11
CAPS
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Physical Sciences 3-in-1
GRADE 8 - 12
ALL MAJOR SUBJECTS IN
ENGLISH & AFRIKAANS
Physical Sciences
CLASS TEXT & STUDY GUIDE
Retha Louw
Retha Louw
3-in-1
11
GRADE
CAPS
Grade 11 Physical Sciences 3-in-1 CAPS
CLASS TEXT & STUDY GUIDE
This Grade 11 Physical Sciences 3-in-1 study guide simplifies the theory of Grade 11 Physical Sciences and builds confidence
through clear explanations supported by revision questions. It allows you to rapidly improve your problem-solving skills.
Key features:
• Comprehensive, explanatory notes and worked examples per topic
• Exercises and exam questions per topic
• Detailed answers with explanations and handy hints
This study guide provides reliable guidance through Grade 11, while building a solid platform for the upcoming
Grade 12 curriculum.
11
GRADE
CAPS
Physical Sciences
Retha Louw
3-in-1
THIS CLASS TEXT & STUDY GUIDE INCLUDES
1
Comprehensive Notes
2
Exercises and Exam Questions
3
Detailed Memos with Explanations
(available in a separate booklet)
2013 publication | ISBN: 978-1-920558-31-4
E-book
available
120221 | NOVUS
CONTENTS
Questions ............................................................................................ Q33 - Q34
i
10%
35%
15%
3
60
Matter and Materials
(Module 2)
150
20
Chemical Systems
(Module 6)
Chemistry
focus
Module 6: Chemical Systems .......................................................... 6.1 - 6.10
Paper 2:
Paper
Questions ............................................................................................ Q27 - Q33
Physics
focus
Module 5: Electricity and Magnetism ............................................ 5.1 - 5.29
Paper 1:
Content
Questions ............................................................................................ Q20 - Q27
Mechanics
(Module 1)
Questions ............................................................................................ Q16 - Q19
Module 4: Chemical Change............................................................ 4.1 - 4.33
40%
40%
35%
70
Module 3: Waves, Sound and Light ............................................... 3.1 - 3.16
Chemical Change
(Module 4)
Questions ............................................................................................. Q7 - Q15
50
Module 2: Matter and Materials ..................................................... 2.1 - 2.40
Electricity and Magnetism
(Module 5)
Questions ................................................................................................ Q1 - Q7
15%
Module 1: Mechanics ....................................................................... 1.1 - 1.28
3
NOTES with QUESTIONS:
150
The Periodic Table of Elements (back of the book).............................................. v
32
Physical Constants and Formulae (back of the book) ......................................... iv
Waves, Sound and Light
(Module 3)
Detailed contents per term according to CAPS.................................................... ii
68
Weighting of questions
Total
across cognitive levels
marks Duration
Marks
per
(hours) Level Level Level Level
paper
1
2
3
4
The Grade 11 November Exam ............................................................................. i
10%
THE GRADE 11 NOVEMBER EXAM
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4
NOTES
4: CHEMICAL CHANGE
We already did simple stoichiometric calculations in Grade 10 and now we will
expand on that to reactions containing a limiting reagent or where the percentage
yield is calculated.
QUANTITATIVE ASPECTS
OF CHEMICAL CHANGE
REVISION OF IMPORTANT CONCEPTS
In Grade 10, we looked at different types of chemical reactions. For example,
synthesis, decomposition, ion-exchange and electron-transfer (redox) reactions
were covered.
The mole concept and Avogadro’s constant
A chemical reaction can be represented by a chemical equation. During any
chemical reaction, the principle of conservation of atoms applies. Therefore, a
chemical reaction is balanced so that the number of reactant atoms and the
number of product atoms are equal, e.g.:
`
+
3H2
t
2NH3
N2
2 N-atoms
6 H-atoms
The mole is the SI unit used to indicate amount of substance.
It is the amount of a substance containing as many particles as
there are atoms in exactly 12 g of carbon-12. This number,
i.e. 6,02 % 1023, is known as Avogadro’s constant (NA).
