chapter 9 chemical equations & reaction stoichiometry
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Chapter 9Chemical Equations &
Reaction Stoichiometry
2
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
3
Chemical Equations
• Attempt to show on paper what is happening at the laboratory and molecular levels.
4
Chemical Equations
• Look at the information an equation provides:
Fe O + 3 CO 2 Fe + 3 CO2 3 2
5
Chemical Equations
• Look at the information an equation provides:
reactants yields productsFe O + 3 CO 2 Fe + 3 CO2 3 2
6
Chemical Equations
• Look at the information an equation provides:
reactants yields products
1 formula unit 3 molecules 2 atoms 3 molecules
Fe O + 3 CO 2 Fe + 3 CO2 3 2
7
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
Fe O + 3 CO 2 Fe + 3 CO2 3 2
8
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 moles159.7 g 84.0 g 111.7 g 132g
Fe O + 3 CO 2 Fe + 3 CO2 3 2
9
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
10
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
11
Calculations Based on Chemical Equations
• Example 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 ?
3232
12
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 ?
13
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?
OFe units formula 1
atoms Fe 2
OFe units formula 102.50=atoms Fe ?
32
325
14
Calculations Based on Chemical Equations
• Example 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?
atoms Fe 105.00 OFe units formula 1
atoms Fe 2
OFe units formula 102.50=atoms Fe ?
5
32
325
15
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 ?
16
Calculations Based on Chemical Equations
• Example 3: What mass of CO is required to react with 146 g of iron (III) oxide?
3232
3232 OFe mol 1
CO mol 3
OFe g 7.159
OFe mol 1OFe g 146 = CO g ?
17
Calculations Based on Chemical Equations
• Example 3-3: 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
18
Calculations Based on Chemical Equations
• Example 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 1
CO mol 3OFe mol 540.0CO g ?
32
2322
19
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?
2
2
32
2322 CO mol 1
CO g 0.44
OFe mol 1
CO mol 3OFe mol 540.0CO g ?
20
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?
? 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.
.
21
Calculations Based on Chemical Equations
• Example 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!You do it!
22
Calculations Based on Chemical Equations
3232
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 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?
23
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?
24
Limiting Reactant Concept• Mix Packets Eggs Milk
1 1 dozen 1 gallonlimiting reactant is the muffin mix
2 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
25
Limiting Reactant Concept
• Example 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
26
Limiting Reactant Concept
• Example 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
1 mol 3 mol 1 mol 2 mol
27
Limiting Reactant Concept
• Example 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?
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
28
Limiting Reactant Concept
• Example 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?
g 76.2
CS mol 1CS g 6.95 SO mol ?
SO 2 CO O 3 CS
222
2222
29
Limiting Reactant Concept
22
2
2
2
222
2222
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 g ?
SO 2 CO O 3 CS
What do we do next?You do it!You do it!
30
Limiting Reactant Concept
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 g ?
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 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.
31
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 yield100%
32
Percent Yields from Reactions
• Example 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?
33
Percent Yields from Reactions
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
34
Percent Yields from Reactions
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
35
Percent Yields from Reactions
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
52
52352523
2523523
36
Percent Yields from Reactions
%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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