1 CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY CHEMICAL EQUATIONS & REACTION STOICHIOMETRY

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CHAPTER THREE

CHEMICAL EQUATIONS & REACTION CHEMICAL EQUATIONS & REACTION STOICHIOMETRYSTOICHIOMETRY

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Chapter Three Goals

1.1. Chemical EquationsChemical Equations

2.2. Calculations Based on Chemical EquationsCalculations Based on Chemical Equations

3.3. Percent Yields from Chemical ReactionsPercent Yields from Chemical Reactions

4.4. The Limiting Reactant ConceptThe Limiting Reactant Concept

5.5. Concentrations of SolutionsConcentrations of Solutions

6.6. Dilution of solutionsDilution of solutions

7.7. Using Solutions in Chemical ReactionsUsing Solutions in Chemical Reactions

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Chemical Equations

A chemical process is represented by a chemical equation

1. Reaction of methane with O2:

CH4 + 2O2 CO2 + 2H2O

reactantsreactants productsproducts

2. reactants on left side of reaction3. products on right side of equation4. relative amounts of each using stoichiometric coefficients

4

Chemical Equations

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Chemical Equations

Look at the information an equation provides:

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 132 g

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

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

OH 4 CO 3 O 5 HC 22283

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NH3 burns in oxygen to form NO & water

OH 6 + NO 4 O 5 + NH 4

correctlyor

OH 3 + NO 2 O + NH 2

223

2225

3

8


C7H16 burns in oxygen to form carbon dioxide and water. You do it! You do it!

Balancing equations is a skill acquired only with lots of practice– work many problems

OH 8 + CO 7 O 11 + HC 222167

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Chemical Equations

Look at the information an equation provides:

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 132 g

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

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Calculations Based on Chemical Equations

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

25 Fe2O3 + ? CO Product

1Fe2O3 needs 3 CO

25Fe2O3 needs ? CO

11


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

Fe2O3 + excess CO 2Fe + 3CO2 1Fe2O3 gives 2Fe

2.5 X 105 Fe2O3 gives ? Fe

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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

OFe mol 1OFe g 146 = CO g ?

3232

3232

Fe2O3 + 3CO Product

MW(Fe2O3) needs 3MW(CO)

146 g needs ?g CO

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What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? Fe2O3 + excess CO 2Fe + 3CO2

1mol Fe2O3 gives 3 mol CO2

0.540 mol Fe2O3 gives ? mol CO2

? mol CO2 = ? g CO2/MW(g/mol) CO2

? 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.

.

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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!

Fe2O3 + excess CO 2Fe + 3CO2

? g Fe2O3 = 9.57 g Fe2O3

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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 from a reaction.

% yield = actual yield

theoretical yield100%

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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?

CH3COOH + C2H5OH CH3COOC2H5 + H2O

MW MW

10.0 g X (Theoretical Yield)

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%5.77%100HCOOCCH g 19.1

HCOOCCH g 14.8= yield %

yield.percent theCalculate .2

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

523

523

523

52

52352523

2523523

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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 C7H6O3 + C4H6O3 2 C9H8O4 + H2O salicylic acid acetic anhydride aspirin

29 billion tablets are consumedby Americans each year

MW = 138 g/mol MW = 180 g/mol

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2 C7H6O3 + C4H6O3 2 C9H8O4 + H2Osalicylic acid acetic anhydride aspirin

2MW 2MW X 191.08 (Theoretical Yield)

Answer: X = 146.5 g

actual Yield (150 g)

Theoretical Yield (g)78.5 = x 100

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Limiting Reactant Concept

1. In a given reaction, there is not enough of one reagent to

use up the other reagents completely.

2. The reagent in short supply LIMITS the quantity of the

product that can be formed.

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

1 frame + 2 wheels 1 bike

excess limiting

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

Hg + Br2 HgBr2

Thus the limiting reagent is

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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?

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

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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 mol ?

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 mol ?

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.