2 N-atoms + 6 H-atoms
Number of particles to/from mole
From this follows the Law of Conservation of Matter, i.e., that the mass of
substances in the reaction mixture is conserved during a chemical reaction.
In a complete reaction, the mass of the reactants = mass of the products.
number of moles (n) =
This allows us to do quantitative calculations to determine the mass of the
reactants and products during a chemical reaction.
â number of particles = nNA
CHEMICAL CHANGE
number of
particles
n
A chemical equation is balanced by placing
coefficients in front of the formulae of the reactants
and products. Do not change the formulae themselves.
NA
Example:
Determine the number of moles of ammonia (NH3) comprising 1,806 % 1024
molecules. Determine the number of atoms present in this amount of ammonia.
STOICHIOMETRY
Solution:
number of particles
n =
NA
Stoichiometry is the study of the quantitative composition of chemical
substances. The stoichiometric ratio for a chemical reaction is the relationship
between the reacting substances that react fully with each other. In this way, it
can be determined how much of each reactant is required to react fully (so that no
unreacted reactants remain) and how much of each of the products are formed.
This has a very important application in industries where substances are
produced on a large scale.
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number of particles
NA
=
1,806  10 24
6,02  10 23
= 3
4.1
Number of moles of atoms = 3 % 4 = 12 mol
Number of atoms = nNA = 12 % 6,02 % 1023
= 7,22 % 1024
1
12
Mass of a substance to/from mole
of the mass of the carbon-12 atom.
mass
molar mass
m
n =
M
Number of moles =
A mass of 12 units is assigned to the carbon-12 atom, and the masses of
atoms of other elements are determined relative to this.
It is only a comparative figure and has no units, e.g.:
m
n
M
â m = nM
A r (Mg) = 24
A r (He) = 4
NOTES
1 mole of oxygen gas (O2) = 6,02 % 1023 molecules
= 32 g
The relative atomic mass (A r ) of an element is a
figure comparing the real mass of one atom of the
element with
4
Therefore: 1 mole of oxygen atoms = 6,02 % 1023 atoms
= 16 g
The mole concept and mass of a substance
Therefore, an Mg atom is about twice as heavy
as a C-12 atom and an He atom’s mass is
about one third of that of a C-12 atom.
Example 1:
Determine the molar mass of A´2(SO4)3.
Solution:
The average relative atomic mass of the different isotopes of an element
is determined according to its percentage abundance in nature. These
values are shown on the Periodic Table as the atomic masses of the
different elements.
M(A´2(SO4)3) = (2 % 27) + 3[32 + 4(16)] = 54 + 288
= 342 g·mol-1
Example 2:
Remember:
Isotopes are atoms of the same element (with the same atomic number),
but with different numbers of neutrons, i.e. different mass numbers.
Calculate the number of moles in:
1) 213 g of chlorine gas
2) 39,5 g KMnO4
m
M
213
=
71
The average 1) relative molecular mass (M r ) of molecules or
2) relative formula mass (M r ) of ionic compounds are
obtained by determining the sum of the relative atomic masses
of the atoms in the compound. These quantities have no units.
1) n =
M(C´2) = 2 % 35,5
= 71 g·mol-1
= 3 mol
m
M
39,5
=
158
2) n =
The relative atomic mass of an element or the relative
molecular/formula mass of a compound in grams indicates one
mole of the specific element or compound. This is called the
molar mass (M) of the substance. The unit for this is g·mol-1.
M(KMnO4) = 39 + 55 + 4(16)
= 158 g·mol-1
= 0,25 mol
4.2
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CHEMICAL CHANGE
Solution:
NOTES
4 MOLAR VOLUME OF GASES; CONCENTRATION
Volume of a gas at STD to/from mole
OF SOLUTIONS
n =
Molar volume of gases
V
n
In many chemical reactions, the reactants are gases, or gases are formed
during the reaction.
