Stoichiometry

Stoichiometry
Dr. Damodar Koirala
Shree Amarsingh Model Secondary School
Pokhara, Nepal

Dr. Damodar Koirala | koirala2059@gmail.com
Content
✤ Dalton’s atomic theory
✤ Laws of stoichiometry
✤ Avogadro’s hypothesis
✤ Mole concept
✤ Calculation based on balanced chemical equation
✤ Limiting reagent
✤ Theoretical yield and percentage yield
2

Dalton’s atomic theory
✤ Proposed by John Dalton in 1808 A.D
✤ Postulates:
1. Matter consists of individual atoms
2. Atom can neither be created nor destroyed
3. All the atoms of a given element have identical properties including identical
mass. Atoms of different elements differ in mass
4. Atoms can combine in small whole number ratio to give compound atom
5. During chemical reaction, atoms of an element does not convert to atom of
another element
3

Drawbacks of Dalton’s atomic theory
1. Atoms is not the smallest particle since it is made up of
sub-atomic particles
2. All atoms of given element can have different mass called
isotopes. Like 1H, 2H, 3H are hydrogen isotopes with
different atomic mass
3. Atoms of different element can have same mass called
isobars. 40Ca and 40Ar
4. By nuclear reaction, one atom can be changed into another
atom
4

Laws of stoichiometry
✤ Branch of physical chemistry that deals with weight or volume
relationship in chemical reaction
1. Law of conservation of mass
2. Law of definite proportion
3. Law of multiple proportion
4. Law of reciprocal proportion
5. Gay-Lussac’s law of gaseous volumes
5

1. Law of conservation of mass
✤ “The total mass of reactant before reaction equals to
the total mass of product after reaction”
✤ No mass is lost or gained during chemical reaction
✤ Also, known as indestructibility of matter
6

1. Illustration
✤ Take a H-shaped tube and take known mass of sodium chloride and
silver nitrate
✤ Close the opening of tube and shake well. Reactants come in contact
and get reacted
✤ Formation of white ppt signal the reaction had occurred
✤ Measure the weight of the product. Following is observed:
Wt before addition = Wt after reaction
7
NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3
White ppt
NaCl AgNO3

1. Numerical
✤ 4 gram of hydrogen reacts with 12 gram of carbon to
form 16 gram of methane (CH4). How does this
illustrate the law of conservation of mass.
8

2. Law of definite proportion
✤ “Same compound always contains the same elements
combined together in the same proportion by weight
regardless of origin or made of formation of compound”
✤ Also called law of constant composition
✤ Suppose, if A and B combine to form AB.
✤ Case I: ‘a’ gram of A combine with ‘b’ gram of B
✤ Case II: ‘x’ gram of A combine with ‘y’ gram of B
✤ Then:
9
a
=
x
b y

2. Illustration
Variation of source and method of preparation does not
vary the composition of compound
✤ Water can be obtained from different sources like lake,
river, pond etc.
✤ Also, by reaction of H2 and O2.
✤ In all case, the ratio of mass of hydrogen to the ratio of
mass of oxygen is 1:8
10

2. Illustration
✤ 4 gram of metal reacts with 32 gram metal oxide and
also 32 gram of same metal reacts with oxide to
produce 256 gram metal oxide. How does this
illustrate law of constant composition.
✤ In an experiment, 0.12g of carbon react with oxygen to
form 0.28 g oxide. In second experiment, 0.30g of
carbon react with oxygen to form 0.70g oxide. Show it
shows the law of constant proportion.
11

3.Law of multiple proportion
✤ “When one element combines with a second element
to form two or more different compounds, then the
weight of one element which combine with a constant
weight of one element which combine with a constant
weight of another, bear a simple ratio to one another”
12

3. Illustration
In H2O
1 parts by wt of H combine with 8 parts by wt of O
In H2O2
1 parts by wt of H combine with 16 parts by wt of O
Hence, in H2O and H2O2, 1 parts by weight of H combine with 8:16 i.e. 1:2
parts by wt of oxygen respectively and it is simple whole number ratio
13
Consider,
Hydrogen + Oxygen
H2O (water)
H2O2 (Hydrogen peroxide)

