Download CHAPTER 3 STOICHIOMETRY:

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CHAPTER 3
STOICHIOMETRY:
CHEMICAL EQUATIONS
Stoichiometry: The area of study that examines the
quantities of substances consumed and produced in
chemical reactions.
Remember: Atoms are neither created nor destroyed
during any chemical reaction or physical process.
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CHEMICAL EQUATIONS
Chemical equation represents the chemical recation .
The arrow is called the yield sign.
2
BALANCING EQUATIONS
A chemical equation must have an equal
number of atoms of each element on each side
of the arrow.
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RULES/HINTS FOR BALANCING EQUATIONS
1) Use correct formulas.
2) The coefficients must be the lowest whole-numbers
possible.
3) Do not change or add subscripts (number of atoms
of each element in a molecule).
4) Balance H and O last.
5) If a polyatomic ion stays together on both sides of
the
arrow, then keep it together when balancing.
2HCl + Na2CO3 → 2NaCl + 2H2O + CO2
C2H5OH + 3O2 → 2CO2 + 3H2O
SAMPLE EXERCISE 3.1
The following diagram represents a chemical reaction in which the
red spheres are oxygen atoms and the blue spheres are nitrogen
atoms.
a) Write the chemical formulas for the reactants and products.
O2+ NO → NO2
(unbalanced)
(b) Write a balanced equation for the reaction.
O2+ 2NO → 2NO2 (balanced)
(c) Is the diagram consistent with the law of conservation of mass?
Yes; because there are equal numbers of both N and O atoms in
the two boxes
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PRACTICE EXERCISE
In the following diagram, the white spheres represent hydrogen
atoms, and the blue spheres represent nitrogen atoms. To be
consistent with the law of conservation of mass, how many NH3
molecules should be shown in the right box?
Answer: Six NH3 molecules.
STATES OF MATTER
We use the symbols (g) for gas, (l) for liquid, (s) for solid,
and (aq) for aqueous to identify the states of matter of the
reactants and products.
Catalysts, photons (hν), or heat (∆) may stand above the
reaction arrow (→).
CaCO3(s) + 2HCl(aq) → Ca2+(aq) + 2Cl–(aq) + CO2(g) + H2O(l)
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SAMPLE EXERCISE 3.2
Balance this equation:
Na(s) + H2O(l)
NaOH(aq) + H2(g)
Answer:
2Na(s) + 2 H2O(l) →2 NaOH(aq) + H2(g)
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PRACTICE EXERCISE:
Balance the following equations:
a. Fe(s) + O2(g)
Fe2O3(s)
b. C2H4(g) + O2(g)
CO2(g) + H2O(g)
c. Al(s) + HCl(aq)
AlCl3(aq) + H2(g)
SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITY
1) Combination Reactions:
In synthesis/combination reactions two
substances react to form one product.
Generic Reaction:
A+B
AB
Real Reaction:
2Mg + O2
2MgO
Examples of Combination Reactions:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
C3H6 (g) + Br2 (l) → C3H6Br2 (l)
٥
or
more
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Examples of Combination Reactions:
Decomposition Reactions:
In a decomposition reaction one substance undergoes a
reaction to produce two or more products.
Generic Reaction:
AB
A+B
Real Reaction:
CaCO3
CaO + CO2
Examples:
CaCO3 (s) → CaO (s) + CO2 (g)
2 KClO3 (s) → 2 KCl (s) + O2 (g)
2 NaN3 (s) → 2 Na (s) + 3 N2 (g)
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SAMPLE EXERCISE 3.3
Write the balanced equations for the following reactions:
a. The synthesis reaction that occurs when lithium metal
and fluorine gas react.
2 Li(s)
+
F2(g)
→
2 LiF(s)
b. The decomposition reaction that occurs when solid
barium carbonate is heated.
BaCO3(s)
٧
→
BaO(s)
+
CO2(g)
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PRACTICE EXERCISE
Write the balance equation for the following reactions:
a. Solid mercury (II) sulfide decomposes into its elements
when heated.
b. The surface of aluminum metal undergoes a combination
reaction with oxygen in the air.
Answer: (a) HgS(s) → Hg(l) + S(s)
(b)4 Al(s) + 3 O2(g) →2 Al2O3(s)
COMBUSTION REACTIONS
Combustion reactions are rapid reactions that produce a
flame. They require air (O2) as a reactant.
When hydrocarbons are combusted completely, the
products are CO2 and H2O.
