1. 1.2 MOLE CONCEPT
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2. Learning 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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3. (c) Determine empirical and molecular
formulae from mass composition or
combustion data.
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4. (d) Define and perform calculation for each
of the following concentration
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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5. (e) Determine the oxidation number of an
element in a chemical formula.
(f) Write and balance :
i) chemical equation by inspection
method
ii) redox equation by ion-electron
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6. (g) Define limiting reactant and percentage yield.
(h) Perform stoichiometric calculations
using mole concept including reactant
and percentage yield.
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7. 1.2 Mole Concept
A mole is defined as the amount of substance which contains equal number of particles (atoms / molecules / ions) as there are atoms in exactly g of carbon-12.
Pair = 2
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8. One mole of carbon-12 atom has a mass of exactly 12
One mole of carbon-12 atom has a mass of exactly grams and contains 6.02 x 1023 atoms.
The value 6.02 x 1023 is known as Avogadro Constant.
NA = x 1023 mol-1
Dozen = 12
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9. Example 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
= 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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10. Molar Mass
The mass of one mole of an element or one mole 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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11. Number of Mole
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12. Example 1
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14. Example 1 (cont…)
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15. Example 2
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16. Example 3
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17. 1.2.1 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 = K
P = 1 atm
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18. 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 22.4 dm3 mol-1
1 mole of gas has a volume of 24.0 dm3 at room temperature
At room temperature,
volume of gas (dm3) = number of mole X 24.0 dm3 mol-1
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19. Example 1
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20. Cont… from example 1
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21. Exercise A sample of CO2 has a volume of 56 cm3 at STP. Calculate:
The number of moles of gas molecules
mol
The number of molecular
1.506 x 1021 molecules
The number of oxygen atoms in the sample
3.011x1021atoms
Note:
1 dm3 = 1000 cm3
1 dm3 = 1 L
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22. Empirical And Molecular Formulae
Empirical formula is a chemical formula that shows the simplest ratio of all elements in a molecule.
Molecular formula is a formula that show the actual number of atoms of each element in a molecule.
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23. The relationship between empirical formula and molecular formula is :
Molecular formula = n ( empirical formula )
Where ;
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24. Example
A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56. Determine the empirical formula and molecular formula of the compound.
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25. Solution : Empirical formula = CH2 C H mass 85.7 14.3 Number of mol 12
7.1417 1
Simplest ratio 2
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26. n = 56 14
= 4
molecular formula = C4H8
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27. 1.2.2 Concentration of Solution
When an amount of solute dissolved completely in a solvent and it will form a homogeneous mixture.
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28. Exercise
A combustion of g of an organic sample that contains carbon, hydrogen and oxygen produce 0.361g carbon dioxide and g water. If the relative molecular mass of the sample is 148, what is the molecular formula.
Ans : C6H12O4
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29. Units of concentration of a solution: A. Molarity B. Molality
C. Mole Fraction
D. Percentage by Mass
E. Percentage by Volume
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30. A. Molarity (M)
The number of moles of solute per cubic decimetre (dm3) or litre (L) of solution.
Note:
1 dm3 = 1000 cm3
1 L = 1000 mL
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31. Example
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32. Cont…
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33. Exercises
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34. B. Molality (m)
Molality is the number of moles of solute dissolved in 1 kg of solvent
Note:
Mass of solution = mass of solute +
mass of solvent
Volume of solution ≠ volume of solvent
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35. Example 1
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37. Example 2
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39. Exercises
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41. C. Mole Fraction (X)
Mole fraction is the ratio of the number of moles of one component to the total number of moles of all component present.
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42. 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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43. Example 1
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47. Example 2
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49. D. Percentage by Mass (%w/w)
Percentage by mass is defined as the percentage of the mass of solute per mass of solution.
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50. Example 1
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51. Example 2
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53. Exercises
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54. E. Percentage By Volume (%V / 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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55. Example
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56. 1.2.3 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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57. Example:
x A + y B z C + w D
Reactants
Products
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58. The methods to balance an equation: Inspection Method
The total number of atoms of each element is the same on both sides in a balanced equation.
The number x, 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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59. Inspection Method
Write down the unbalanced equation. Write the correct formulae for the reactants and products.
Balance the metallic element, followed by non-metallic atoms.
Balance the hydrogen and oxygen atoms.