3. Illustration
14
Consider, the compounds of nitrogen and oxygen
Weight of
***
Whole
number
Nitrogen Oxygen
Nitrous oxide N2O 28 16 16 1
Nitric oxide NO 14 16 32 2
Nitrogen trioxide N2O3 28 48 48 3
Nitrogen peroxide N2O4 28 64 64 4
Nitrogen pentaoxide N2O5 28 80 80 5
Nitrogen dioxide NO2 14 32 64 4
*** weight of oxygen that react with 28 part by weight of nitrogen
Hence, the ratio is 1:2:3:4:5:2, which is simple whole number ratio

✤ A certain element X1 forms three different binary compounds with
chlorine containing 59.68%, 68.95% and 74.75% chlorine respectively.
Show how these data illustrate law of multiple proportion.
15

4. Law of reciprocal proportion
✤ “When two different elements combine with the same
quantity of a third element the proportion in which
they do so will be the same as or some simple multiple
of proportions in which they unite with each other”
✤ Also called law of equivalent proportion because
weight of two elements which combines with fixed wt
of third element are called equivalent wt.
16

4. Illustration
Consider three element carbon, oxygen and hydrogen.
Lets take the compound methane (CH4), hydrogen peroxide (H2O2) and carbondioxide (CO2)
Case I: In CH4
4 parts by wt of H combines with 12 parts by wt of C
1 parts by wt of H combines with 12/4 = 3 parts by wt of C
CaseII: In H2O2
2 parts by wt of H combines with 32 parts by wt of O
1 parts by wt of H combines with 32/2 = 16 parts by wt of O
Therefore, from Case I and Case II, if carbon and oxygen happen to combine together they must do
either in C : O = 3 : 16 or in its simple multiple of 3 : 16
Case III: In CO2
12 parts by wt of C combines with 32 parts by wt of O
Hence, C : O = 12 : 32 = 3 : 8
3/8 * 1/2 = 3/16, which is as expected.
17
C
H O
CH4
H2O
CO2

4. Numerical
1. A, B and C are three elements. 1.00 g of A combines with 1.33
gram of B, 1 g of B combines with 0.125g of C, 1g of C combines
with 6 g of A. Show that the above results are in accordance with
law of reciprocal proportion.
2. Show the following data illustrate law of reciprocal proportion
a. 0.46g of metal produces 0.77g of metal oxide
b. 0.80g of same metal displaced 760cc of H2 gas at STP from HCl
c. 1.26g of same was formed by union of 1.12g of oxygen with
hydrogen (1L hydrogen = 0.089g)
18

Dr. Damodar Koirala | koirala2059@gmail.com 19
1. A, B and C are three elements. 1.00 g of A combines with 1.33
gram of B, 1 g of B combines with 0.125g of C, 1g of C combines
with 6 g of A. Show that the above results are in accordance with
law of reciprocal proportion.

2. Show the following data illustrate law of reciprocal proportion
a. 0.46g of metal produces 0.77g of metal oxide
b. 0.80g of same metal displaced 760cc of H2 gas at STP from HCl
c. 1.26g of same was formed by union of 1.12g of oxygen with
hydrogen (1L hydrogen = 0.089g)

✤ “At similar temperature and pressure, when two gases
combine to give a product which is also a gases
combine to give a product which is also a gas, the ratio
by their volume in which they do so bear a simple
whole number ratio to one another”
21

✤ 15 cc of a gaseous mixture containing ethylene and
acetylene require 45cc of oxygen for complete
combustion. Determine the percentage composition of
the mixture.
22

Avogadros hypothesis
✤ “Equal volume of all gases contain the equal number of
molecules under similar condition of temperature and
pressure”
✤ He made a distinction between atoms and molecules
✤ Dalton and other believed at that time that atoms of the
same kind can not combine and the molecule oxygen or
hydrogen containing two asme atoms did not exist
✤ It was only after about 50 years in 1860 A.D, this theory was
accepted
23