â V = nVm
Vm
where: n = number of moles
V = real volume of gas
Vm = molar gas volume = 22,4 dm3
The amount of gas formed can be determined by weighing it, i.e. determining
the mass, or by collecting it in a gas syringe and determining the volume.
Avogadro’s law: Equal volumes of different gases (at the
same temperature and pressure) contain the same number
of moles of gas, i.e. the same number of molecules.
Volume of gases at STP
In the unit on gases, the relationship between p, V, T and the number of moles of
gas (n) was indicated by means of the ideal gas law, i.e.:
Therefore, gases react in the same simple mole or volume ratio with each other
as can be seen in the following balanced equation:
pV = nRT
nRT
âV =
p
3H2(g)
3 moles
3 % 22,4
= 67,2 dm3
â for 1 mole of gas at STP:
1  8,31  273,15
101,3  103
+
N2(g)
t
1 mole
1 % 22,4
= 22,4 dm3
2NH3(g)
2 moles
2 % 22,4
= 44,8 dm3 (at STP)
Therefore:
The volume of a gas is directly proportional to the number of molecules of the
gas at a specific temperature and pressure.
= 0,0224 m3
= 22,4 dm3
In symbols:
One mole of any gas at STP has a volume of 22,4 dm3.
This is known as the molar gas volume (Vm), Vm = 22,4 dm3.
CHEMICAL CHANGE
at STP
Volume of gases at the same temperature and pressure
It is easier to determine the volume and then convert it to number of moles
and mass by using Avogadro’s findings, i.e.:
V =
V
V
=
Vm
22,4
V } n
V = kn
V
= k
n
V1
V
= k = 2
n1
n2
Therefore:
where: V1, V2 = volume of gas1 and gas2
N2(g)
H2(g)
1 mol = 28 g
= 22,4 dm
NH3(g)
1 mol = 2 g
3
= 22,4 dm
n1, n2 = number of moles of gas1 and gas2
1 mol = 17 g
3
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The volume of a gas is influenced by its temperature and
pressure. Therefore, the volumes of gases may only be
compared at the same constant temperature and pressure.
= 22,4 dm3
4.3
Concentration of solutions to/from mole
number of moles
Concentration =
volume
Therefore:
Example:
200 cm3 nitrogen gas is added to an excess of hydrogen gas to form ammonia.
The reaction takes place at a temperature of 500 K and pressure of 800 kPa.
c =
n
Calculate the volume of ammonia produced.
â n = cV
V
c
Solution:
where: c = concentration of the solution in mol·dm-3
n = number of moles of solute
V = volume of solution (dm3)
unit of concentration = mol·dm-3
3H2(g) + N2(g) t 2NH3(g)
V1
â
=
n2
n1
n1 (N2) = 1
n2 (NH3) = 2
V2
2
=
200
1
â V2 = 400 cm3
from balanced
equation
1 dm3 = 1 000 cm3 = 1 000 m ´ = 1 ´
V1 (N2) = 200 cm3
V2 (NH3) = ?
A concentration expressed in moles per dm3 (mol·dm-3)
or moles per litre is called the molarity or molar
concentration (M) of a solution, e.g. a 0,5 M KOH solution
is diluted KOH with a concentration of 0,5 mol·dm-3.
NB: The volume can be in any unit, but it must
be the same on either side of the reaction.
Example:
1) Determine the concentration of an NaOH solution if 5 g NaOH is dissolved in
250 cm3 water.
Shortcut:
2) What mass of copper(II)sulphate must be dissolved in 200 m´ water to yield a
0,4 mol·dm-3 solution?
The volume of NH3 will be twice that of N2,
because it has twice the number of moles.
Solution:
1) n =
Concentration of solutions
c =
Calculation of the concentration of solutions
If a soluble substance is dissolved in a solvent, a homogeneous solution is
obtained, e.g. sugar in water. The more sugar is dissolved in the same amount
of water, the sweeter it tastes and the higher the concentration of the solution.
2)
The concentration of a solution is determined by the amount (number
of moles) of solute in a certain volume of solvent. Water is usually used
as the solvent.
