05 persamaan kimia dan stoikiometri.ppt
TRANSCRIPT
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Chapter 4 1
Chemical Equations andStoichiometry
Chapter 4
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Chapter 4 2
2H2(g) + O2(g) 2H2O(g)
Chemical Equations
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Chapter 4 3
2H2(g) + O2(g) 2H2O(g)
• The materials you start with are called Reactants.
Chemical Equations
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Chapter 4 4
2H2(g) + O2(g) 2H2O(g)
• The materials you start with are called Reactants.
• The materials you make are called Products.
Chemical Equations
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Chapter 4 5
2H2(g) + O2(g) 2H2O(g)
• The materials you start with are called Reactants.
• The materials you make are called Products.
• The numbers in front of the compounds (H2 andH2O) are called stoichiometric coefficients .
– Coefficients are multipliers, in this equation 2 in front of
the H2 indicates that there are 2 molecules of H2 in the
equation.
Chemical Equations
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Chapter 4 6
2H2(g) + O2(g) 2H2O(g)
• Notice that the number of hydrogen atoms and
oxygen atoms on the reactant side and the product
side is equal.
Law of Conservation of Mass
Matter cannot be created or lost in any chemical
reaction.
Chemical Equations
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Balancing Chemical Reactions
___NH4NO3(s) ___N2O(g) + ___H2O(g)
Chemical Equations
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Balancing Chemical Reactions
___NH4NO3(s) ___N2O(g) + _ 2 _H2O(g)
Chemical Equations
Reactants Products
N 2 N 2
H 4 H 2 4
O 3 O 2 3
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Chapter 4 10
Balancing Chemical Reactions
___Mg3N2(s) + ___H2O(l) ___Mg(OH)2(s) + ___NH3(aq)
Chemical Equations
Reactants Products
Mg 3 Mg 1
N 2 N 1
H 2 H 5
O 1 O 2
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Chapter 4 11
Balancing Chemical Reactions
___Mg3N2(s) + ___H2O(l) _ 3 _Mg(OH)2(s) + ___NH3(aq)
Chemical Equations
Reactants Products
Mg 3 Mg 1 3
N 2 N 1
H 2 H 5 9
O 1 O 2 6
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Chapter 4 12
Balancing Chemical Reactions
___Mg3N2(s) + ___H2O(l) _ 3 _Mg(OH)2(s) + _ 2 _NH3(aq)
Chemical Equations
Reactants Products
Mg 3 Mg 1 3
N 2 N 1 2
H 2 H 5 9 12
O 1 O 2 6
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Chapter 4 13
Balancing Chemical Reactions
___Mg3N2(s) + _ 6 _H2O(l) _ 3 _Mg(OH)2(s) + _ 2 _NH3(aq)
Chemical Equations
Reactants Products
Mg 3 Mg 1 3
N 2 N 1 2
H 2 12 H 5 9 12
O 1 6 O 2 6
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Chapter 4 14
2 H2(g) + O2(g) 2 H2O(g)
• The coefficients in a balanced equation represent both
the number of molecules and the number of moles in a
reaction.
•The coefficients can also be used to derive ratiosbetween any two substances in the chemical reaction.
2 H2 : 1 O2
2 H2 : 2 H2O
1 O2 : 2 H2O
Quanti tative I nformation
•The ratios can be used to predict
•The amount of product formed
•The amount of reactant needed
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Chapter 4 15
Quanti tative I nformation
Contoh:
A B
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Chapter 4 16
Quanti tative I nformation
2 C4H
10(l ) + 13 O
2(g) 8 CO
2(g) + 10 H
2O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
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Chapter 4 17
Quanti tative I nformation
2 C4H
10(l ) + 13 O
2(g) 8 CO
2(g) + 10 H
2O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
1. Moles of C4H10
F.W. 58.124g
g
mol g H C moles
124.58
100.1
104
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Chapter 4 18
Quanti tative I nformation
2 C4H
10(l ) + 13 O
2(g) 8 CO
2(g) + 10 H
2O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
1. Moles of C4H10
F.W. 58.124g
mol g
mol g H C moles 0172.0
124.58
100.1
104
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Chapter 4 19
Quanti tative I nformation
2 C4H
10(l ) + 13 O
2(g) 8 CO
2(g) + 10 H
2O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
2. Ratio of C4H10:CO2
2 C4H10 : 8 CO2 or
104
2
2
8
H C
CO
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Chapter 4 20
Quanti tative I nformation
2 C4H
10(l ) + 13 O
2(g) 8 CO
2(g) + 10 H
2O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
3. Set-up ratio and proportion between known and unknownquantities
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Chapter 4 21
Quanti tative I nformation
2 C4H
10(l ) + 13 O
2(g) 8 CO
2(g) + 10 H
2O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
3. Set-up ratio and proportion between known and unknownquantities
104104
2
0172.02
8
H C mol
x
H C
CO
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Chapter 4 22
Quanti tative I nformation
2 C4H
10(l ) + 13 O
2(g) 8 CO
2(g) + 10 H
2O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
3. Set-up ratio and proportion between known and unknownquantities
2
104104
2
0688.0
0172.02
8
COmol x
H C mol
x
H C
CO
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Chapter 4 23
Quanti tative I nformation
2 C4H
10(l ) + 13 O
2(g) 8 CO
2(g) + 10 H
2O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
4. Convert the moles of unknown substance into the desiredunits
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Chapter 4 24
Quanti tative I nformation
2 C4H
10(l ) + 13 O
2(g) 8 CO
2(g) + 10 H
2O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
4. Convert the moles of unknown substance into the desiredunits
FW of CO2: 44.011g/mol
mol g mol CO g /011.440688.02
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Chapter 4 25
Quanti tative I nformation
2 C4
H10
(l ) + 13 O2
(g) 8 CO2
(g) + 10 H2
O(g)
How many grams of CO2 are formed if 1.00g of butane (C4H10) is
allowed to react with excess oxygen?