Berzelius hypothesis : Starting of Avogadro's hypothesis
✤ Equal volume of all gases contain the equal number of atom under
similar condition of temperature and pressure
✤ Experimentally: it is found that 1 vol of hydrogen reacts with 1 vol of
chlorine to produce 2 vol of hydrogen chloride
Hydrogen + Chlorine ! Hydrogen chloride
1 vol 1 vol 2 vol
Let , 1 vol = n atoms, according to Berzelius hypothesis, then
n atoms n atoms 2n atoms
1 atom 1 atom 2 atoms
1/2 atoms 1/2 atoms 1 atom
✤ It shows that 1 compound atom of hydrogen chloride is obtained by
reaction of 1/2 atom of hydrogen and 1/2 atom of chlorine.
✤ This is against Dalton’s atomic theory since 1/2 atom is not possible.
Hence, Berzelius hypothesis failed. 24

Avogadro's hypothesis
✤ Equal volume of all gases contain the equal number of molecule under
similar condition of temperature and pressure
✤ Experimentally: it is found that 1 vol of hydrogen reacts with 1 vol of
chlorine to produce 2 vol of hydrogen chloride
Hydrogen + Chlorine ! Hydrogen chloride
1 vol 1 vol 2 vol
Let , 1 vol = n molecules, according to Avogadro’s hypothesis, then
n molecules n molecules 2n molecules
1 molecule 1 molecule 2 molecules
1/2 molecule 1/2 molecule 1 molecule
✤ It shows that 1 compound atom of hydrogen chloride is obtained by
reaction of 1/2 molecule of hydrogen and 1/2 molecule of chlorine.
✤ Since 1/2 molecule possible, it supports Dalton’s atomic theory
25

Avogadros hypothesis:Application
1.Deduction of atomicity of the elementary gasses.
✤ The number of atom present in a molecule of a gas
in called atomicity
✤ eg: Consider the reaction between hydrogen and
chlorine gas. It is found experimentally hydrogen
and chlorine react in equal volume to produce twice
the volume of hydrogen chloride (HCl). i.e.
26

1.Deduction of atomicity of the elementary gasses.
Hydrogen + Chlorine ! Hydrogen chloride (HCl)
1 vol 1 vol 2 vol
n molecules n molecules 2n molecules
Now, we know that 1 molecule of hydrogen chloride (HCl) contains 1 hydrogen
atom and 1 chlorine atom
From above experimental data,
1/2 molecule of hydrogen = 1 hydrogen atom
or, 1 molecule of hydrogen = 2 hydrogen atom
Similarly,
1/2 molecule of chlorine = 1 chlorine atom
or, 1 molecule of chlorine = 2 chlorine atom
Hence, the atomicity of both hydrogen and chlorine is 2. 27

2. Determination of molecular formula of gaseous
compound
✤ Experimentally it is found that one volume of
nitrogen (N2) reacts with three volume of hydrogen
(H2) to form two volume of ammonia
28

2.Determination of molecular formula of gaseous compound
Hydrogen + Nitrogen ! Ammonia
3 vol 1 vol 2 vol
3n molecules n molecules 2n molecules
Now, we know that hydrogen is H2, and nitrogen is N2
From above experimental data,
1 molecule of ammonia = 3/2 hydrogen molecule
= 3/2 * 2 hydrogen atoms = 3 hydrogen atoms
Similarly,
1 molecule of ammonia = 1/2 nitrogen molecule
= 1/2 * 2 nitrogen atoms = 1 nitrogen atom
Hence, the molecular formula of ammonia is NH3. 29

3. Derivation of relationship between molar mass and vapour density
( M.Wt. = 2* VD)
✤ Vapour density of a gas is defined as the ratio between the mass of
certain volume of gas, to the mass of the same volume of hydrogen gas
under similar condition of temperature and pressure
30
V.D =
Mass of Vml of substance in gaseous state
Mass of Vml of hydrogen under similar condition
V.D =
Mass of ’n’ molecule of substance in gaseous state
Mass of ’n' of hydrogen under similar condition
Let , Vml = n molecules, according to Avogadro’s hypothesis, then

3. Derivation of relationship between molar mass and vapour
density ( M.Wt. = 2* VD)
or, M.Wt = 2* V.D 31
V.D =
Mass of 1 molecule of substance in gaseous state
Mass of 1 of hydrogen under similar condition
V.D =
2* Mass of 1 atom of hydrogen under similar condition
1 molecule of hydrogen contain 2 atoms, hence
Also,
M Wt =
Mass of 1 atom of hydrogen under similar condition
Hence, V.D =
M.Wt
2