4.4
m
5
=
= 0,125 mol
M
40
n
0,125
=
= 0,5 mol·dm-3
V
0,25
c =
m
MV
0,4 =
m
(159,5)(0,2)
0,4 =
m
31,9
m = 5g
M(NaOH) = 23 + 16 + 1
= 40 g·mol-1
V = 250 cm3
= 0,25 dm3
c = 0,4 mol·dm-3
V = 200 m´
= 0,2 dm3
M(CuSO4) = 63,5 + 32 + 4(16)
= 159,5 g·mol-1
m = 0,4 % 31,9 = 12,76 g
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CHEMICAL CHANGE
V2
m
n
m 
or c =
 substitute n with n = 
V
MV 
M
4
NOTES
This constant ratio between volume and number of moles for gases at the same
temperature and pressure can be used to calculate an unknown volume of gas.
NOTES
4 Preparation of a standard solution
Stoichiometric calculations with concentration and volume
The number of moles of a substance participating in a chemical reaction can
be calculated from n = cV if its concentration and volume are known.
A standard solution is a solution of
which the exact concentration is known.
The number of moles of any other reactant or product can be determined
from the mole ratio of the substances in the balanced equation.
If the volume of this reactant or product is known, but the concentration is
unknown (or vice versa), the unknown value can be calculated from
n = cV. Therefore:
Steps to make up a standard solution
 Calculate the amount of solute that that has to be dissolved in a
specific amount of solvent to give the required concentration.
Example:
Prepare a 250 cm3 NaOH solution with a concentration
of 0,5 mol·dm-3.
calculation:
c = m
MV
m = cMV
= (0,5)(40)(0,25)
= 5g
For substance a:
actual number of moles = ca Va
For substance b:
actual number of moles = cbVb
c aVa
na
â
=
nb
c bVb
M(NaOH) = 23 + 1 + 16
= 40 g·mol-1
ca, cb = concentration of a and b
Va , Vb = volume of a and b during the reaction
na
= mole ratio from balanced equation
nb
 Accurately weigh the correct amount of solute in a watch glass
on a scale.
Substitute all the known values and calculate an unknown volume
or concentration.
 Use a funnel to carefully pour it into a conical flask. Carefully rinse the
last bit of solute from the watch glass through the funnel with a little
distilled water.
 First add a little distilled water (solvent) and gently swirl the flask to
Application: The calculation of an unknown
concentration/volume during a titration reaction
 Carefully add more water, exactly up to the correct volume mark (the
These types of calculations (with concentration and volume) are
generally used during titration reactions of an acid with a base to
calculate the unknown concentration of an acid or base.
dissolve the substance.
bottom of the meniscus of the water must be right on the mark).
 Shake well.
CHEMICAL CHANGE
Steps:
 A standard solution (with known concentration) of the acid or
base is prepared.
 A titration procedure is performed to neutralise the acid with
a base (see p. 4.24).
 The volumes of the acid and base required to neutralise
each other are noted.
 The unknown concentration is calculated from the
above equation.
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4.5
1) M(NH4NO3) = (2 % 14) + (4 % 1) + (3 % 16) = 80 g·mol-1
12 cm of a 0,4 mol·dm KOH solution is used to neutralise
25 cm3 of a dilute sulphuric acid solution. Potassium sulphate and
water form during the reaction. Calculate the concentration of the
sulphuric acid solution.
Solution:
2KOH + H2SO4 t K2SO4 + 2H2O
c bVb
=
na
nb
ca = 0,096 mol·dm-3
j 0,1 mol·dm-3
%O =
3(16)
% 100 = 60%
80
2) Mass of N in 40 g =
%H =
4(1)
% 100 = 5%
80
35
% 40 = 14 g
100
Empirical and molecular formulae
Volumes above and below the
line must have the same unit.
The empirical formula of a compound is calculated from its
percentage composition.
The empirical formula of a compound indicates the simplest whole number
ratio in which the different elements are bonded with each other.
