7766719 MATTER 12 Mole Concept[1]

8/9/2019 7766719 MATTER 12 Mole Concept[1]

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1.2 MOLE CONCEPT1.2 MOLE CONCEPT


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Learning OutcomeLearning Outcome

At the end of this topic, students should be

able :

(a) Define mole in terms of mass of

carbon-12 and Avogadro constant, NA.

(b) Interconvert between moles, mass, number of

particles, molar volume of gas at s.t.p. and

room temperature.


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(c)(c) Determine empirical and molecular

formulae from mass composition orfrom mass composition orcombustion data.combustion data.


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(d)(d) Define and perform calculation for eachfor each

of the following concentrationof the following concentrationmeasurements :measurements :

i) molarity (M)ii) molality (m)

iii) mole fraction, X

iv) percentage by mass, % w/w

v) percentage by volume, %V/V


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(e)(e) Determine the oxidation numberof anof an

element in a chemical formula.element in a chemical formula.

(f)(f) Write and balance ::

i) chemical equation byi) chemical equation by inspectioninspectionmethodmethod

ii) redox equation byii) redox equation by ionion--electronelectron

methodmethod


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(g)(g) Define limiting reactant andand percentage

yield.

(h)(h) Perform stoichiometric calculations

using mole concept including reactantusing mole concept including reactantand percentage yield.and percentage yield.


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1.2 Mole Concept1.2 Mole Concept

A mole is defined as the amount of

substance which contains equal number of

particles (atoms / molecules / ions) asthere are atoms in exactly 12.000g of

carbon-12.


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One mole of carbon-12 atom has a mass ofexactly 12.000 grams and contains 6.02 x 1023

atoms.

The value 6.02 x 1023 is known as AvogadroConstant.

NA = 6.02 x 1023 mol-1


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ExampleExample

1.0 mole of chlorine atom = 6.02 x 1023 chlorine atoms

= 35.5 g Cl

1.0 mole of chlorine

molecules

= 6.02 x 1023 chlorine

molecules= 71.1 g Cl2

= 6.022 x 1023 x 2 chlorine

atoms

1.0 mole of NH3 = 6.02x 1023 molecules= 6.02 x 1023 x 4 atoms

= 6.02 x 1023 N atom

= 6.02 x 1023 X 3 H atoms


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Molar MassMolar Mass

The mass of one mole of an element or onemole of compound is referred as molar mass.

Unit : g mol-1

Example:

- molar mass of Mg = 24 g mol-1

- molar mass of CH4 = (12 + 4) gmol-1

= 16 g mol-1


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r f M lr f M l


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E l 1E l 1


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E l 1 ( t)E l 1 ( t)


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E l 2E l 2


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E l 3E l 3


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1.2.1 Mole Concept of Gases1.2.1 Mole Concept of Gases

Molar volume of any gas at STP = 22.4 dm3 mol-1

s.t.p. = Standard Temperature and Pressure

Where,

T = 273.15 K

P = 1 atm


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1 mole of gas has a volume of 22.4 dm3 at s.t.p

At s.t.p,volume of gas (dm3) = number of mole X 22.4dm3 mol-1

1 mole of gas has a volume of 24.0 dm3 at roomtemperature

At room temperature,

volume of gas (dm3) = number of mole X 24.0

dm

3

mol

-1


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E l 1E l 1


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Cont from example 1Cont from example 1

mol0.1

4.22

2.24dm

4.22

)(dmgasofmoleofNumber

2,Solution

13

3

13

3

!

!

!

moldm

moldm

volume


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ExerciseExercise

A sample of CO2 has a volume of 56 cm3 at STP.

Calculate:

a. The number of moles of gas molecules

0.0025 mol

b. The number of molecular

1.506 x 1021 molecules

c. The number of oxygen atoms in the sample

3.011x1021atoms

Note:

1 dm3 = 1000 cm3

1 dm3 = 1 L


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Empirical And Molecular FormulaeEmpirical And Molecular Formulae

-- Empirical formulaEmpirical formula is a chemical formulais a chemical formula

that shows the simplest ratio of allthat shows the simplest ratio of all

elements in a molecule.elements in a molecule.-- Molecular formulaMolecular formula is a formula that showis a formula that show

the actual number of atoms of eachthe actual number of atoms of each

element in a molecule.element in a molecule.