2. What volume of CO2 at STP will be delivered to
extinguish fire from a cylinder of 10L capacity
containing 5 kg CO2 gas.
32

Mole concept
✤Mole is the concept of quantity in terms of mass, volume and
number. It relates the number of atoms, ions or molecules in it.
✤A mole of chemical species is defined as exactly 6.023×10²³
chemical species, which may be atoms, molecules, ions, or
electrons
33
Number of particles Mole Mass
Volume

Mole and number of particles
A mole of chemical species is defined as exactly 6.023×10²³
chemical species, which may be atoms, molecules, ions, or
electrons
✤1 mole of atom = 6.023×10²³ atoms
✤1 mole of molecule = 6.023×10²³ molecules
✤1 mole of radical= 6.023×10²³ radicals
✤1 mole of man = 6.023×10²³ man
6.023×10²³ is called Avogadro number or NA
34

Mole and particles number
Important formula
35
Mole =
No. Of Particles
NA
No. Of particles = Mole * NA

Mole and number of particles
1. How many NaCl molecules are present in 0.3 mol NaCl?
2. How many mole of H2O molecules are present in 20* 1024
H2O molecules?
36

Mole and mass
A mole of molecule is gram molecular weight and a mole
of atom is gram atomic weight
✤1 mole of hydrogen atom = 1 g
✤1 mole of hydrogen molecule = 2 g
✤1 mole of HCl = (35.5 + 1 )g = 36.5 g
✤1 mole of H2SO4 = (2*1 + 1*32 + 4*16) g = 98 g
✤1 mole of sodium = 23 g
37

Mole and mass
Important formula
38
Mole of molecule =
Mass in gram of molecule
Gram molecular wt of moleculer
Mole of element =
Mass in gram of element
Gram atomic wt of element

Mole concept : mole and mass
✤ Q1: How many moles of water are present in 5g of water ?
✤ Q2: How many grams are there in 0.03 mole of CaSO4 ?
✤ Q3: What mass of sodium is present in 0.2 mole of NaOH ?
39

Mole concept : Element
✤ Q1: How many moles are present in 2g of potassium
✤ Q2: How many grams are there in 0.01 mole of carbon ?
40

Mole and volume
The volume occupied by one mole of any substance is called
molar volume. For all gases, molar volume (Vm) = 22.4L at STP.
( STP is when T = 25 C = 298K and P = 760 torr = 1atm).
✤1 mole of hydrogen gas at STP = 22.4 L
✤1 mole of oxygen gas at STP = 22.4 L
✤1 mole of carbondioxide gas at STP = 22.4 L
✤1 mole of chlorine gas at STP = 22.4 L
✤1 mole of nitrogen gas at STP = 22.4 L
41

Mole and volume
✤A gaseous mixture contains 50% CO and 50% CO2 by
volume at NTP. Calculate the percentage by mass of
methane in mixture.
42

Mole and volume
43
Q1: How many moles are present in 2L of oxygen gas at STP?
Q2: How many litres are there in 0.01 mole benzene gas at
STP ?

Mole concept
44
1 Mole
Gram
atomic wt
Avogadro’s
number of
molecule
Avogadro’s
number of
atom
Gram
molecular wt
22.4 L @ NTP

Conversion between units
45
No. of atoms Mole Mass in gm
Mass in gm Mole No. of atoms
No. of molecules Mole Mass in gm
Mass in gm Mole No. of molecules
Vol in L @ STP Mole Mass in gm
Mass in gm Mole Vol in L @STP
÷ gm atomic wt
÷ gm M.Wt
÷ gm M.Wt
÷ NA
÷ NA
÷ 22.4 L
* NA
* NA
* 22.4 L
* gm atomic wt
* gm M.Wt
* gm M.Wt
Note: to mole is division and from mole is multiplication

Let’s be clear on terminologies
✤ amu vs. gram
✤ both can be used as the unit of atomic or molecular mass
✤ amu is NA times smaller than gram
✤ gram is NA times greater than amu
✤ amu is for atom but gram is for mole of atom
✤ amu is for molecule but gram is for mole of molecule
46
1 amu =
1 gram
NA

