Download ppt - CHAPTER 03 Equations and Reaction Stoichiometry

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CHAPTER THREE

CHEMICAL EQUATIONS & REACTION STOICHIOMETRY

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Chapter Three Goals

1. Chemical Equations2. Calculations Based on Chemical Equations3. The Limiting Reactant Concept4. Percent Yields from Chemical Reactions5. Sequential Reactions6. Concentrations of Solutions7. Dilution of solutions8. Using Solutions in Chemical Reactions9. Synthesis Question

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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

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Chemical Equations

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

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Chemical Equations

Look at the information an equation provides:

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

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Chemical Equations


reactants yields products

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

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Chemical Equations



1 formula unit 3 molecules 2 atoms 3 molecules

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

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Chemical Equations




1 mole 3 moles 2 moles 3 moles

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

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Chemical Equations




1 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.– 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

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Law of Conservation of Matter

NH3 burns in oxygen to form NO & water

You do it!You do it!

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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

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2225

3

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C7H16 burns in oxygen to form carbon dioxide and water.

You do it! You do it!

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OH 8 + CO 7 O 11 + HC 222167

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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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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

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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

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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 ?

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OFe units formula 1atoms Fe 2

OFe units formula 102.50=atoms Fe ?

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325

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atoms Fe 105.00 OFe units formula 1

atoms Fe 2OFe units formula 102.50=atoms Fe ?

5

32

325

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

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3232 OFe g 7.159

OFe mol 1OFe g 146 = CO g ?

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3232

3232 OFe mol 1

CO mol 3OFe g 7.159

OFe mol 1OFe g 146 = CO g ?

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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

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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 ?

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2322

25



2

2

32

2322 CO mol 1

CO g 0.44 OFe mol 1

CO mol 3OFe mol 540.0CO g ?

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? 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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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?


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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?

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Example 3-6: How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide?


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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!!!

Now go to your text and work the problems assigned!

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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?

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Mix Packets Eggs Milk1 1 dozen 1 gallon

limiting reactant is the muffin mix2 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

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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

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Look at a chemical limiting reactant situation.Zn + 2 HCl ZnCl2 + H2

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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?

2222 SO 2 CO O 3 CS

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mol 2 mol 1 mol 3 mol 1SO 2 CO O 3 CS 2222

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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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g 76.2CS mol 1CS g 6.95 SO mol ?

SO 2 CO O 3 CS

222

2222

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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 mol ?

SO 2 CO O 3 CS

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

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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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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%

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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?

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yield al theoretic theCalculate 1.OH HCOOCCH OHHC + COOHCH 2523523

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523

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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

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yield.percent theCalculate .2HCOOCCH g 1.19

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


523

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52352523

2523523

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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 ?


523

523

523

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52352523

2523523

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Sequential Reactions

N O2 NH2HNO3H2SO4

SnConc. 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).

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

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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!You do it!

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N O2 NH2HNO3H2SO4

SnConc. HCl


nenitrobenze g 123.0

nenitrobenze mol 1nenitrobenze g 15.8 = aniline g ?

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N O2 NH2HNO3H2SO4

SnConc. HCl


? g aniline = 15.8 g nitrobenzene1 mol nitrobenzene

123.0 g nitrobenzene

1 mol aniline

1 mol nitrobenzene93.0 g aniline1 mol aniline

g aniline

11 9.

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


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

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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 %

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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 ?

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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

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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.


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

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.


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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 ?

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Second common unit of concentration:

mLmmol

Lmoles

solution of liters ofnumber solute of moles ofnumber molarity

M

M

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


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42

424242

SOH g 98.1SOH mol 1

nsol' L 75.1 SOH g 12.5

nsol' LSOH mol ?

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


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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

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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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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

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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

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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.”– 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.

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

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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

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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?


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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

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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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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

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42

42422

4242

SONa g 142SONa mol 1 SOgNa 4.32 BaCl L ?

NaCl 2 + BaSO BaCl + SONa

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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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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?

3 33 3Al NO 3 NaOH Al OH 3 NaNO

You do it!

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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

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


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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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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 an indicator also

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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

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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

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KOH 249.0KOH mL 38.7KOH mmol 9.63

KOH mmol 63.9HCl mmol 1

KOH mmol 1HCl mmol 9.63


M

M

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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

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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

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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) mL 27.2

Ba(OH) mmol 2.27 HCl mmol 2

Ba(OH) mmol 1HCl mmol 54.4


M

M

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Synthesis Question

Nylon is made by the reaction of hexamethylene diamine

CH2CH2CH2

CH2CH2CH2NH2

NH2

COH

O CH2

CH2

CH2

CH2

COH

Owith adipic acid.

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Synthesis Question

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

CNH

OCH2

CH2

CH2

CH2

CH2

CH2

CO

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?

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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

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Synthesis Question

g 106.81 lb

g 454lb) 10 (1.5 poundsmillion 1.5


450,000 g/mol] [226


8

6

8


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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


450,000 g/mol] [226


88

8

6

8


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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

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Synthesis Question

lb 1066.9g 454

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



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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



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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?

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End of Chapter 3