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-- The relationship between empirical formula andThe relationship between empirical formula andmolecular formula is :molecular formula is :

Molecular formula = n ( empirical formula )Molecular formula = n ( empirical formula )

Where ;Where ;

massormulaemprical

massmolecularrelative

n!


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ExampleExample

A sample of hydrocarbon contains 85.7%A sample of hydrocarbon contains 85.7%carbon and 14.3% hydrogen by mass. Itscarbon and 14.3% hydrogen by mass. Its

molar mass is 56. Determine the empiricalmolar mass is 56. Determine the empirical

formula and molecular formula of theformula and molecular formula of the

compound.compound.


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Solution :Solution :

Empirical formula =Empirical formula = CHCH22

CC HH

massmass 85.785.7 14.314.3

Number of molNumber of mol 85.785.7

1212

7.14177.1417

14.314.3

11

14.314.3

Simplest ratioSimplest ratio 11 22


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massformulaempricalmassmolecularrelativen !

4

14

56

!

n = 56

14

= 4

molecular formula = C4H8


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1.2.2 Concentration of Solution1.2.2 Concentration of Solution

Solution

When an amount ofsolute dissolved completely in a solvent andit will form a homogeneous mixture.


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ExerciseExercise

A combustion of 0.202 g of an organic sampleA combustion of 0.202 g of an organic sample

that contains carbon, hydrogen and oxygenthat contains carbon, hydrogen and oxygen

produce 0.361g carbon dioxide and 0.147 gproduce 0.361g carbon dioxide and 0.147 g

water. If the relative molecular mass of thewater. If the relative molecular mass of thesample is 148, what is the molecular formula.sample is 148, what is the molecular formula.

Ans :Ans : CC66HH1212OO44


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Units of concentration of a solution:

A. Molarity

B. MolalityC. Mole Fraction

D. Percentage by Mass

E. Percentage byVolume


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A.A. M l rit (M)M l rit (M)

The number f m les f s lute per cubic decimetre(dm3) r litre (L) f s luti n.

Note:

1 dm3 = 1000 cm3

1 L = 1000 mL


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E lE l


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ContCont

Lmol0.01

L5.0

mol0.005

solutionovolume

sucroseomolesucrosesolutionomolarity

mol0.005

342

g1.71

massmolar

masssucroseomoleo

1-

1

!

!

!

!

!

!

molg

Number


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E r iE r i


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.. M l lit (M l lit (mm))

Molality is the number of moles of solute dissolvedin 1 kg of solvent

Note:

Mass of solution = mass of solute +

mass of solvent

Volume of solution volume of solvent


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Example 1Example 1

]molg98.08OHmass[molar

water?og198in

acidsulphuricog24.4containingsolution

acidsulphuricomolalitythe

1-

42!

Calculate


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m1.26

kg198.0mol2488.0

(kg)solvento

(mol)soluteoOHoMolality

mol0.2488

08.98

4.24

mass

:

42

1

42

!

!

!

!

!

!

mass

moles

molg

g

molar

massn

Solution

SOH


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Example 2Example 2

]molg18.02OHmass[molar

water?omol40.0

inCuClomol0.30dissolvingbyprepared

solutionaoionconcentratmolaltheishat

1-

2

2

!

W


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m0.416

kg7208.0

mol3.0

(kg)solventof

(mol)soluteofofolality

kg7208org720.8

gmol18.02xmol0.40ofmass

mass

:

2

1

2

2

!

!

!

!

!

!

mass

moles

molar

massn

Solution


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E r iE r i


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. M l r ti ( ). M l r ti ( )

M l fr ti is th r ti f th umb r f m l s

f mp t t th t t l umb r f m l s

f ll mp t pr s t.


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It is always smaller than 1

The total mol fraction in a mixture

(solution) is equal to one.