Let’s be clear on terminologies
✤ hydrogen atom vs. hydrogen molecule
✤ both contains hydrogen element only
✤ formula are H and H2 respectively
✤ 1 molecule of hydrogen = 2 atoms of hydrogen
✤ weight: 1 atom = 1amu vs.1 molecule = 2 amu
✤ weight: 1 mole of atom = 1g vs.1 mole of molecule = 2 g
47

Conversion problems
✤ Q: Calculate the number of molecules in
1. 0.25 mole oxygen molecule
2. 0.7 gram of nitrogen
3. 5.3 gram of Na2CO3
48

Conversion problems
✤ Q: Calculate the number of molecules in
1. 1.8 gram of H2O
2. 1.2 ml CO gas @ STP
3. 0.2 mol hydroxychloroquine
49

Conversion problems
✤ Calculate the number of different atoms in
1. 5.0 gram of CaCO3
2. 1.8 gram of H2O2
3. 1.2 L CO2 gas @ STP
50

Conversion problems
✤ Q: Calculate the number of different atoms in
1. 0.2 mol ethene
2. 0.02 mol of blue vitrol
3. 3 g of glucose
51

Conversion problems
✤ One million silver atoms weigh 1.79×l0–16 gram. Calculate the atomic
mass of Ag?
✤ One drop of water weighs 0.04gm. Calculate the number of H2O
molecules in one drop of water.
✤ One atom of an element 'A' weighs 6.644×10-20 g. Calculate the
number of gram atom in 80kg of it.
✤ Calculate the mass of i) two atom of nitrogen ii) one molecule of CO2
✤ How many atoms of hydrogen and oxygen are there in 9 gram of
water?
52

✤ One million silver atoms weigh 1.79×l0–16 gram. Calculate
the atomic mass of Ag?

✤ One atom of an element 'A' weighs 6.644×10-20 g. Calculate
the number of gram atom in 80kg of it.

✤ Calculate the mass of i) two atom of nitrogen ii) one
molecule of CO2

Chemical equation
A chemical equation is the symbolic representation of
a chemical reaction in the form of symbols and formulae,
wherein the reactant entities are given on the left-hand side
and the product entities on the right-hand side.
56
NaOH + HCl NaCl + H2O
NaOH and HCl are reactants.
NaCl and H2O are products.

Chemical reaction to chemical equation
(i) Write the symbols or the formulae of the reactants on the left hand
side, with a (+) sign between them
(ii) Write the symbols or the formulae of the products on the right
hand side, with a (+) sign between them
(iii) Put the sign of an arrow (→) in between the reactant side and the
product side
57
Example:
Chemical reaction:- Sodium hydroxide reacts with hydrogen chloride to
produce sodium chloride and water
Chemical equation:-
NaOH + HCl NaCl + H2O

Balancing chemical equation
(i) Count the number of times an element occurs on both sides.
(ii) An element with the least frequency of occurrence is
balanced first.
(iii) When two or more elements have same frequency, metallic
elements are balanced first.
58
Using Hit andTrial method
Fe + H2O Fe3O4 + H2

Numerical based on chemical equation
Here
✤ The quantity of one molecule (Reactant or Product) is given and the
quantity of other is asked.
✤ In this type of numerical, there is either only one reactant or the
quantity of one reactant is given and the other reactants are present
in excess. (Concept based on limiting reagent will be discussed
latter) OR. The quantity of one of the product is given and the
quantity of others (Reactant or product) is asked.
✤ Example: How many mole of HCl is produced if 4g of hydrogen is
completely reacted with chlorine
59

Information obtained from chemical equation
Qualitative information:
1) Name of the reactant which takes part in the reaction
2) Name of the products formed in the reaction
Quantitative information:
1) The number of molecules or atoms of reactant and products
taking part in the reaction
2) The number of mole of each substance involved in the reaction
 
3) The mass of each substance involved in the reaction
 
4) The mass-mass, mass-volume, volume-volume relationships
between the reactants and products
60