Generic Reaction:
CxHy + O2
CO2 + H2O
Real Reaction:
C3H8 + 5O2
3CO2 + 4H2O
Examples:
CH4 (g) + 2 O2 (g) →
2H2
٨
+
O2
CO2 (g) + 2 H2O (g)
→ 2H2O
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SAMPLE EXERCISE 3.4
Write the balanced equation for the reaction that occurs
when methanol, CH3OH(l), is burned in air.
CH3OH(l) + O2(g) →CO2(g) + H2O(g)
2 CH3OH(l) + 3 O2(g) →2 CO2(g) + 4 H2O(g)
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Practice Exercise:
Write the balance equation for the reaction that occurs
when ethanol, C2H5OH(l), is burned in air.
Answer: C2H5OH(l) + 3 O2(g) →2 CO2(g) + 3 H2O(g)
FORMULA WEIGHTS
The formula weight is a substance is the sum of the atomic
weights of each atom in its chemical formula.
Example:
H2SO4 = 2(1 amu) + 32 amu + 4(16 amu) = 98 amu
For molecular compounds: The formula weight is also
called the molecular weight (MW).
For ionic compounds: No separated molecules so it called
specifically the formula weight (FW).
For Elements: In case of isolated atoms is called atomic
weight (AW).
FW of NaCl = 23 amu + 35.5 amu = 58.5 amu
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SAMPLE EXERCISE 3.5
Calculate the formula weight of
a. sucrose (C12H22O11)
MW or (FW) of sucrose (C12H22O11) =
12(12 amu) + 22(1 amu) + 11(16 amu) = 342 amu
b. Calcium nitrate .
FW Ca(NO3)2 = 40.1 amu) + 2[14 amu+3(16 amu)] = 164.1 amu
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Practice Exercise:
Calculate the formula weight of (a) Al(OH)3 and (b) CH3OH.
Answer: (a) 78.0 amu, (b) 32.0 amu
PERCENT COMPOSITION
Percent Composition is the percentage
contributed by each element in a substance.
Generally:
by
mass
(number of atoms)(atomic weight)
x 100
% element =
(FW of the compound)
Example: For C2H6 (MW = 30 amu)
(2)(12.0 amu)
x 100 =
%C =
(30.0 amu)
x 100 =
(30.0 amu)
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x 100
= 80.0%
x 100
= 20.0%
30.0 amu
(6)(1.01 amu)
%H =
24.0 amu
6.06 amu
30.0 amu
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SAMPLE EXERCISE 3.6
Calculate the percent of carbon, hydrogen, and oxygen (by mass)
in C12H22O11.
Solution:
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Practice Exercise:
Calculate the percentage of nitrogen, by mass, in calcium
Answer: 17.1%
nitrate.
AVOGADRO’S NUMBER AND THE MOLE
The mole is just a number of things.
In daily life:
1 dozen = 12 things
1 pair = 2 things
1 Cross = 144 things
In Chemistry the unit for dealing with the number of
atoms, ions, or molecules, any particles the mole,
abbreviated mol.
1 mol = 6.02 x 1023 representative particles
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SAMPLE EXERCISE 3.7
Arrange the following samples in order of increasing
numbers of carbon atoms:
12 g 12C, 1 mol C2H2, 9 x 1023 molecules of CO2.
Solution:
12 g of 12C = 1 mol of C atoms = 6.02 ×1023 C atoms.
1 mol of C2H2 = 6.02 ×1023 C2H2molecules
= 2 × 6.02 ×1023 C atoms.
9 x 1023 molecules of CO2 = 9 x 1023 C atoms.
The order:
12 g 12C (1 mol C) < 9 ×1023CO2 molecules (1.5 mol C) < 1
mol C2H2 (2 mol C).
PRACTICE EXERCISE
Arrange the following samples in order of increasing number
of O atoms: 1 mol H2O, 1 mol CO2, 3 x 1023 molecules O3.
Answer: 1 mol H2O (6 ×1023O atoms) < 3 ×1023molecules O3(9 ×1023O
atoms) < 1 mol CO2(12 ×1023O atoms)
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Important Notes:
1 molecule of H2SO4 →
1 molecule of H2SO4 →
1 molecule of H2SO4 →
2 atoms of H
1 atom of S
4 atoms of S
1 mol of H2SO4 → (6.022x 1023) molecules of H2SO4
1 mol of H2SO4 → 2 moles of H → 2(6.022x 1023) atoms of H
1 mol of H2SO4 → 1 moles of S → (6.022x 1023) atoms of S
1 mol of H2SO4 → 4 moles of S → 4(6.022x 1023) atoms of O
SAMPLE EXERCISE 3.8
Calculate the number of H atoms in 0.350 mol of C6H12O6.