XA + XB + XC = 1


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Example 1Example 1

]molg18.02mass[molar

aterofmol40.0

inu lofmol0.30dissolvingbyprepared

solutionainu loffractionmoletheishat

1-

2

2

2

!

W


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0.007

400.3

3.0

n

:

total

2

2

!

!

!CuCl

CuCl

nX

Solution


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0.993

400.3

40

ntotal

2

2

!

!

!OH

OH

nX


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0.093

007.01X

1XX

O2H

O2H2CuCl

!

!

!


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Example 2Example 2

79.9]Br1.01,12.01,[Ar

componenteachoffractionmoletheisWhat

Br.nebromobenzeofg55and

toluene,ofg55mixingbypreparedissolution

5687

!!!

A


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. r t M ( / ). r t M ( / )

r t m i d fi d th p r t of th

m of olut p r m of olutio .


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E l 1E l 1

solution?in themass

bypercentageisWhatwater.ofg54.3indissolved

isKClchloride,potassiumofg0.892ofsampleA


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Example 2Example 2

solution.massbypercent

16.2aofnpreparatioin theureaofg5.00toaddedbe

mustthatgrams)(inwaterofamounttheCalculate


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E r iE r i


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E. Percentage By Volume (%E. Percentage By Volume (%V / VV / V))

Percentage by volume is defined as the percentage of

volume of solute in milliliter per volume of solution in

milliliter.


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E lE l

solution?inthislcoholofolume%theisWhat

alcohol.ofmLcontainserfumeofmLA


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1.2.3 Balancing Chemical Equation1.2.3 Balancing Chemical Equation

A chemical equation shows a chemical

reaction using symbols for the reactants

and products.

The formulae of the reactants are written

on the left side of the equation while the

products are on the right.


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The total number of atoms of each

element is the same on both sides in abalanced equation.

The numberx, y, z and w, showing the

relative number of molecules reacting,are called the stoichiometric coefficients.

The methods to balance an equation:

Inspection Method


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Inspection MethodInspection Method

a. Write down the unbalanced equation. Write thecorrect formulae for the reactants and products.

b. Balance the metallic element, followed by non-

metallic atoms.

c. Balance the hydrogen and oxygen atoms.

d. Check to ensure that the total number of atoms ofeach element is the same on both sides of equation.


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ExampleExample

Balance the chemical equation by applying the

inspection method.

NH3 + CuO Cu + N2 + H2O


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ExerciseExercise

1. Balance the chemical equation below by applying

inspection method.

a. Fe(OH)3 + H2SO4 Fe2(

SO4)3 + H2O

b.C6H6 + O2 CO2 + H2O

c. N2H4 + H2O2 HNO3 + H2O

d.ClO2 + H2O HClO3 + HCl


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1.2.4 Redox Reaction1.2.4 Redox Reaction

Redox reaction is a reaction that involves

both reduction and oxidation.


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Oxidation

The substance loses one or more

elactrons.

Increase in oxidation number

Act as an reducing agent (reductant)


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Reduction

The substance gains one or more

elactrons.

decrease in oxidation number

Act as an oxidising agent (oxidant)


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Oxidation numbers of any atoms can be

determined by applying the following rules:

1. In a free element , as an atom or a molecule the

oxidation number is zero.

Example:Na = 0 Cl2 = 0

Br2 = 0 O2 = 0

Mg = 0


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2. Formonoatomic ion, the oxidation

number is equal to the charge on the

ion.

Example:

Na+ = +1 Mg2+= +2

Al3+ = +3 S2- = -2


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3. Fluorine and other halogens always have

oxidation number of-1 in its compound.Only have a positive number when

combine with oxygen.

Example:

Oxidation number of F in NaF = -1

Oxidation number of Cl in HCl = -1

Oxidation number of Cl in Cl2

O7

= +7


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4. Hydrogen has an oxidation number of+1 in its

compound except in metal hydrides which hydrogen

has an oxidation number of-1

Example:

Oxidation number of H in HCl = +1

Oxidation number of H in NaH = -1

Oxidation number of H in MgH2 = -1


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5. Oxygen has an oxidation number of-2 in

most of its compound.