Quantitative relationship from chemical equation
61
Quantitative relationship:
Quantitatively a balanced chemical equation specifies numerical
relationship among the quantities of its reactants and products.
These relationships can be expressed in terms of :
(i) Microscopic quantities: atoms and molecules of reactants
and products
(ii)Macroscopic quantities: moles, masses and volumes (in case
of gaseous substances) of reactants and products

a) Mole relationship
62
The numerical coefficient present in front of reactants / products
provides the mole relationship
CH4 + 2O2 2H2O + CO2

63
CaCO3 CaO + CO2

64
Find the mole relationship is following reactions:
3Fe + 4H2O Fe3O4 + 4H2
4NH2 + 5O2 6H2O + 4NO

b) Mass relationship
65
Step 1: Determine the mole relationship
Step 2: Convert the required mole into mass
2Na + 2H2O 2NaOH + H2
How many gram of Na is required to produce 0.3mole of NaOH ?

H2 + Cl2 2HCl
66
How many gram of HCl is produced from 10gram of H2 ?

67
Calculate the mass of oxygen required to produce a)10gram CO2
from first reaction and b) 10gram of SO3 from second reaction.
C2H4 + 3O2 2H2O + 2CO2
2SO2 + O2 2SO3

c)Volume relationship
68
Step 2: Convert the required mole into volume
Remember @ NTP: 1 mole of gas = 22.4L of gas
H2(g) + Cl2(g) 2HCl(g)

d) Microscopic relationship
69
The numerical coefficient present in front of reactants / products provides
the microscopic relationship (no. of molecules or no. of atoms)
Fe2O3 + 3CO 2Fe + 3CO

Q. What information do we get from following reactions ?
70
Solution:

71
Solution:

72
Solution:

Numerical
1. What mass of CO2 is produced upon complete oxidation of 6.4gram
of methane?
2. How many mole of CaO is produced when 10g of CaCO3 is heated?
3. In Kipps apparatus, 14g of FeS was reacted completely with excess
dilute sulphuric acid. How many mole of H2S is produced?
4. How many gram of nitrogen is required to produce 510gram of
ammonia via Haber process.
5. How many mole of sulphur trioxide is required to produce 300
gram of oleum using following reaction.
73
SO3 + H2SO4 ! H2S2O7
conc. Oleum

Conversion problems
✤ One atom of an element 'A' weighs 6.644×10-20 g.
Calculate the number of gram atom in 80kg of it.
74

Numerical
1. What mass of CO2 is produced upon complete oxidation
of 6.4gram of methane?
75
CH4 + O2 H2O + CO2

Numerical
3. In Kipps apparatus, 14g of FeS was reacted completely with excess
dilute sulphuric acid. How many mole of H2S is produced?
76
FeS + H2SO4 FeSO4 + H2S

Numerical
4. How many gram of nitrogen is required to produce
510gram of ammonia via Haber process.
77

Limiting reagent
✤ Chemical reaction is like making egg sandwich. Not
everything is used up completely
78

Numerical
✤ A chemical reaction may have more than 1 reactant.
✤ For simplicity we will focus on up to 2 reactants.
✤ The reactant that is used up completely is called limiting
reagent.
✤ The reactant that is not used up completely (part of it is left)
is called excess reagent.
✤ The mass of excess reagent that is not reacted when reaction
is complete is called leftover mass.
79

Limiting reagent related problem
Question types:
✤ Define limiting reagent?
✤ Which reagent is the limiting reagent?
✤ What is the quantity (mass, mole, particles number, volume)
of excess reagent that is leftover?
✤ What is the quantity of (mass, mole, particles number,
volume) of product(s) ?
✤ What quantity of X compound is required to react with thus
formed product?
80

Numerical
1. A chemical reaction is carried out by mixing 25g of pure calcium carbonate
and 0.75 mole of pure hydrochloric acid to give CaCl2, H2O and CO2 ?
A. Which one is limiting reactant and why ?
B.Calculate the mass of CaCl2 produced ?
C.How many number of water molecules are formed?
D.What mass of NaOH is required to absorb the whole CO2 produced in the
reaction ?
81

Finding limiting reagent
1.Determine the balanced chemical equation for the chemical reaction.
2.Convert all given information into moles (most likely, through the use of molar mass as
a conversion factor).
3.Calculate the mole ratio from the given information. Compare the calculated ratio to
the actual ratio.
82
Given Equation
Mole of A
?
Mole of A
Mole of B Mole of B