The molecular formula could be the same as the empirical formula, or a
multiple of it, e.g. the molecular formula of acetic acid is CH3COOH
(or C2O2H4) and its empirical formula is COH2.
REVISION OF STOICHIOMETRIC CALCULATIONS
(GRADE 10)
Application: Determine the empirical and
molecular formulae of a compound
Quantitative analysis
Steps:
Empirical formula:
 Take the percentage composition of the compound and express
the percentage of each element as a mass, in grams. The total
mass is 100 g.
A quantitative analysis of a compound indicates what elements the compound
consists of, as well as the percentage/amount of each.
Percentage composition
 Determine the number of moles of each element: n = m
The percentage of each element in the compound (percentage composition of the
compound) can be calculated by expressing the molar mass of each element as a
percentage of the molar mass of the compound.
% element =
2(14)
% 100 = 35%
80
Take a = acid (H2SO4)
b = base (KOH)
c  25
1
=
â a
2
0,4  12
50 % ca = 4,8
%N =
NOTES
-3
M
 Write down the mole ratio of the elements.
 Calculate the whole number mole ratio (divide by the smallest value
throughout) and use these values to write down the empirical formula.
molar mass of element
100
×
molar mass of compound
1
Molecular formula:
 Calculate the molar mass of the empirical compound.
Example:
Divide this into the given molar mass of the compound to find the
whole number ratio between the masses.
 Multiply the number of atoms of each element in the empirical
formula by this value.
1) Calculate the % composition of NH4NO3.
2) Calculate the mass of nitrogen in 40 g NH4NO3.
4.6
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CHEMICAL CHANGE
3
c a Va
4
Solution:
Example:
NOTES
4 Example:
Substance A
A certain compound of nitrogen and oxygen with a general formula Nx O y,
contains 36,8% nitrogen. Determine the empirical formula of the compound.
mass
Solution:
% N = 36,8% â % O = 63,2%
âN
:
O
Step 1:
Step 2:
Step 3:
Step 4:
=
36,8 g
:
63,2 g
=
=
=
2,63
1
2
:
:
:
3,95
1,5
3
36,8
63,2
mol :
mol
=
14
16
â Empirical formula N2O3
n =
m
Number of moles (n) =
M
number of particles
n =
m
M
M(N) = 14 g·mol-1
M(O) = 16 g·mol-1
number of particles
NA
volume
(gases at STP)
n =
concentration
(solutions)
n = cV
V
3
Vm / 22,4 dm
Mol (n) substance A
conversion using mole
ratio in equation
Simplified mole ratio:
Divide both sides by 2,63
(smallest value).
Mol (n) substance B
Stoichiometric calculations with mass, number of particles,
volume (of gases) and concentration (of solutions)
mass
number of particles
volume
(gases at STP)
concentration
(solutions)
Application: Stoichiometric calculations
Substance B
Calculate the quantity of a certain substance (B) that is needed
or formed during a chemical reaction if the quantity of another
substance (A) participating in the reaction is known.
NB: For gases at other temperatures and pressures, the ideal gas
pV 
equation is used  n =
 to calculate the number of moles or volume.
RT 

CHEMICAL CHANGE
Steps:
 Write a balanced equation for the reaction.
 Underline the substances (A and B). It is sometimes necessary
to determine their molar mass. Substances in excess may usually
be ignored.
 Convert the given quantity of substance A (mass/number of
particles, etc.) to mole.
 Determine the number of moles of substance B required/formed
using the mole ratio of the two substances in the equation.
Convert the calculated number of moles (of B) to the mass/number
of particles or volume (gases), depending on the question.
Example:
Nitrogen dioxide is prepared in the laboratory by reacting copper with
concentrated nitric acid. The reaction equation is:
Cu(g) + 4HNO3(´) t Cu(NO3)2(aq) + 2H2O(´) + 2NO2(g)
1) Calculate the mass of HNO3 required to react fully with 12,7 g Cu.