Solution:
moles C6H12O6 →molecules C6H12O6→atoms H
Another solution:
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PRACTICE EXERCISE
How many oxygen atoms are in:
a. 0.25 mol of calcium nitrate?
Answer: 9.0 ×1023.
b. 1.50 mol of sodium carbonate?
Answer: 2.71 ×1024
MOLAR MASS
A mole is always the same number (6.02 x 1023).
But 1 mole sample of different substances will have
different masses.
A mol of an element = AW (amu) in grams of that
element.
The mole in grams of substance called the molar mass
of the substance (g/mol).
Each is 1 mole of
different
elements
have
different
masses
but
all
contain
same
number of atoms
= 6.02 x 1023
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SAMPLE EXERCISE 3.9
What is the mass in grams of 1.000 mol of glucose,
C6H12O6?
Solution:
MW glucose C6H12O6 =
6(12 amu) + 12(1 amu) + 6(16 amu) = 180 amu
So 1 mole of glucose = 180 g
And the molar mass = 180g/mol
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Practice Exercise:
Calculate the molar mass of calcium nitrate.
Answer: 164.1 g/mol
SAMPLE EXERCISE 3.10
Calculate the number of moles of glucose in 5.380g of
glucose.
?? mol Glucose → 5.380 g Glucose
Conversion factor:
1mol Gl ↔ 180 g Gl
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Practice Exercise:
How many moles of sodium bicarbonate are in 508g of sodium
bicarbonate (also known as sodium hydrogen carbonate)?
Answer: 6.05 mol NaHCO3
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SAMPLE EXERCISE 3.11
Calculate the mass, in grams. Of 0.433 mol of calcium
nitrate.
?? g of Ca(NO3)2 → mol Ca(NO3)2
mol Ca(NO3)2 → 164.1 g Ca(NO3)2
PRACTICE EXERCISE
What is the mass, in grams, of
a. 6.33 mol sodium bicarbonate?
b. 3.0 x 10-5 mol of sulfuric acid?
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Answer: 532 g
Answer: 2.9 ×10–3g
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SAMPLE EXERCISE 3.12
a) How many molecules are in 5.23g of glucose?
?? molecules → 5.23 g glucose
b. How many oxygen atoms are in this sample?
?? O atoms → 5.23 g glucose
PRACTICE EXERCISE
a) How many molecules are in 4.20g of nitric acid?
Answer: 4.01 ×1022 molecules HNO3
b) How many O atoms are in this sample?
Answer: 1.20 ×1023atoms O
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EMPIRICAL FORMULAS (EF) FROM ANALYSES
An empirical formula is the lowest whole number ratio of
elements in chemical formula (written as subscripts).
1. When given percentages, assume 100g so that the
percentages can be grams.
2. Convert grams into moles.
3. Divide all answers by the lowest number to get
subscripts.
4. If you get a 0.5, or 1.5 then multiply all numbers by 2.
If you get a 0.33 or 0.66, then multiply all numbers by
3 (in order to obtain smallest whole numbers).
SAMPLE EXERCISE 3.13
Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H,
and 54.5% O by mass. What is the empirical formula of
ascorbic acid?
Solution: Assume 100 g sample for simplicity:
Divide on the smallest value: (3.406)
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Multiply by 3:
The EF:
PRACTICE EXERCISE
A 5.325g sample of methyl benzoate, a compound used in
perfumes, contains 3.758g C, 0.316g H, and 1.251g of O. What
is the empirical formula of this substance?
Answer: C4H4O
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MOLECULAR FORMULA (MF)
The molecular formula is the actual formula of a molecule.
Sometimes it can be the same as the empirical formula.
MF can be determined from the EF if we have the actual
molecular weight.
MF = n(EF)
where n is whole number
C2H6 = 2(CH3)
Example:
The whole number obtained by:
MW = 30 for C2H6
EW = 15 for CH3
So
= 30/15 = 2
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In the previous example 3.13: If the actual MW of
ascorbic acid = 176 what is the MF of the acid?
EF was : C3H4O3
EW → [3(12) +4(1)+3(16) ]= 88
Actual MW → 176
n = 176/88 = 2
MF = 2 (C3H4 O3) = C6H8O6
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SAMPLE EXERCISE 3.14
Mesitylene, a hydrocarbon in crude oil, has an empirical
formula of C3H4. The molecular weight of the substance is
121 amu. What is the molecular formula of mesitylene?