4. Convert the moles of unknown substance into the desiredunits
FW of CO2: 44.011g/mol
g mol g mol CO g 03.3/011.440688.02
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Chapter 4 26
“What runs out first”
2 C8H18 + 25 O2 16 CO2 + 18 H2O
• If you have 2 moles of C8H18 and 20 moles of O2
all the O2 will be used and the reaction will stop• O2 is call the l imiting reagent (reactant)
L imiting Reactant – The reagent present in the smallest
stoichiometric quantity in a mixture of reactants.
L imiting Reactants
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Chapter 4 27
Example
2 C8H18 + 25 O2 16 CO2 + 18 H2O
Determine the limiting reagent of this reaction if 10.0
grams of C8H18 and 25.0 grams of O2 are allowed to
react.1. Convert grams to moles
FW(C8H18) 114.268g/mol FW(O2) = 32.00g/mol
L imiting Reactants
g
mol g Omoles
g mol g H C moles
00.32
10.25
268.11410.10
2
188
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Chapter 4 28
Example
2 C8H18 + 25 O2 16 CO2 + 18 H2O
Determine the limiting reagent of this reaction if 10.0
grams of C8H18 and 25.0 grams of O2 are allowed to
react.1. Convert grams to moles
FW(C8H18) 114.268g/mol FW(O2) = 32.00g/mol
L imiting Reactants
22
188188
781.000.32
10.25
0875.0268.114
10.10
Omol g
mol g Omoles
H C mol g
mol g H C moles
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Chapter 4 29
Example
2 C8H18 + 25 O2 16 CO2 + 18 H2O
Determine the limiting reagent of this reaction if 10.0
grams of C8H18 and 25.0 grams of O2 are allowed to
react.2. Divide each reagent by its own coefficient
L imiting Reactants
25
781.0
2
0875.0
2
188
O
H C
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Chapter 4 30
Example
2 C8H18 + 25 O2 16 CO2 + 18 H2O
Determine the limiting reagent of this reaction if 10.0
grams of C8H18 and 25.0 grams of O2 are allowed to
react.2. Divide each reagent by its own coefficient
L imiting Reactants
0313.025
781.0
0438.0
2
0875.0
2
188
O
H C
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Chapter 4 31
Example
2 C8H18 + 25 O2 16 CO2 + 18 H2O
Determine the limiting reagent of this reaction if 10.0
grams of C8H18 and 25.0 grams of O2 are allowed to
react.3. The substance with the smallest calculated value will
be the limiting reagent. In this case, O2 is the limiting
reagent.
L imiting Reactants
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Chapter 4 32
Theoretical Yield
- The calculated amount of product based on thelimiting reactant (Theoretical yield).
L imiting Reactants
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Chapter 4 33
2 C8H18 + 25 O2 16 CO2 + 18 H2O
Determine the theoretical yield of CO2 for this reaction if
10.0 grams of C8H18 and 25.0 grams of O2 are allowed to
react.- already know that O2 is the limiting reactant.
