05 persamaan kimia dan stoikiometri.ppt

7/21/2019 05 Persamaan Kimia dan Stoikiometri.ppt

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Chapter 4 1

Chemical Equations andStoichiometry

Chapter 4



Chapter 4 2

2H2(g) + O2(g) 2H2O(g)

Chemical Equations



Chapter 4 3

2H2(g) + O2(g) 2H2O(g)

• The materials you start with are called Reactants.

Chemical Equations



Chapter 4 4

2H2(g) + O2(g) 2H2O(g)


• The materials you make are called Products.

Chemical Equations



Chapter 4 5

2H2(g) + O2(g) 2H2O(g)


• 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



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


http://slidepdf.com/reader/full/05-persamaan-kimia-dan-stoikiometrippt 7/54Chapter 4 7

Balancing Chemical Reactions

___NH4NO3(s) ___N2O(g) + ___H2O(g)

Chemical Equations


http://slidepdf.com/reader/full/05-persamaan-kimia-dan-stoikiometrippt 9/54Chapter 4 9


___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



Chapter 4 10


___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



Chapter 4 11


___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



Chapter 4 12


___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



Chapter 4 13


___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



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



Chapter 4 15


Contoh:

A B



Chapter 4 16


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?



Chapter 4 17


2 C4H

10(l ) + 13 O

2(g) 8 CO

2(g) + 10 H

2O(g)



1. Moles of C4H10

F.W. 58.124g

g

mol g H C moles

124.58

100.1

104



Chapter 4 18


2 C4H

10(l ) + 13 O

2(g) 8 CO

2(g) + 10 H

2O(g)



1. Moles of C4H10

F.W. 58.124g

mol g

mol g H C moles 0172.0

124.58

100.1

104



Chapter 4 19


2 C4H

10(l ) + 13 O

2(g) 8 CO

2(g) + 10 H

2O(g)



2. Ratio of C4H10:CO2

2 C4H10 : 8 CO2 or

104

2

2

8

H C

CO



Chapter 4 20


2 C4H

10(l ) + 13 O

2(g) 8 CO

2(g) + 10 H

2O(g)



3. Set-up ratio and proportion between known and unknownquantities



Chapter 4 21


2 C4H

10(l ) + 13 O

2(g) 8 CO

2(g) + 10 H

2O(g)




104104

2

0172.02

8

H C mol

x

H C

CO



Chapter 4 22


2 C4H

10(l ) + 13 O

2(g) 8 CO

2(g) + 10 H

2O(g)




2

104104

2

0688.0

0172.02

8

COmol x

H C mol

x

H C

CO



Chapter 4 23


2 C4H

10(l ) + 13 O

2(g) 8 CO

2(g) + 10 H

2O(g)



4. Convert the moles of unknown substance into the desiredunits



Chapter 4 24


2 C4H

10(l ) + 13 O

2(g) 8 CO

2(g) + 10 H

2O(g)




FW of CO2: 44.011g/mol

mol g mol CO g /011.440688.02



Chapter 4 25


2 C4

H10

(l ) + 13 O2

(g) 8 CO2

(g) + 10 H2

O(g)




FW of CO2: 44.011g/mol

g mol g mol CO g 03.3/011.440688.02



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



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



Chapter 4 28

Example

2 C8H18 + 25 O2 16 CO2 + 18 H2O



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



Chapter 4 29

Example

2 C8H18 + 25 O2 16 CO2 + 18 H2O



react.2. Divide each reagent by its own coefficient

L imiting Reactants

25

781.0

2

0875.0

2

188

O

H C



Chapter 4 30

Example

2 C8H18 + 25 O2 16 CO2 + 18 H2O



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



Chapter 4 31

Example

2 C8H18 + 25 O2 16 CO2 + 18 H2O



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



Chapter 4 32

Theoretical Yield

- The calculated amount of product based on thelimiting reactant (Theoretical yield).

L imiting Reactants



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



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



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



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



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.



Chapter 4 38

Combustion Analysis




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


2

2

011.44

12829.0

CO g COmol g



Chapter 4 40

Combustion Analysis




Mass of Carbon



22

2

1

1

011.44

12829.0

COmol C mol

CO g COmol g



Chapter 4 41

Combustion Analysis




Mass of Carbon



C mol C g

COmol C mol

CO g COmol g

1

011.12

1

1

011.44

12829.0

22

2



Chapter 4 42

Combustion Analysis




Mass of Carbon



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



Chapter 4 43

Combustion Analysis




Mass of Hydrogen

mass H2O moles H2O moles H grams H


O H g O H mol g 2

2

02.18

11159.0



Chapter 4 44

Combustion Analysis




Mass of Hydrogen



O H mol H mol

O H g O H mol g

22

2

1

2

02.18

11159.0



Chapter 4 45

Combustion Analysis




Mass of Hydrogen



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



Chapter 4 46

Combustion Analysis




Mass of Hydrogen



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



Chapter 4 47

Combustion Analysis




Mass of Oxygen

mass O = mass of sample – (mass C +mass H)


g g g Oxygenmass 01299.007721.01005.0

f



Chapter 4 48

Combustion Analysis




Mass of Oxygen

mass O = mass of sample – (mass C +mass H)


g

g g g Oxygenmass01030.0

01299.007721.01005.0

E i i l F l f A l



Chapter 4 49

Combustion Analysis




Now we can determine the empirical formula

Mass of elements:

C 0.07721g

H 0.01299g

O 0.01030g


E i i l F l f A l



Chapter 4 50

Combustion Analysis





Moles of elements:

C 0.07721g/12.011g/mol =

H 0.01299g/1.01g/mol =

O 0.01030g/16.00g/mol =


E i i l F l f A l



Chapter 4 51

Combustion Analysis





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


0006428.0012861.0006428.0 O H C

E i i l F l f A l



Chapter 4 52

Combustion Analysis





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


0006428.0

0006428.0

0006438.0

012861.0

0006428.0

006428.0 O H C

E i i l F l f A l



Chapter 4 53

Combustion Analysis





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


12010 O H C

P ti P bl



Practice Problems

4, 10, 14, 22, 26, 28, 38, 46, 50, 54

05 persamaan kimia dan stoikiometri.ppt

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