 1.64CS mol 1

SO mol 2g 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.

41

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 yieldtheoretical yield

× 100%

42

Percent Yields from Reactions

• Example 3-9: 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?

43


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

=

×

+→

44


523

52

52352523

2523523

HCOOCCH g 1.19 OHHC g 0.46

HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?


=

×

+→

45


yield.percent theCalculate .2HCOOCCH g 1.19

OHHC g 0.46 HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?


523

52

52352523

2523523

=

×

+→

46


%5.77%100HCOOCCH g 19.1

HCOOCCH g 14.8= yield %

yield.percent theCalculate .2HCOOCCH g 1.19

OHHC g 0.46 HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?


523

523

523

52

52352523

2523523

=×

=

×

+→

47

Sequential Reactions N O2 NH2

HNO3

H2SO4

Sn

Conc. HCl

benzene nitrobenzene aniline

• Example 3-10: Starting with 10.0 g of benzene (C6H6), calculate the theoretical yield of nitrobenzene (C6H5NO2) and of aniline (C6H5NH2).

48

Sequential Reactions

××benzene g 0.78benzene mol 1benzene g 10.0 = nenitrobenze g ?

49


nenitrobenze g 8.15nenitrobenze mol 1nenitrobenze g 0.123

benzene mol 1nenitrobenze mol 1

benzene g 0.78benzene mol 1benzene g 10.0 = nenitrobenze g ?

=×

××

• Next calculate the mass of aniline produced. You do it!

50


HNO3

H2SO4

Sn

Conc. HCl


××nenitrobenze g 123.0

nenitrobenze mol 1nenitrobenze g 15.8 = aniline g ?

51


HNO3

H2SO4

Sn

Conc. HCl


? g aniline = 15.8 g nitrobenzene 1 mol nitrobenzene123.0 g nitrobenzene

1 mol aniline1 mol nitrobenzene

93.0 g aniline1 mol aniline

g aniline

× ×

× = 119.

52


• If 6.7 g of aniline is prepared from 10.0 g of benzene, what is the percentage yield?

You do it!

%56%100aniline g 11.9aniline g 6.7 = yield % =×

53

Concentration of Solutions • Solution is a mixture of two or more substances

dissolved in another. – Solute is the substance present in the smaller amount. – Solvent is the substance present in the larger amount. – In aqueous solutions, the solvent is water.

• The concentration of a solution defines the amount of solute dissolved in the solvent. – The amount of sugar in sweet tea can be defined by its

concentration. • One common unit of concentration is:

w/w% symbol thehas solute of massby %solvent of mass + solute of mass =solution of mass

%100solution of masssolute of mass = solute of massby % ×

54

Concentration of Solutions

• Example 3-11: What mass of NaOH is required to prepare 250.0 g of solution that is 8.00% w/w NaOH?

NaOH g 0.20solution g 100.0NaOH g 8.00solution g 0.250 =

55


• Example 3-12: Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.

nsol' g .400NaOH g 8.00

solution g 100.0NaOH g 32.0 =solution g ?

=

×

56


• Example 3-13: 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!

NaOH g 2.26nsol' g 100

NaOH g 8.00

nsol' mL 1nsol' g 1.09nsol' mL 300.0 = NaOH g ?

=

××

57

Concentrations of Solutions

• Example 3-14: 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!

58


• Example 3-14: What volume of 12.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL.

solution mL .300solution g 1.11solution mL 1

KOH g 12.0solution g 100.0KOH g 40.0 =solution mL ?

=

××

59


• Second common unit of concentration:

mLmmol

Lmoles

solution of liters ofnumber solute of moles ofnumber molarity

=

=

=

M

M

60


• Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.

You do it!

61



42

424242

SOH g 98.1SOH mol 1

nsol' L 75.1 SOH g 12.5

nsol' LSOH mol ?

×=

62



42

42

42

424242

SOH 0728.0L

SOH mol 0728.0SOH g 98.1SOH mol 1

nsol' L 75.1 SOH g 12.5

nsol' LSOH mol ?

M=

=

×=

63


• Example 3-16: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 .

You do it!