Check to ensure that the total number of atoms of each element is the same on both sides of equation.
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60. Example NH3 + CuO → Cu + N2 + H2O
Balance the chemical equation by applying the
inspection method.
NH3 + CuO → Cu + N2 + H2O
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61. Exercise
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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62. 1.2.4 Redox Reaction
Redox reaction is a reaction that involves both reduction and oxidation.
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63. The substance loses one or more elactrons.
Oxidation
The substance loses one or more elactrons.
Increase in oxidation number
Act as an reducing agent (reductant)
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64. The substance gains one or more elactrons.
Reduction
The substance gains one or more elactrons.
decrease in oxidation number
Act as an oxidising agent (oxidant)
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65. Oxidation numbers of any atoms can be determined by applying the following rules:
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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66. For monoatomic 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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67. Oxidation number of F in NaF = -1 Oxidation number of Cl in HCl = -1
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 Cl2O7 = +7
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68. Oxidation number of H in HCl = +1 Oxidation number of H in NaH = -1
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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69. 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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70. 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
Oxidation number of O in OF2 = +2
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71. Oxidation number of H2O = 0 Oxidation number of HCl = 0
In neutral molecule, the sum of the oxidation number of all atoms that made up the molecule is equal to zero.
Example:
Oxidation number of H2O = 0
Oxidation number of HCl = 0
Oxidation number of KMnO4 = 0
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72. Oxidation number of KMnO4- = -1
For polyatomic ions, the total oxidation number of all atoms that made up the polyatomic 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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73. Assign the oxidation number of Cr in Cr2O72-. Solution :
Example :
Assign the oxidation number of Cr in Cr2O72-.
Solution :
Cr2O7 = -2
2 Cr + 7 (-2) = -2
2 Cr = + 12
Cr = + 6
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74. Exercise
Assign the oxidation number of Mn in the following chemical compounds.
i. MnO2 ii. MnO4-
Assign the oxidation number of Cl in the following chemical compounds.
i. KClO3 ii. Cl2O72-
Assign the oxidation number of following:
i. Cr in K2Cr2O7
ii. U in UO22+
iii. C in C2O42-
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75. 1.2.4.1 Balancing Redox Reaction
Redox reaction may occur in acidic and basic solutions.
Follow the steps systematically so that equations become easier to balance.
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76. Balancing Redox Reaction In Acidic Solution
Fe2+ + MnO4- → Fe3+ + Mn2+
Divide the equation into two half reactions, one involving oxidation and the other reduction
i. Fe2+ → Fe3+
ii. MnO4- → Mn2+
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77. Balance each half-reaction
a. first, balance the element other than oxygen and hydrogen
i. Fe2+ → Fe3+
ii. MnO4- → Mn2+
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78. and hydrogen by adding H+ i. Fe2+ → Fe3+ ii. MnO4- + 8H+ → Mn2+ + 4H2O
b. second, balance the oxygen atom by adding H2O
and hydrogen by adding H+
i. Fe2+ → Fe3+
ii. MnO H+ → Mn H2O
c. then, balance the charge by adding electrons to the
side with the greater overall positive charge.
i. Fe2+ → Fe e
ii. MnO H e → Mn H2O
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79. ____________________________________________
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+ → Fe e)
5Fe2+ → 5Fe e
ii. MnO H e → Mn H2O
4. Add the two half-reactions and simplify where possible by canceling species appearing on both sides of the equation.
i. 5Fe2+ → 5Fe e
____________________________________________
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn H2O
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80. 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.
5Fe2+ + MnO4- + 8H → Fe3+ + Mn H2O
Total charge reactant
= 5(+2) + (-1) + 8(+1) =
= +17
Total charge product
= 5(+3) + (+2) + 4(0)
= (+2)
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81. Example: In Acidic Solution
C2O42- + MnO4- + H+ → CO2 + Mn2+ + H2O
Solution:
i. Oxidation : C2O42- → CO2
ii. Reduction : MnO4- → Mn2+
2. i. C2O42- → 2CO2
ii. MnO4- + 8H+ → Mn H2O
3. i. C2O42- → 2CO2 + 2e
ii. MnO4- + 8H+ + 5e→ Mn H2O
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82. ii. 2 x (MnO4- + 8H+ + 5e→ Mn2+ + 4H2O)
4. i. 5 x (C2O42- → 2CO2 + 2e)
→ 5C2O42- → 10CO2 + 10e
ii. 2 x (MnO4- + 8H+ + 5e→ Mn H2O)
→ 2MnO H+ + 10e→ 2Mn H2O
5. i. 5C2O42- → 10CO2 + 10e
ii. 2MnO H+ + 10e→ 2Mn H2O
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5C2O MnO H+ → 10CO2 + 2Mn H2O
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83. Balancing Redox Reaction In Basic Solution
Firstly balance the equation as in acidic solution .
Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O.
The number of hydroxide ions (OH-) added is equal to the number of hydrogen ions (H+) in the equation.
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84. Example: In Basic Solution
Cr(OH)3 + IO3- + OH- → CrO32- + I- + H2O
Solution:
i. Oxidation : Cr(OH)3 → CrO32-
ii. Reduction : IO3- → I-
2. i. Cr(OH)3 → CrO H+
ii. IO H+ → I H2O
3. i. Cr(OH)3 → CrO H+ + 1e
ii. IO H e → I H2O
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85. i. 6 x (Cr(OH)3 → CrO32- + 3H+ + 1e) → 6Cr(OH)3 → 6CrO32- + 18H+ + 6e
ii. IO H e → I H2O
5. i. 6Cr(OH)3 → 6CrO H+ + 6e
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6Cr(OH)3 + IO3- → 6CrO32- + I H H2O
6. 6Cr(OH)3 + IO OH- → 6CrO32- + I H+ + 3H2O + 12OH-
7. 6Cr(OH)3 + IO OH- → 6CrO32- + I H2O
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86. Exercise Balance the following redox equations: a. In Acidic Solution
i. Cu + NO3 + H+→ Cu2+ + NO2 + H2O
ii. MnO H2SO3 → Mn SO H2O + H+
iii. Zn + SO H+ → Zn SO2 + H2O
b. In Basic Solution
i. ClO- + S2O32- → Cl- + SO42-
ii. Cl2 → ClO Cl-
iii. NO2 → NO3 + NO
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87. 1.2.5 Stoichiometry Stoichiometry is the quantitative study of
reactants and products in a chemical reaction.
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88. CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)
Example:
CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)
1 mole of CaCO3 reacts with 2 moles of HCl to yield 1 mole of CaCl2, 1 mole of CO2 and 1 mole of H2O.
Stoichiometry can be used for calculating the species we are interested in during a reaction.
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89. Example 1
How many moles of hydrochloric acid, HCl do we need to react with 0.5 moles of zinc?
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90. Example 2
How many moles of H2O will be formed when 0.25 moles of C2H5OH burns in oxygen?
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91. Exercise 1
A mL M KMnO4 solution is needed to oxidise 20.00mL of a FeSO4 solution in an acidic medium. What is the concentration of the FeSO4 solution? The net ionic equation is:
5Fe 2+ + MnO4- +8H Mn 2+ +5Fe 3+ +4H2O
Answer : M
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92. Exercise 2
How many mililitres of M HCl will react exactly with the sodium carbonate in 21.2 mL of M Na2CO3 according to the following equation?
2HCl(aq)+Na2CO3(aq) NaCl(aq)+CO2(g)+H2O(l)
Answer : 56.8 mL
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93. Limiting Reactant
A limiting reactant is the reactant that is completely consumed in a reaction and limits the amount of products formed.
An excess reactant is the reactant that is not completely consumed in a reaction and remains at the end of the reaction.
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94. Example 1
S + 3F2 → SF6
If 4 mol of S reacts with 10 mol of F2 , which of the two reactants is the limiting reagent?
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95. Example 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.
Which is the limiting reactant?
Calculate the number of mole(s) of C?
How much of the excess reactant (in mol) is left at the end of the reaction?
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99. Percentage yield
The percentage yield is the ratio of the actual yield (obtained from experiment) to the theoretical yield (obtained from stoichiometry calculation) multiply by 100%
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100. Percentage yield = actual yield x 100% theoretical yield
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101. Exercise
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 CCl SbF CCl2F SbCl3
[ Ar Sb = 122, F = 19, C= 12, Cl = 35.5 ]
What was the theoretical yield of CCl2F2 in grams ?
b) What was the percentage yield of CCl2F2 ?
Ans : a) 11.6 g b) %
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