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If 25.0 g of each reactant were used in performing the

following reaction, which would be the limiting reactant?

(a) PbO2 (b) H2O (c) K2SO4 (d) PbSO4 (e) Cr2(SO4)3

3PbO2 + Cr2(SO4)3 + K2SO4 + H2O 3PbSO4 + K2Cr2O7 + H2SO4

If 20.0 g of each reactant were used in performing the

following reaction, which would be the limiting reactant?

(a) MnO2 (b) KOH (c) O2 (d) Cl2 (e) KMnO4

2MnO2 + 4KOH + O2 + Cl2 2KMnO4 + 2KCl + 2H2O

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Concentration of Solutions

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, H2O:

NaCl: solute, H2O: solvent

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Concentration =Amount of solute

Mass or Volume of solutionRelative 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

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1- Percent by mass:

w/w% symbol thehas solute of massby %

solvent of mass + solute of mass =solution of mass

%100solution of mass

solute of mass = solute of massby %

Note: if the question says the solution is aqueous oe does notSpecify the solvent, the solvent is water, H2O.

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Calculate the mass of potassium nitrate, KNO3 required to prepare 250.0 g of solution that is 20.0 % KNO3 by mass.

What is the mass of water in the solution?

(a) % by mass = g KNO3

g solutionx 100

20.0 % = g KNO3

250.0 gx 100

g KNO3 = 100

20.0 % x 250.0 g= 50.0 g

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(b) mass of solution = mass of KNO3 + mass H2O

mass H2O = mass of solution - mass of KNO3

mass H2O = 250.0 g - 50.0 g

mass H2O = 200.0 g

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

nsol' g .400

NaOH g 8.00

solution g 100.0NaOH g 32.0 =solution g ?

32


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!You do it!

NaOH g 2.26nsol' g 100

NaOH g 8.00

nsol' mL 1

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

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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.


solution mL .300

solution g 1.11

solution mL 1

KOH g 12.0

solution g 100.0KOH g 40.0 =solution mL ?

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2- Second common unit of concentration: Molarity

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2- Second common unit of concentration: Molarity

mL

mmolL

molessolution of liters ofnumber

solute of moles ofnumber molarity

M

M

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


42

42

42

424242

SOH 0728.0L

SOH mol 0728.0

SOH g 98.1

SOH mol 1

nsol' L 75.1

SOH g 12.5

nsol' L

SOH mol ?

M

37


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

38


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

M x L = moles solute

and

M x mL = mmol solute

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

HCl 11.80HCl g 36.46

HCl mol 1

nsol' g 100

HCl g 31.36

solution L

solution g 1185 = HCl/L mol ?

1185g/Lor g/mL 1.185=density

us tells1.185 =gravity specific

M

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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.

The relationship M1V1 = M2V2 is appropriate for dilutions,

but not for chemical reactions.

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

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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

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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

MM

MM

20.1mL 100.0

mL 0.100.12

mL 100.0mL 0.10 0.12

VV

2

2

2211

44


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

M

M

M

M

MM

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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.

46


What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4?

L 0.0608 BaCl mol 0.500

BaCl L 1

SONa mol 1

BaCl mol 1

SONa g 142

SONa mol 1 SOgNa 4.32 BaCl L ?

NaCl 2 + BaSO BaCl + SONa

2

2

42

2

42

42422

4242

47


(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

48


(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

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

3

3

3

33

3

33

33

333

Al(OH) g 780.0

Al(OH) mol 1

Al(OH) g 0.78

)Al(NO mol 1

Al(OH) mol 1

nsol' )Al(NO L1

)Al(NO mol 0.200mL 1000

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

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Homework AssignmentHomework Assignment

One-line Web Learning (OWL):One-line Web Learning (OWL):

Chapter 3 Exercises and Tutors – Chapter 3 Exercises and Tutors – OptionalOptional

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End of Chapter 3

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1 CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY CHEMICAL EQUATIONS & REACTION STOICHIOMETRY