64


• Example 3-16: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 .

? g Ca(NO L 0.800 mol Ca(NO

L164 g Ca(NO mol Ca(NO

g Ca(NO

33

3

33

) .)

))

)

22

2

22

3 50

1459

= × ×

=

65


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

M x L = moles solute and

M x mL = mmol solute

66


• Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?

g/L 1185or g/mL 1.185=density us tells1.185 =gravity specific

67



××nsol' g 100

HCl g 31.36solution L

solution g 1185 = HCl/L mol ?

1185g/Lor g/mL 1.185=density us tells1.185 =gravity specific

68



HCl 11.80HCl g 36.46

HCl mol 1nsol' g 100

HCl g 31.36solution L

solution g 1185 = HCl/L mol ?

1185g/Lor g/mL 1.185=density us tells1.185 =gravity specific

M=

××

69

Dilution of Solutions

• To dilute a solution, add solvent to a concentrated solution. – One method to make tea “less sweet.” – How fountain drinks are made from syrup.

• The number of moles of solute in the two solutions remains constant.

• The relationship M1V1 = M2V2 is appropriate for dilutions, but not for chemical reactions.

70


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

71


• Example 3-18: 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?

M

MM

MMMM

20.1mL 100.0

mL 0.100.12mL 100.0mL 0.10 0.12

VV

2

2

2211

=

×=

×=×=

72


• Example 3-19: 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!

73


• Example 3-19: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?

mL 333or L 0.333 18.0

2.40 L 2.50V

V V

V V

1

1

221

2211

=

×=

×=

=

MM

MMMM

74

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.

75

Using Solutions in Chemical Reactions • Example 3-20: What volume of 0.500 M BaCl2

is required to completely react with 4.32 g of Na2SO4?

NaCl 2 + BaSO BaCl + SONa 4242 →

76



××=

→

42

42422

4242

SONa g 142SONa mol 1 SOgNa 4.32 BaCl L ?

NaCl 2 + BaSO BaCl + SONa

77



L 0.0608 BaCl mol 0.500

BaCl L 1SONa mol 1

BaCl mol 1SONa g 142SONa mol 1 SOgNa 4.32 BaCl L ?

NaCl 2 + BaSO BaCl + SONa

2

2

42

2

42

42422

4242

=×

××=

→

78

Using Solutions in Chemical Reactions • Example 3-21: (a)What volume of 0.200

M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate, Al(NO3)3?

( ) ( )You do it!

3333 NaNO 3OHAlNaOH 3NOAl +→+

79

Using Solutions in Chemical Reactions • Example 3-20: (a)What volume of 0.200 M

NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate?

( )

nsol' NaOH mL 150or L 0.150 NaOH mol 0.200

NaOH L 1)Al(NO mol 1

NaOH mol 3 nsol' )Al(NO L 1

n sol' )Al(NO mol 0.200mL 1000

L 1nsol' )Al(NO mL 50.0 = NaOH mL ?

NaNO 3Al(OH)NaOH 3NOAl

3333

33

33

3333

=

××

×

+→+

80

Using Solutions in Chemical Reactions • (b)What mass of Al(OH)3 precipitates in

(a)? You do it!

81

Using Solutions in Chemical Reactions • (b) What mass of Al(OH)3 precipitates in

(a)?

3

3

3

33

3

33

33

333

Al(OH) g 780.0Al(OH) mol 1Al(OH) g 0.78

)Al(NO mol 1Al(OH) mol 1

nsol' )Al(NO L1 )Al(NO mol 0.200

mL 1000 L1nsol' )Al(NO mL 50.0 Al(OH) g ?

=

××

×=

82

Using Solutions in Chemical Reactions • Titrations are a method of determining the

concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry. – Requires special lab glassware

• Buret, pipet, and flasks – Must have an indicator also

83

Using Solutions in Chemical Reactions • Example 3-22: What is the molarity of a KOH

solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

OH + KCl HCl + KOH 2→

84



HCl mmol 9.63 = HCl 0.223 mL 43.2OH + KCl HCl + KOH 2

M×→

85



KOH mmol 63.9HCl mmol 1

KOH mmol 1HCl mmol 9.63


=×

×→

M

86



KOH 249.0KOH mL 38.7KOH mmol 9.63

KOH mmol 63.9HCl mmol 1

KOH mmol 1HCl mmol 9.63


M

M

=

=×

×→

87

Using Solutions in Chemical Reactions • Example 3-23: What is the molarity of a barium

hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution?