EF : C3H4
EW → [3(12) +4(1)]=40
Actual MW → 121
n = 121/40 = 3.02
MF = 3 (C3H4) = C9H12
PRACTICE EXERCISE
Ethylene glycol, the substance used in antifreeze, is composed of
38.7%C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol.
a) What is the empirical formula of ethylene glycol?
Answers: CH3O
b. What is the molecular formula for the previous question?
Answers: C2H6O2
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COMBUSTION ANALYSIS
Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like
this
C is determined from the mass of CO2 produced
H is determined from the mass of H2O produced
O is determined by difference after the C and H have
been determined
SAMPLE EXERCISE 3.15
Isopropyl alcohol, a component of rubbing alcohol, is composed
of C, H, and O. Combustion of 0.255g of isopropyl alcohol
produces 0.561g of CO2 and 0.306g H2O. Determine the
empirical formula.
Solution:
We need to determine:
1) g C
2) g H
3) g O = mass of sample – (g C + g H)
4) Change all grams to moles
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PRACTICE EXERCISE
(b) Caproic acid, the smell in dirty socks, is composed of C, H, and O
atoms. Combustion of a 0.225g sample of this compound produces
0.512g CO2, 0.209g H2O. What is the empirical formula of caproic
acid?
Answers: C3H6O
(b) Caproic acid has a molar mass of 116 g/mol. What is the
molecular formula?
Answer: C6H12O2
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QUANTITATIVE INFORMATION FROM BALANCED EQUATIONS
The coefficients in a balanced equation indicate both the
relative numbers of moles in the reaction.
A mole ratio is the ratio of the coefficients of two
substances in a chemical equation.
You must use the mole ratio to convert from one
substance to another.
C2H5OH + 3O2 → 2CO2 + 3H2O
Conversion factors:
1 mol C2H5OH
3 mol O2
↔
1 mol C2H5OH
2 mol CO2
↔
1 mol C2H5OH
3 H2O
↔
3 mol O2
2 mol CO2
↔
3 mol O2
3 mol H2O
↔
2 mol CO2 ↔
3 mol H2O
SAMPLE EXERCISE 3.16
How many grams of water are produced in the oxidation of
1.00g of glucose (MW=180)?
6CO2(g) + 6H2O(g)
C6H12O6(s) + 6O2(g)
1.00 g
?? g
1mol C6H12O6(s) ↔
٢٤
6 mol H2O(g)
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PRACTICE EXERCISE
How many grams of O2 can be prepared from 4.50g KClO3?
2KClO3(s)
2KCl(s) + 3O2(g)
Answer: 1.77 g
SAMPLE EXERCISE 3.17
Lithium hydroxide reacts with carbon dioxide to form lithium
carbonate and water. How many grams of carbon dioxide (MW =
44) can be absorbed by 1.00g of lithium hydroxide (FW = 23.95)?
Solution:
2 LiOH(s)
1.00g
+
CO2(g) → Li2CO3(s) + H2O(l)
??g
2 mol LiOH(s) ↔ 1 mol CO2(g)
Grams LiOH →moles LiOH →moles CO2 →grams CO
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PRACTICE EXERCISE
What mass of O2 is consumed in the combustion of 1.00g of
propane (C3H8)?
Answer: 3.64 g
LIMITING REAGENTS (LR)
Limiting reactant or (limiting reagent): The reactant
that is completely consumed in a reaction.
The other reactants are called the excess reactants or
reagents.
The amount of product is always based on the limiting
reagent.
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DETERMINE THE (LR)
For the balanced equation:
2H2
Stiochimetry:
Excess reactant:
+ O2 →
2H2O
2 mol
1mol
2 mol
5 mol
1mol
2 mol
The reaction stops when O2 consumed; so O2 is called LR
Excess of Reactant : 5 mol H2 – 2 mol H2 = 3 mol H2
Guide for determination of the (LR):
1.
2.
3.
4.
5.
Reactions contain more than given amounts of
reactants are LR reactions.
Convert grams of reactants to moles.
Divide the results of each mole on its coefficient in the
balanced equation.
The smallest value is for the LR substance.
Your calculations should be based on the LR.
SAMPLE EXERCISE 3.18
How many moles of NH3 can be formed from 3.0 mol of
N2 and 6.0 mol of H2?
Given:
٢٧
N2(g) +
3 mol
3.0/1 = 3
3H2(g)
6 mol
6/3 = 2
LR
2NH3(g)
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PRACTICE EXERCISE
A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to
react.
2AlCl3(s)
2Al(s) + 3Cl2(g)
a) Which is the limiting reagent?
Answer: Al
b) How many moles of AlCl3 are formed?
