Chemical and Physical Properties - Armstrong State · PDF fileChemical and Physical Properties • Chemical Properties - chemical changes – rusting or oxidation – chemical reactions

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CHAPTER FOUR

• CHEMICAL EQUATIONS & STOICHIOMETRY

2

Chemical and Physical Properties

• Chemical Properties - chemical changes

– rusting or oxidation

– chemical reactions

• Physical Properties - physical changes

– changes of state

– density, color, solubility, melting, boiling

• Extensive Properties - depend on quantity

• Intensive Properties - do not depend on quantity

3

States of Matter

• Change States –Physical Change– heating

– cooling

4

Physical Change vs. Chemical Change

5

Physical Change vs. Chemical Change

6

Chemical Equations

• Symbolic representation of a chemical reaction (chemical change) that shows:

1. reactants on left side of reaction

2. products on right side of equation

3. relative amounts of each using coefficients

7

Chemical Equations

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

8

Chemical Equations

• Look at the information an equation provides:

reactants yields products

1 formula unit 3 molecules 2 atoms 3 molecules

States of matter also listed

)(CO 3 + Fe(s)2 CO(g) 3 +(s) OFe 232 g→∆

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Chemical Equations

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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.

OH 4 CO 3 O 5 HC 22283 +→+∆

Balancing equations is a skill acquired only with lots of practice

�work many problems

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Balancing Equations

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Combination Reactions

• Combination reactions occur when two or more substances combine to form a compound.

• There are three basic types of combination reactions.

1. Two elements react to form a new compound

2. An element and a compound react to form one new compound

3. Two compounds react to form one compound

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( ) ( ) ( )sg2s NaCl ?Cl?Na ? →+

( ) ( ) ( )sg2s MgO?O? Mg? →+

( ) ( ) ( )s32s AlBr ?Br?Al ? →+l

14


( ) ( ) ( )s104g2s4 O?P O ??P →+

( ) ( ) ( )l3g2s4 PCl_ Cl __P →+

( ) ( ) ( )g3g2g2 SO_O__SO →+

( ) ( ) ( )g2g2g CO_O__CO →+

104264 OP_O_O_P →+

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Decomposition Reactions

• Decomposition reactions occur when one compound decomposes to form:

1. Two elements

2. One or more elements and one or more compounds

3. Two or more compounds

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( ) ( ) ( )g2g2g2 O_N _ON _ +→∆

( ) ( ) ( )l2s

h

s B_Ag _AgBr _ r+→ν

17


( ) ( ) ( )g22aq22 O_O H_O_H +→l

( ) ( ) ( ) ( )g2g2g3s34 COO HNHHCONH ++→∆

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Displacement Reactions

•• Displacement reactionsDisplacement reactions occur when one element displaces another element from a compound. (there are many types)

( ) ( ) (s)aq3(s) aq3 Ag _ _CuNO _Cu +_AgNO +→

( ) ( ) ( )g2aq342aq42(s) H_ + )(SO_Al SO_H+ _Al →

( ) ( ) ( ) (aq)s2aqg2 _NaCl_I NaI _ + _Cl +→

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)(2aq)(23(aq)3(aq)2 O_H+ )Ca(NO__HNO + _Ca(OH)l

→

(s)3)aq(3aq)(32(aq)23 CaCO _+_KNO CO_K + )_Ca(NO →

( )2(s)43)aq(aq)(43(aq)2 POCa _+NaCl _ PO_Na + CaCl _ →

Displacement Reactions

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Reactions Overview

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Law of Conservation of Matter

• NH3 burns in oxygen to form NO & water

22


• 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

→

→

∆

∆

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

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

OH 8 + CO 7 O 11 + HC 222167 →∆

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Calculations Based on Chemical Equations

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• Example 3-1: 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 ?

32

32

=

×

Fe O + 3 CO 2 Fe + 3 CO2 3 2∆

→

27


• 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?

32

5OFe units formula 102.50=atoms Fe ? ×

28



=×

×

OFe units formula 1

atoms Fe 2

OFe units formula 102.50=atoms Fe ?

32

32

5

29



atoms Fe 105.00 OFe units formula 1

atoms Fe 2

OFe units formula 102.50=atoms Fe ?

5

32

32

5

×=×

×

30


• 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 ? ×

31



3232

3232

OFe mol 1

CO mol 3

OFe g 7.159

OFe mol 1OFe g 146 = CO g ? ××

32



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

=×

××

33


• Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with carbon monoxide?

OFe mol 1

CO mol 3OFe mol 540.0CO g ?

32

2322 ×=

34



2

2

32

2322

CO mol 1

CO g 0.44

OFe mol 1

CO mol 3OFe mol 540.0CO g ? ××=

35



? 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

= × ×054044 0

1.

.

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• Example 3-5: What mass of iron (III) oxide reacted with carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

You do it!You do it!

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32

32

32

2

32

2

2232

O Feg 5.10O Femol 1

O 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 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?

You do it!You do it!

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CO lb 7.65CO g 454

CO lb 1

CO mol 1

CO g 28

OFe mol 1

CO mol 3

OFe g 7.159

OFe mol 1

OFe lb 1

OFe g 454OFe lb 125 = CO lb ?

3232

32

32

3232

=×

×××

×

YOU MUST BE PROFICIENT WITH THESE

TYPES OF PROBLEMS!!!

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Limiting Reactant Concept

• In a chemical reaction, the reaction proceeds forward according to the stoichiometry of the reaction until one reagent (or both) is consumed. The reagent consumed first is the limiting reactant

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• 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 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

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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 determined

sets 55 is make can number we maximum the

sets 99nut 1

set 1nuts 99

sets 55 washers2

set 1 washers110

sets 87bolt 1

set 1bolts 87

=

=

=

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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 1

SO 2 CO O 3 CS 2222 +→+

46



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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• 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?– Determine which mass makes the most product

g 76.2

CS mol 1CS g 6.95 SO mol ?

SO 2 CO O 3 CS

222

2222

×=

+→+

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2

2

2

2

2

222

2222

SO g 161SO mol 1

SO g 1.64

CS mol 1

SO mol 2

g 76.2


SO 2 CO O 3 CS

=××

×=

+→+

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2

2

2

2

2

2

222

2

2

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


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.

50

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 yield

theoretical yield× 100%

51


• 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?

yieldal theoretic theCalculate 1.

O HHCOOCCH OH HC +COOH CH 2523523 +→

52


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

=

×

+→

53


yield.percent theCalculate .2

HCOOCCH g 1.19

OHHC g 0.46




523

52

52352523

2523523

=

×

+→

54


%5.77%100HCOOCCH g 19.1

HCOOCCH g 14.8= yield %

yield.percent theCalculate .2

HCOOCCH g 1.19

OHHC g 0.46




523

523

523

52

52352523

2523523

=×

=

×

+→

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• 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∆

→

56


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

Fe O + 3 CO 2 Fe + 3 CO2 3 2∆

→

57


• Example: 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

=×

××

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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?

59


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

? 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

= × ×054044 0

1.

.

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Chemical and Physical Properties - Armstrong State · PDF fileChemical and Physical Properties • Chemical Properties - chemical changes – rusting or oxidation – chemical reactions