HCl mmol 4.54 = HCl) HCl)(0.103 mL (44.1OH 2 + BaCl HCl 2 + Ba(OH) 222

M→

88



HCl mmol 2 Ba(OH) mmol 1HCl mmol 54.4

HCl mmol 4.54 = HCl) HCl)(0.103 mL (44.1OH 2 + BaCl HCl 2 + Ba(OH)

2

222

×

→M

89



2

2

222

Ba(OH) mmol 2.27 HCl mmol 2

Ba(OH) mmol 1HCl mmol 54.4


=

×

→M

90



22

2

2

2

222

Ba(OH) 0593.0Ba(OH) mL 3.38Ba(OH) mmol 27.2

Ba(OH) mmol 2.27 HCl mmol 2

Ba(OH) mmol 1HCl mmol 54.4


M

M

=

=

×

→

91

Synthesis Question

• Nylon is made by the reaction of hexamethylene diamine

CH2CH2CH2

CH2CH2CH2NH2

NH2

C

OH

O CH2

CH2

CH2

CH2

C

OH

O

with adipic acid.

92

Synthesis Question

in a 1 to 1 mole ratio. The structure of nylon is:

CNH

O

CH2

CH2

CH2

CH2

CH2

CH2

C

O

CH2

CH2

CH2

NH ** n

where the value of n is typically 450,000. On a daily basis, a DuPont factory makes 1.5 million pounds of nylon. How many pounds of hexamethylene diamine and adipic acid must they have available in the plant each day?

93

Synthesis Question

moleculesnylon g/mol 10 1.02 450,000 g/mol] [226

000,450 ])162()142()221( 12) (12[ moleculenylon 1 of massMolar

8

units of #atoms Oatoms Natoms Hatoms C

×=

=

×+×+×+×=

94

Synthesis Question

g 106.81 lb

g 454lb) 10 (1.5 poundsmillion 1.5



8

6

8


×=

×=

×=

=

×+×+×+×=

95

Synthesis Question

( )nylon of mol 6.68

g 101.02nylon mol 1 g 106.81 moleculesnylon of mol #

g 106.81 lb

g 454lb) 10 (1.5 poundsmillion 1.5



88

8

6

8


=

×

×=

×=

×=

×=

=

×+×+×+×=

96

Synthesis Question

:requiresnylon of mol 6.68 make toformed,nylon of moleper 450,000 diamine enehexamethyl of mole 1 plus

450,000 acid adipic of mole 1 usesreaction formation nylon theBecause×

×

97

Synthesis Question

( ) lb 1066.9g 454

lb 1g 104.39 g/mol 146450,000 6.68 - acid adipic


450,000 acid adipic of mole 1 usesreaction formation nylon theBecause

58 ×=

×=××

××

98

Synthesis Question

( )

( ) lb 1068.7g 454

lb 1g 1049.3 g/mol 116450,000 6.68 - diamine enehexamethyl

lb 1066.9g 454

lb 1g 104.39 g/mol 146450,000 6.68 - acid adipic


450,000 acid adipic of mole 1 usesreaction formation nylon theBecause

58

58

×=

×=××

×=

×=××

××

99

Group Activity

Manganese dioxide, potassium hydroxide and oxygen react in the following fashion:

OH 2 KMnO 4 O 3 + KOH 4 + MnO 4 2422 +→ A mixture of 272.9 g of MnO2, 26.6 L of 0.250

M KOH, and 41.92 g of O2 is allowed to react as shown above. After the reaction is finished, 234.6 g of KMnO4 is separated from the reaction mixture. What is the per cent yield of this reaction?

3 Chemical Equations &

Reaction Stoichimoetry

chemical equations & 3 reaction stoichiometry

Documents