Example:

Oxidation number of O in MgO = -2

Oxidation number of O in H2O = -2


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However there are two exceptional cases:

- in peroxides, its oxidation number is -1

Example:

Oxidation number of O in H2O2 = -1

- When combine with fluorine, posses a

positive oxidation number

Example:

Oxidation number of O in OF2 = +2


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6. In neutral molecule, the sum of the

oxidationnumberof all atoms thatmade up the molecule is equal to zero.

Example:

Oxidation number of H2O = 0Oxidation number of HCl = 0

Oxidation number of KMnO4 = 0


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7. Forpolyatomic ions, the total oxidation

number of all atoms that made up thepolyatomic ion must be equal to the nett

charge of the ion.

Example:

Oxidation number of KMnO4- = -1

Oxidation number of Cr2O72- = -2

Oxidation number of NO3- = -1


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Example :Example :

Assign theAssign the oxidation number of Crin Crin Cr22OO7722--

..

Solution :Solution :

CrCr22OO77 == --222 Cr + 7 (2 Cr + 7 (--2) =2) = --22

2 Cr = + 122 Cr = + 12

Cr = + 6Cr = + 6


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ExerciseExercise

1. Assign the oxidation number ofMn in the followingchemical compounds.

i. MnO2 ii. MnO4-

2. Assign the oxidation number ofCl in the followingchemical compounds.

i. KClO3 ii. Cl2O72-

3. Assign the oxidation number of following:

i. Cr in K2Cr2O7ii. U in UO2

2+

iii.C in C2O42-


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1.2.4.1 Balancing Redox Reaction1.2.4.1 Balancing Redox Reaction

Redox reaction may occur in acidic and basic

solutions.

Follow the steps systematically so thatequations become easier to balance.

B l i R d R ti IB l i R d R ti I


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Balancing Redox Reaction InBalancing Redox Reaction In

Acidic SolutionAcidic Solution

Fe2+ + MnO4- Fe3+ + Mn2+

1. Divide the equation into two half reactions, oneinvolving oxidation and the other reductioni. Fe2+ Fe3+

ii.MnO4- Mn2+


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2. Balance each half-reaction

a. first, balance the element other

than oxygen and hydrogen

i. Fe2+ Fe3+

ii. MnO4- Mn2+


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b. second, balance the oxygen atom by adding H2O

and hydrogen by adding H+

i. Fe2+ Fe3+

ii. MnO4-

+ 8H+

Mn2+

+ 4H2O

c. then, balance the charge by adding electrons to the

side with the greater overall positive charge.

i. Fe2+ Fe3+ + 1eii. MnO4

- + 8H+ + 5e Mn2+ + 4H2O


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3. Multiply each half-reaction by an interger, so that number of

electron lost in one half-reaction equals the number gained in the

other.i. 5 x (Fe2+ Fe3+ + 1e)

5Fe2+ 5Fe3+ + 5e

ii. MnO4- + 8H+ + 5e Mn2+ + 4H2O

4. Add the two half-reactions and simplify where possible by

canceling species appearing on both sides of the equation.

i. 5Fe2+ 5Fe3+ + 5e

ii. MnO4- + 8H+ + 5e Mn2+ + 4H2O

____________________________________________5Fe2+ + MnO4

- + 8H+ 5Fe3+ + Mn2+ + 4H2O


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5. Check the equation to make sure that thereare the same number of atoms of each kind

and the same total charge on both sides.

