7766719 matter 12 mole concept[1]
TRANSCRIPT
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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
!
!
@
p
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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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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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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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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
2
2
22
@
!
!
@
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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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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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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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Amol..8-re ct nte cessmountT e
Amol8.
5
ofmoleswitproduceBofmol
AofmoleswitproduceBofmol5
e uation,t eFrom
.reactante cesst eisA
:Solution
!!
!
!
@
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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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Percentage yield = actual yield x 100%
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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 %