L imiting Reactants
Theoretical Yield
22 781.000.32
10.25 Omol
g
mol g Omoles
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Chapter 4 34
2 C8H18 + 25 O2 16 CO2 + 18 H2O
•Calculate moles of oxygen
L imiting Reactants
Theoretical Yield
22 781.0
00.32
10.25 Omol
g
mol g Omoles
•Calculate moles of CO2
2
22
2
500.0
781.025
16
COmoles x
Omol
x
O
CO
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Chapter 4 35
2 C8H18 + 25 O2 16 CO2 + 18 H2O
•Calculate moles of CO2
L imiting Reactants
Theoretical Yield
22 0.221
0.44500.0 CO g mol
g mol COmoles
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Chapter 4 36
Percent Yield
- Calculation which indicates how much of thetheoretical yield was obtained.
L imiting Reactants
100 l yield Theoretica
ld Actual yie% Yield
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Chapter 4 37
Combustion Analysis
Empir ical Formulas from Analyses
Typical example:2 C2H6(g) + 7 O2 4 CO2(g) + 6 H2O(g)
- The combustion of any hydrocarbon produces CO2
and water.
- This observation can be used to determine the
empirical formula of the reactant.
Combustion Reaction: The “burning” of any substance
in oxygen.
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Chapter 4 38
Combustion Analysis
Empir ical Formulas from Analyses
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Chapter 4 39
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Carbon
mass CO2 moles CO2 moles C grams C
Empir ical Formulas from Analyses
2
2
011.44
12829.0
CO g COmol g
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Chapter 4 40
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Carbon
mass CO2 moles CO2 moles C grams C
Empir ical Formulas from Analyses
22
2
1
1
011.44
12829.0
COmol C mol
CO g COmol g
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Chapter 4 41
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Carbon
mass CO2 moles CO2 moles C grams C
Empir ical Formulas from Analyses
C mol C g
COmol C mol
CO g COmol g
1
011.12
1
1
011.44
12829.0
22
2
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Chapter 4 42
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Carbon
mass CO2 moles CO2 moles C grams C
Empir ical Formulas from Analyses
C g C mol
C g COmol
C mol CO g
COmol g 07721.01
011.12
1
1
011.44
12829.0
22
2
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Chapter 4 43
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Hydrogen
mass H2O moles H2O moles H grams H
Empir ical Formulas from Analyses
O H g O H mol g 2
2
02.18
11159.0
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Chapter 4 44
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Hydrogen
mass H2O moles H2O moles H grams H
Empir ical Formulas from Analyses
O H mol H mol
O H g O H mol g
22
2
1
2
02.18
11159.0
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Chapter 4 45
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Hydrogen
mass H2O moles H2O moles H grams H
Empir ical Formulas from Analyses
H mol H g
O H mol H mol
O H g O H mol g
1
01.1
1
2
02.18
11159.0
22
2
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Chapter 4 46
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Hydrogen
mass H2O moles H2O moles H grams H
Empir ical Formulas from Analyses
g H mol H g
O H mol H mol
O H g O H mol g 01299.0
1
01.1
1
2
02.18
11159.0
22
2
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Chapter 4 47
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Oxygen
mass O = mass of sample – (mass C +mass H)
Empir ical Formulas from Analyses
g g g Oxygenmass 01299.007721.01005.0
f
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Chapter 4 48
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Oxygen
mass O = mass of sample – (mass C +mass H)
Empir ical Formulas from Analyses
g
g g g Oxygenmass01030.0
01299.007721.01005.0
E i i l F l f A l
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Chapter 4 49
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Mass of elements:
C 0.07721g
H 0.01299g
O 0.01030g
Empir ical Formulas from Analyses
E i i l F l f A l
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Chapter 4 50
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Moles of elements:
C 0.07721g/12.011g/mol =
H 0.01299g/1.01g/mol =
O 0.01030g/16.00g/mol =
Empir ical Formulas from Analyses
E i i l F l f A l
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Chapter 4 51
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Moles of elements:
C 0.07721g/12.011g/mol = 0.006428 mol
H 0.01299g/1.01g/mol = 0.012861 mol
O 0.01030g/16.00g/mol = 0.0006428mol
Empir ical Formulas from Analyses
0006428.0012861.0006428.0 O H C
E i i l F l f A l
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Chapter 4 52
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Moles of elements:
C 0.07721g/12.011g/mol = 0.006428 mol
H 0.01299g/1.01g/mol = 0.012861 mol
O 0.01030g/16.00g/mol = 0.0006428 mol
Empir ical Formulas from Analyses
0006428.0
0006428.0
0006438.0
012861.0
0006428.0
006428.0 O H C
E i i l F l f A l
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Chapter 4 53
Combustion Analysis
Menthol, the substance we can smell in mentholated cough drops,is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Moles of elements:
C 0.07721g/12.011g/mol = 0.006428 mol
H 0.01299g/1.01g/mol = 0.012861 mol
O 0.01030g/16.00g/mol = 0.0006428 mol
Empir ical Formulas from Analyses
12010 O H C
P ti P bl