Answer:1.50 mol
How many moles of the excess reactant remain at the end of the
reaction?
Answer: 0.75 mol Cl2
SAMPLE EXERCISE 3.19
Suppose a fuel cell is set up with 150g of hydrogen gas and
1500g of oxygen gas. A)How many grams of water can be
formed?
2H2(g)
150 g
For H2 : 75/2 = 37.5
H2 is LR
٢٨
+
O2(g)
1500 g
2H2O(g)
??g
For O2: 47/1 =47
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B) How many grams of reactant present in excess
remain after the reaction completed?
2H2(g)
+
O2(g)
150 g
2H2O(g)
?? g reacted
Excess O2 = origin – reacted = 1500-1200 = 300 g
PRACTICE EXERCISE
A strip of zinc metal with a mass of 2.00g is placed in an
aqueous solution containing 2.50g of silver nitrate, causing the
following reaction to occur:
Zn(s)
+
2AgNO3(aq)
a. Which reactant is limiting?
٢٩
2Ag(s) +
Zn(NO3)2(aq)
١٢/٦/١٤٣٦
b. How many grams of Ag will form?
c. How many grams of Zn(NO3)2 will form?
d. How many grams of the excess reagent will be left?
PERCENT YIELD
An important indicator of the efficiency of a particular
laboratory or industrial reaction.
The percent yield of a reaction relates the actual yield
or experimental yiel (done in the lab) to the
theoretical yield (calculated using stoichiometry).
Actual yield
× 100% = percent yield
Theoretica l yield
Actual yield < Theoretical yield , because:
1- Incomplete reaction.
2- Side reactions.
3- Some product lost (gas or deposited solids).
4- personal error.
Note: Percent yield should never exceed 100%.
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SAMPLE EXERCISE 3.20
Adipic acid, H2C6H8O4, is used to produce nylon. Adipic acid is
made by the following reaction:
2C6H12(l) + 5O2(g) 2H2C6H8O4(l) + 2H2O(g)
a. Assume that you carry out this reaction starting with 25.0g of
cyclohexane and 100.0 g Oxygen. What is the theoretical
yield of the adipic acid?
2C6H12(l)
+
5O2(g)
2H2C6H8O4(l) +
2H2O(g)
25.0g
100.0 g
(g ???) Th. Yield
For C6H12 : 0.3/2 = 0.15
C6H12 is LR
For O2: 3.12/5 = 0.63
b. If you obtain 33.5g of adipic acid from your reaction,
what is the percent yield of adipic acid?
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PRACTICE EXERCISE
a. If you start with 150g of Fe2O3 as the limiting reagent, what
is the theoretical yield of Fe?
Fe2O3(s)
+
3CO(g)
2Fe(s)
+
3CO2(g)
Answer: 105 g Fe
b. If the actual yield of Fe in your test was 87.9g, what is the
percent yield?
Answer: 83.7%
HW PROBLEMS
1,
29,
69,
٣٢
4,
35,
76,
6,
38,
81,
8,
41,
90,
15,
46,
98,
19, 22,
52, 54,
104.
24,
58,
27,
63,
‫‪١٢/٦/١٤٣٦‬‬
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‫‪ ,‬دي‪:‬‬
‫ا م وو ‪ +‬ر‬
‫ﺳ ﺣ ا م‪ ،‬و ن اؤه أن ‪ U‬ا ;‪ ,‬دي إ ‪ 6‬ا‬
‫و ‪ + S4 DE ، @+‬و ‪@+‬‬
‫@‪ 4‬آ‪1‬ف أ‪ E‬ى‪ ،‬اﺳ \ ب ا‬
‫ا ‪ F‬ط( ‪ # =W ًV‬ل ‪ DE :‬ا * ‪ ،S (G‬و‬
‫د ا م !ة آ‪#‬ف د‪ %‬ر‪،‬‬
‫ر ﻣ‬
‫ا ‪ ، F‬و ل ‪ :‬و ‪ ،‬ل ا دي‪ :‬أ‬
‫و ل ‪ :‬اط!ح ﻣ ( أ ) * ا ‪!+‬م‪ , ،‬د ‪0* ، (-.‬ن ردھ إ ‪ 2-‬ﻣ و ھ *‪5‬د*‪ 4‬ا ل‬
‫‪ .6‬إ ‪ 0* -‬أﻣ‪. -‬‬
‫ً‬
‫ﺣ ‪”P4 1 9‬‬
‫‪ 5‬و ز‪#‬‬
‫^ﷲ >‬
‫‪ #‬ل ﷲ @ ‪” :6‬و‬
‫‪٣٣‬‬