2) Calculate the volume of nitrogen dioxide gas formed at STP if 12,7 g Cu and
sufficient HNO3 react with each other.
NB: The mole, as a unit to indicate amount of substance, is central to
stoichiometric calculations in chemistry. Other amounts of
substance, whether mass, number of particles, volume (of
gases) or concentration of solutions, can be converted to or from
the mole. See the accompanying schematic representation.
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3) Calculate the volume of nitrogen dioxide gas that forms if 150 cm3 HNO3
with a concentration of 0,5 mol·dm-3 reacts with an excess of copper:
3.1) at STP
4.7
3.2) at a pressure of 200 kPa and a temperature of 25ºC.
1) Step 1 & 2: Cu + 4HNO3 t Cu(NO3)2 + 2H2O + 2NO2
Step 4:
Calculate the number of moles in 12,7 g Cu:
m
n =
M
M(Cu) = 63,5 g·mol-1
12,7
m = 12,7 g
= 0,2 mol
=
63,5
n = cV
= 0,075 mol
HNO3 : NO2
=
4 : 2
=
2 : 1
= 0,075 : 0,0375
Step 5:
Convert number of moles of HNO3 to mass:
m = n % M = 0,8 % 63
M(HNO3) = 1 + 14 + 3(16)
= 50,4 g
= 63 g·mol-1

moles (Cu)

moles (HNO3)

Therefore:
mass (HNO3)
c(HNO3)
use mole ratio between
Cu and HNO3 in equation
n(Cu) = 0,2 mol (already calculated)
Step 4:
Calculate the number of moles of NO2 formed according
to mole ratio:
V =
=
Cu : NO2
1 : 2
(from equation)
= 0,2 : 0,4
moles (NO2)

volume (NO2)
nRT
p
0,0375  8,31  298
2  10 5
p = 200 kPa = 2 % 105 Pa
T = 25 + 273 = 298 K
MORE COMPLEX STOICHIOMETRIC CALCULATIONS
V
Vm
Limiting reagent
â V = n % Vm = 0,4 % 22,4 = 8,96 dm3
Therefore:
When the reactants in a chemical reaction are not
added in the correct stoichiometric ratios, one reactant
will be used up first and some of the other reactants will
remain after the reaction has run its course.
V = n Vm
moles (Cu)

= 4,64 % 10-4 m3
Convert n(NO2) to volume:
mass (Cu)
moles (HNO3)
pV = nRT
Step 3:


3.2) The number of moles of NO2 has already been calculated in Question 3.1,
i.e. n(NO2) = 0,0375
2) Steps 1 & 2: Cu + 4HNO3 t Cu(NO3)2 + 2H2O + 2NO2
n =
Convert n(NO2) to volume(NO2)
V = nVm
= 0,0375 % 22,4
= 0,84 dm3 NO2 gas
m = nM
mass (Cu)
Step 5:
Calculate n(NO2) formed from mole ratio
Step 4:
Calculate the number of moles of HNO3 required according to
the mole ratio in the equation:
Therefore:
c(HNO3) = 0,5 mol·dm-3
V(HNO3) = 150 cm3
= 0,15 dm3
= 0,5 % 0,15
Cu : HNO3
= 1 : 4
= 0,2 : 0,8
Step 5:
Calculate number of moles of HNO3

moles (NO2)
4

volume (NO2)
4.8
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CHEMICAL CHANGE
Step 3:
Step 3:
Cu + 4HNO3 t Cu(NO3)2 + 2H2O + 2NO2
NOTES
3.1) Steps 1 & 2:
Solution:
NOTES
reactant/reagent used up first is called the limiting reagent. This reactant
4 The
determines when the reaction stops and how much product is formed.
Example:
During the combustion reaction of Mg in oxygen, 42 g of magnesium ribbon is
burnt in 24 g O2 gas. Determine whether Mg or O2 is the limiting reagent.
