CHAPTER THREE - Department of Chemistry · CHAPTER THREE zCHEMICAL EQUATIONS & REACTION ... Chapter Three Goals 1. Chemical Equations 2. ... Dilution of solutions 7. Using Solutions

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CHAPTER THREE

CHEMICAL EQUATIONS & REACTION CHEMICAL EQUATIONS & REACTION STOICHIOMETRYSTOICHIOMETRY

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Chapter Three Goals

1.1. Chemical EquationsChemical Equations2.2. Calculations Based on Chemical EquationsCalculations Based on Chemical Equations3.3. Percent Yields from Chemical ReactionsPercent Yields from Chemical Reactions4.4. The Limiting Reactant ConceptThe Limiting Reactant Concept5.5. Concentrations of SolutionsConcentrations of Solutions6.6. Dilution of solutionsDilution of solutions7.7. 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

5

Chemical Equations

Look at the information an equation provides:

reactants yields products

1 formula unit 3 molecules 2 atoms 3 molecules1 mole 3 moles 2 moles 3 moles159.7 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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Law of Conservation of Matter

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

⎯→⎯

⎯→⎯

∆

∆

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Law of Conservation of Matter

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 molecules1 mole 3 moles 2 moles 3 moles159.7 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?

25 Fe2O3 + ? CO Product1Fe2O3 needs 3 CO

25Fe2O3 needs ? CO

CO of molecules 75unit formula OFe 1

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

32

=

×

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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 2OFe units formula 102.50=atoms Fe ?

5

32

325

×=×

×

Fe2O3 + excess CO 2Fe + 3CO21Fe2O3 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 8.76COmol1CO g 28.0

OFe mol 1CO mol 3

OFe g 7.159OFe 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 01

..

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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 yieldtheoretical yield

× 100%

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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 .2HCOOCCH 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 + H2Osalicylic acid acetic anhydride aspirin

MW = 180 g/molMW = 138 g/mol

29 billion tablets are consumedby Americans each year

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actual Yield (150 g)Theoretical Yield (g)

78.5 = x 100

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

2MW 2MWX 191.08 (Theoretical Yield)

Answer: X = 146.5 g

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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 bikeexcess 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 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

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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 mol ?

SO g 161SO mol 1

SO g 1.64CS mol 1

SO mol 2g 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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3PbO2 + Cr2(SO4)3 + K2SO4 + H2O 3PbSO4 + K2Cr2O7 + H2SO4

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

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

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

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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, KNO3required 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 NaOHsolution that contains 32.0 g of NaOH.

nsol' g .400NaOH 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 1nsol' 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 .300solution g 1.11solution mL 1

KOH g 12.0solution 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

mLmmol

Lmoles

solution 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 0728.0SOH g 98.1SOH mol 1

nsol' L 75.1 SOH g 12.5

nsol' LSOH mol ?

M=

=

×=

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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

= × ×

=

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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 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=

××

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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 soluteM 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

MMMM

20.1mL 100.0

mL 0.100.12mL 100.0mL 0.10 0.12

VV

2

2

2211

=

×=

×=×=

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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

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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.

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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 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

=×

××=

→

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(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 +→+

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(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 780.0Al(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 ?

=

××

×=

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Homework AssignmentHomework Assignment

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

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

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CHAPTER THREE - Department of Chemistry · CHAPTER THREE zCHEMICAL EQUATIONS & REACTION ... Chapter Three Goals 1. Chemical Equations 2. ... Dilution of solutions 7. Using Solutions