5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O

Total charge reactant

= 5(+2) + (-1) + 8(+1)

= + 10 - 1 + 8

= +17

Total charge product

= 5(+3) + (+2) + 4(0)

= + 15 + (+2)

= +17


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Example: In Acidic SolutionExample: In Acidic Solution

C2O42- + MnO4

- + H+ CO2 + Mn2+ + H2O

Solution:

1. i. Oxidation : C2O42- CO2

ii. Reduction : MnO4- Mn2+

2. i. C2O42- 2CO2

ii. MnO4- + 8H+ Mn2+ + 4H2O

3. i. C2O42- 2CO2 + 2e

ii. MnO4- + 8H+ + 5e Mn2+ + 4H2O


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4. i. 5 x (C2O42- 2CO2 + 2e)

5C2O42- 10CO2 + 10e

ii. 2 x (MnO4- + 8H+ + 5e Mn2+ + 4H2O)

2MnO4- + 16H+ + 10e 2Mn2+ + 8H2O

5. i. 5C2O42- 10CO2 + 10e

ii. 2MnO4- + 16H+ + 10e 2Mn2+ + 8H2O

_________________________________________________

5C2O42- + 2MnO4

- + 16H+ 10CO2 + 2Mn2+ + 8H2O

Balancing Redox Reaction In BasicBalancing Redox Reaction In Basic


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Balancing Redox Reaction In BasicBalancing Redox Reaction In Basic

SolutionSolution1. Firstly balance the equation as in acidic solution .

2. Then, add OH- to both sides of the equation so that itcan be combined with H+ to form H2O.

3. The number of hydroxide ions (OH-) added is equal tothe number of hydrogen ions (H+) in the equation.


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Example: In Basic SolutionExample: In Basic Solution

Cr(OH)3 + IO3- + OH- CrO3

2- + I- + H2O

Solution:

1. i. Oxidation : Cr(OH)3 CrO32-

ii. Reduction :IO3

-

I-

2. i. Cr(OH)3 CrO32- + 3H+

ii. IO3- + 6H+ I- + 3H2O

3. i. Cr(OH)3 CrO32- + 3H+ + 1e

ii. IO3- + 6H+ + 6e I- + 3H2O


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4. i. 6 x (Cr(OH)3 CrO32- + 3H+ + 1e)

6Cr(OH)3 6CrO32- + 18H+ + 6e

ii. IO3- + 6H+ + 6e I- + 3H2O

5. i. 6Cr(OH)3 6CrO32- + 18H+ + 6e

ii. IO3- + 6H+ + 6e I- + 3H2O

________________________________________________6Cr(OH)3 + IO3

- 6CrO32- + I- + 12H+ + 3H2O

6. 6Cr(OH)3 + IO3- + 12OH- 6CrO3

2- + I- + 12H+ + 3H2O + 12OH-

7. 6Cr(OH)3 + IO3-

+ 12OH-

6CrO32-

+ I-

+ 15H2O


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ExerciseExercise

Balance the following redox equations:

a. In Acidic Solution

i. Cu + NO3 + H+ Cu2+ + NO2 + H2O

ii. MnO4- + H2SO3 Mn

2+ + SO42- + H2O + H

+

iii. Zn + SO42- + H+ Zn2+ + SO2 + H2O

b. In Basic Solution

i. ClO- + S2O32- Cl- + SO4

2-

ii. Cl2 ClO3- + Cl-

iii. NO2 NO3 + NO


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1.2.5 Stoichiometry1.2.5 Stoichiometry

Stoichiometry is the quantitative study of

reactants and products in a chemicalreaction.


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Example:

CaCO3 (s) + 2HCl (aq) CaCl2 (aq) + CO2 (g) + H2O (l)

1 mole of CaCO3 reacts with 2 moles of HCl to yield 1mole of CaCl2, 1 mole of CO2 and 1 mole of H2O.

Stoichiometry can be used for calculating thespecies we are interested in during a reaction.


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Example 1Example 1

How many moles of hydrochloric acid, HCl do weneed to react

with 0.5 moles of zinc?

HClmol1

HClofmol1

2x0.5hreact witZnofmole0.5

HClofmol2withreactsZnofmole1

equation,theFrom

(g)H(s)ZnCl(l)2HCl(s)Zn:olution22

@

p


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Example 2Example 2

How many moles of H2O will be formed when 0.25 moles of

C2H5OH burns in oxygen?