Consider the reaction that takes place when a piece of magnesium ribbon
burns in oxygen in the air:
2Mg + O2 t 2MgO
Solution:
2 moles of Mg will react fully with 1 mole of O2 gas, i.e. 48 g of Mg requires
32g of O2 gas for a complete reaction. There is an excess of oxygen present
in the air; therefore the reaction will stop as soon as the magnesium ribbon
has burnt up. The magnesium, thus, is the limiting reagent.
Step 1:
2Mg + O2 t 2MgO
Step 2:
n(Mg) =
m
M
42
=
24
= 1,75 mol
= 0,75 mol
Mg (actual) = 1,75 mol
O2 (actual) = 0,75 mol
Application: Identify the limiting reagent
Steps:
 Write down the balanced equation for the reaction.
Steps 3 & 4: O2 required:
 Convert the given mass, number of particles, volume (for gases) or
Mg : O2
concentration (in solutions) of reactants A and B to the number of
moles (nA actual and nB actual).
=
:
1
reactants to theoretically determine how much of the other is required
according to the ratio a : b, e.g.:
Step 5:
1,75
 1
2
O2 (required) > O2 (actual)
0,875 > 0,75
Take the actual number of moles of reactant A (nA actual) and
determine the number of moles of B required, (nB required).
â O2 is the limiting reagent
OR
If nB (required) < nB (actual), then B is in excess and A is the
limiting reagent.
nMg actual
If nB (required) > nB (actual), then B is the limiting reagent.
OR
nO 2 actual
=
1,75
0,75
Mg
2
=
O2
1
= 2,33
= 2
but 2,33 > 2
n actual
a
to :
Compare A
b
nB actual
â O2 is the limiting reagent
n actual
a
If A
<
t A limiting reagent
b
nB actual
n A actual
a
>
t B limiting reagent
b
nB actual
Use the limiting reagent to determine the amount of products or
unreacted reactants.
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(from equation)
= 1,75 : 0,875
 Use the given value of the actual number of moles of one of the
CHEMICAL CHANGE
2
= 1,75 :
 Write down the mole ratio of A : B from the chemical equation, e.g. a : b.
If
m
M
24
=
32
n(O2) =
4.9
m(Mg) = 42 g
m(O2) = 24 g
4
Collect the precipitate (PbO2):
- place a test tube in a test tube rack
- filter the warm mixture through a funnel lined with filter paper;
collect the PbO2 precipitate in the filter paper
Practical Investigation 1: Determine the mass of PbO2
prepared from Pb(NO3)2
- wash the precipitate by running water and dilute nitric acid
(3 % 5 m´ - measured with pipette) over it
NaOC´ (Jik)
- place the filter paper on a watch glass and let it dry overnight
rod
- accurately weigh the PbO2 on a scale
- carefully add it to a 150 m´ conical flask
- measure 50 m´ water in a measuring
cylinder and add to the flask
- stir the contents in the flask with a
stirring rod until dissolved
- calculate the theoretical yield the reaction should produce from the
balanced equation
NaOH + Pb(NO3)2
gauze wire
- calculate the percentage yield
heating stand
Results and conclusion:
The reaction takes place in two steps:
Bunsen burner
Prepare Pb(NO3)2 solution:
 lead nitrate reacts with the sodium hydroxide and forms lead(II)
- accurately weigh 4 g Pb(NO3)2 in a clean watch glass on a scale
hydroxide and sodium nitrate
- carefully add it to a 250 m´ flask
Pb(NO3)2(aq) + 2NaOH(aq) t Pb(OH)2(aq) + 2NaNO3(aq)
- measure 25 m´ water in a measuring cylinder and add to the flask
 lead(II)hydroxide reacts with sodium hypochlorite and forms
lead(IV)oxide (brownish black precipitate), sodium chloride and water
- dissolve the contents of the flask in the water
Pb(OH)2(aq) + NaOC´(aq) t PbO2(s) + NaC´(aq) + H2O
To make up a solution with a specific concentration:
the substance to be dissolved is weighed, added to a
volumetric flask and then filled with water to the volume
required. (The water is not measured separately, see p. 4.5.)