Determine the limiting reagent?
a. 5 mole of H2SO4 is reacted with 3 mole of NaOH ?
b. A chemical reaction is carried out by mixing 1.3 mole of pure calcium carbonate and 2.5
mole of pure hydrochloric acid to give CaCl2, H2O and CO2 ?
83

a. 1 mole of ammonia is reacted with 45 gram of oxygen to produce NO and H2O ?
b. 2 gram hydrogen is treated with 2 mole chlorine to produce HCl ?
84

a. 2 gram of magnesium is burnt in a closed vessel containing 3 gram of oxygen?
b. 2.4 gram of Calcium is treated with 50gram of HCl to give CaCl2 and H2?
85

Limiting reagent and theoretical yield
✤ The quantity of product formed depends solely on
quantity of limiting reagent.
✤ The quantity of product to be formed for given quantity
of limiting reagent is called theoretical yield.
✤ Theoretical yield can be calculated using directly
proportional relation between limiting reagent and
product, and the balanced chemical reaction.
86

Theoretical yield ?
a. 5 mole of H2SO4 is reacted with 3 mole of NaOH ? What mass of water is produced?
b. A chemical reaction is carried out by mixing 1.3 mole of pure calcium carbonate and
2.5 mole of pure hydrochloric acid to give CaCl2, H2O and CO2 ? How many mole of
CO2 is produced ?
87

Theoretical yield ?
c. Determine the volume of ammonia gas formed at NTP if 4 gram of hydrogen gas is
treated with 55 gram of nitrogen gas ?
d. CH4 is oxidized to form carbondioxide and water. What volume of CO2 is produces at
STP if 17gram of CH4 is treated with 33 gram of oxygen ?
88

Theoretical yield ?
c. Determine the volume of ammonia gas formed at NTP if 4 gram of hydrogen gas is
treated with 55 gram of nitrogen gas ?
d. CH4 is oxidized to form carbondioxide and water. What volume of CO2 is produces at
STP if 17gram of CH4 is treated with 33 gram of oxygen ?
89

Limiting reagent
For a reaction
The reaction is carried out by mixing 7 gram of Ca(OH)2
with 7 gram of NH4Cl.
A. Determine the limiting reagent ?
B. Find the mole of unreacted reactant left over ?
C. How many gram of CaCl2 are formed ?
D. What volume of NH3 is produced at STP ?
90
Ca(OH)2 + 2NH4Cl ! CaCl2 + 2NH3(g) + H2O

Limiting reagent
Solution: here
A) (limiting reagent?)
91
Ca(OH)2 + 2NH4Cl ! CaCl2 + 2NH3(g) + H2O

Limiting reagent
B)(mole of left over)
92

Limiting reagent
C) (CaCl2 Produced)
93

Limiting reagent
D) (Ammonia at STP?)
94

Percentage purity
✤ It is the percentage of pure compound in an impure
sample
95
% purity =
Mass of pure compound in sample
X 100 %
Total mass of impure sample

Percentage purity
1. How many gram of HCl is present in 60 gram of 38% HCl solution ?
2. What mass of 60% if CaCO3 is required to react with 50 gram of HCl
to produce?
3. How many carbon atoms are present in 30g of 20% CO?
4. What mass of CO2 is produced if 30 gram of 60% CaCO3 is heated?
5. How many gram of 10% SO2 gas is required to produce 2 gram of
20% SO3 using following reaction ?
96
SO3(g) + NO2(g) ! SO3(g) + NO(g)

5. How many gram of 10% SO2 gas is required to
produce 2 gram of 20% SO3 using following reaction ?
SO3(g) + NO2(g) ! SO3(g) + NO(g)

A. In a chemical reaction, 80 gram of 60% H2SO4 is
reacted with 60 gram of 80% NaCl.
2NaCl + H2SO4 ! Na2SO4 + 2HCl
i. Determine the limiting reagent ?
ii. Find the mole of unreacted reactant left over ?
iii. How many gram of HCl are formed ?

i. Determine the limiting reagent ?

ii. Find the mole of unreacted reactant left over ?

iii. How many gram of HCl are formed ?

Stoichiometry

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