OHmol7.0

1

3x0.2X

OHofmol sXgiv sOHHCofmol0.2

OHofmol s3giv sOHHCofmol1

quatio ,thFrom

O3H2CO3OOHHC

:olutio

2

22

22

2222

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p


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8/30/20108/30/2010 MATTERMATTER 9191

A 16.50 mL 0.1327 M KMnO4 solution is neededto oxidise 20.00mL of a FeSO4 solution in anacidic medium. What is the concentration of theFeSO4 solution? The net ionic equation is:

5Fe 2+ + MnO4- +8H+ Mn2+ +5Fe 3+ +4H2O

Answer : 0.5474 M

Exercise 1Exercise 1


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8/30/20108/30/2010 MATTERMATTER 9292

How many mililitres of 0.112 M HCl will react

exactly with the sodium carbonate in 21.2 mL of

0.150 M Na2CO3 according to the following

equation?

2HCl(aq)+Na2CO3(aq) 2NaCl(aq)+CO2(g)+H2O(l)

Answer : 56.8 mL

Exercise 2Exercise 2


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8/30/20108/30/2010 MATTERMATTER 9393

1.2.5.1 Limiting Reactant1.2.5.1 Limiting Reactant

A limiting reactant is the reactant that iscompletely consumed in a reaction and limitsthe amount of products formed.

An excess reactant is the reactant that is notcompletely consumed in a reaction andremains at the end of the reaction.

Example 1Example 1


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8/30/20108/30/2010 MATTERMATTER 9494

Example 1Example 1

S + 3F2 SF6

If 4 mol of S reacts with 10 mol of F2 , which of the tworeactants is the limiting reagent?

reactant.limitingtheisFlimit,inisFquestion.in the

availablemol)(10nhewith tmol)(12neededntheCompare

Fmol121

3x4X

FofmolesXwithreactsofmol4

Fofmoles3withreactsofmol1

equation,theFrom

:olution

22

FF

2

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8/30/20108/30/2010 MATTERMATTER 9595

Example 2Example 2

C is prepared by reacting A and B :

A + 5B C

In one process, 2 mol of A react with 9 mol of B.

a. Which is the limiting reactant?

b. Calculate the number of mole(s) of C?c. How much of the excess reactant (in mol) is left at the end of

the reaction?


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8/30/20108/30/2010 MATTERMATTER 9696

re ct nt.limitint eisBlimit,inisBestion.t ein

il blemol)(nt ewitmol)(needednt eomp re

mol

5

Bofmoleswitre ctsAofmol

Bofmoles5witre ctsAofmol

e tion,t eFrom

:ASol tion

BB

@

!

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8/30/20108/30/2010 MATTERMATTER 9797

mol8.5

ofmoleswitprod ceBofmol

ofmoleswitprod ceBofmol5

e tion,t eFrom

re ct nt.limitint e

B,ofmolest eofreliesformedprod ctofmountT e

:BSolution

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8/30/20108/30/2010 MATTERMATTER 9898

Amol..8-re ct nte cessmountT e

Amol8.

5

ofmoleswitproduceBofmol

AofmoleswitproduceBofmol5

e uation,t eFrom

.reactante cesst eisA

:Solution

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8/30/20108/30/2010 MATTERMATTER 9999

Percentage yieldPercentage yield

The percentage yield is the ratio of theThe percentage yield is the ratio of the

actual yield (obtained from experiment) toactual yield (obtained from experiment) to

the theoretical yield (obtained fromthe theoretical yield (obtained from

stoichiometry calculation) multiply bystoichiometry calculation) multiply by100%100%


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8/30/20108/30/2010 MATTERMATTER 100100

Percentage yield = actual yield x 100%

theoretical yield


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ExerciseExercise

In a certain experiment, 14.6g of SbF3 was allowed to react

with CCl4 in excess. After the reaction was finished, 8.62g of

CCl2F2 was obtained.

3 CCl4 + 2 SbF3 3 CCl2F2 + 2 SbCl3

[ Ar Sb = 122, F = 19, C= 12, Cl = 35.5 ]

a) What was the theoretical yield of CCl2F2 in grams ?

b) What was the percentage yield of CCl2F2 ?

Ans : a) 11.6 g b) 74.31 %

Documents

7766719 MATTER 12 Mole Concept[1]