The weighed mass of the lead oxide precipitate is less than that
calculated theoretically. This could be because the sample of lead
nitrate was not pure or because some of the reactants remained on
the filter paper.
Theoretical yield:
Prepare PbO2 solution:
m
M
10
=
40
n(NaOH) =
- carefully add the 50 m´ NaOH solution to the Pb(NO3)2 in the
250 m´ flask
- place the flask on gauze wire on a heating stand over
a Bunsen burner
= 0,25 mol
- set the flame to a low heat and boil the solution for 5-10 minutes while
stirring continuously
m
M
4
=
331
n(Pb(NO3)2) =
- add 30 m´ of a 3,5% NaOC´ solution (household bleach) to the
mixture, while heating it further; stir continuously
m(NaOH) = 10 g
M(NaOH) = 23 + 16 + 1
= 40 g·mol-1
V = 50 m´ = 0,05 dm3
m(Pb(NO3)2) = 4 g
M(Pb(NO3)2) = 207 + 2(14) + 6(16)
= 331 g·mol-1
V = 25 m´ = 0,025 dm3
= 0,012 mol
- a brownish black precipitate should form after 5 minutes
4.10
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CHEMICAL CHANGE
Apparatus and method:
Prepare NaOH solution:
- accurately weigh 10 g NaOH in a
watch glass on a scale
NOTES
Percentage yield
4
Limiting reagent:
NOTES
Pb(NO3)2 : NaOH
Apparatus and method:
Crush a seashell into small pieces using a pestle and mortar.
=
=
Place the crushed seashell on a piece of filter paper and find the
mass of the shell (subtract the mass of the filter paper).
1
:
2
0,012 : 0,024
â Pb(NO3)2 is the limiting reagent, NaOH is in excess
Place the shell in a 100 m´ beaker and cover it with sufficient
hydrochloric acid with a concentration of 3 mol·dm-3. Gas bubbles
are observed due to the release of CO2.
Pb(NO3)2 : Pb(OH)2
=
1
= 0,012
:
:
1
0,012
Wait until the bubbles stop and the reaction is complete.
Filter the mixture to remove the unreacted residue; use a scale and
determine the mass of the residue/impurities.
Assume that an excess of NaOC´ (bleach) is added:
Pb(OH2) :
=
1
= 0,012
:
:
PbO2
Allow the filtrate to dry out and determine the mass of the dry filtrate.
M(PbO2) = 207 + 2(16)
= 239 g·mol-1
1
0,012
CO2 gas
â 0,012 mol PbO2 precipitate should form
hydrochloric acid (HC´)
m(PbO2) = n % M = 0,012 % 239 = 2,87 g
The percentage yield =
crushed seashell
actual/weighed yield
100

theoretical yield
1
Results and conclusion:
The balanced equation for the reaction is as follows:
CaCO3(s) + 2HC´ (aq) t CaC´ 2(s) + CO2(g) + H2O(´)
Practical Investigation 2: Determine the percentage
The mass of the dry filtrate (the CaC´ 2 salt formed), can be used
to determine the mass of CaCO3 in the shell, by means of a
stoichiometric calculation, i.e.:
CaCO3 in a seashell
Calcium carbonate (CaCO3) is commonly found in nature in rock
strata as well as in shells, skeletons and exoskeletons of various
marine or terrestrial organisms. It hardens these structures.
- convert mass (CaC´ 2) to moles
CHEMICAL CHANGE
- calculate the number of moles of CaCO3 according to mole ratio
in equation:
CaC´ 2 : CaCO3
= 1
:
1
- convert the number of moles of CaCO3 to mass
OR
- mass of CaCO3 = initial mass of shell - mass of residue
Calculate the percentage CaCO3 as follows:
Calcium carbonate readily reacts with a strong acid to form
a salt, carbon dioxide and water.
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% CaCO3 =
4.11
mass CaCO3
mass shell
%
100
1