Download 2. Solution Guide to Supplementary Exercises

Topic
3
Metals
Part A Unit-based exercise
Unit 10 Occurrence and extraction of metals
Fill in the blanks
1
tungsten
2
aluminium; iron
3
a) oxide; lead(II) oxide
b) oxide
lead(II) oxide; lead
4
a) coke
b) air
c) Carbon monoxide; iron
carbon monoxide; iron
5
bauxite; aluminium oxide; electrolysis
True or false
6
F
Electric wire is made of pure copper, but NOT of bronze.
Bronze is an alloy of copper containing tin. The addition of tin decreases the electrical conductivity of
copper.
7
T Gold was mainly used for ornaments as it was too soft for making tools and weapons.
8
T Nickel-cadmium rechargeable cells are used to power professional cameras.
9
F
Iron is the second most abundant metal in the Earth’s crust. Aluminium is the most abundant metal in
the Earth’s crust.
10 T The main metallic compound in cinnabar is mercury(II) sulphide.
11 F
Galena consists mainly of lead(II) sulphide.
12 T
13 F
144
Iron is extracted from its ore by reduction with carbon.
14 F
The temperature of Bunsen flame is about 600 °C. Carbon can remove oxygen from zinc oxide at 1 000 °C
or above.
15 T
16 T Copper was known since 6000 B.C.
17 T
18 T The addition of tin to copper produces a much harder alloy called bronze.
19 F
Aluminium was used later than iron as it is difficult to extract.
20 F
Brass is an alloy of copper containing zinc. It is harder than pure copper.
Multiple choice questions
21 C
22 D Option C — Iron can be recycled.
23 C
24 D Option A — Titanium is used to make tooth implants because it is light but very strong, resists corrosion,
can be easily shaped and biocompatible.
Options B and C — Titanium is found in many consumer products, such as jewellery, watch cases,
spectacles, bicycles and clocks.
25 C Option A — Duralumin is an alloy of aluminium containing copper, magnesium and manganese.
Option B — A fuse is a safety device that protects electric circuits from the effects of excessive electric
currents. It commonly consists of a current-conducting wire of low melting point. If we try
to pass a current higher than the rated value through the wire, it will heat up so much that
it melts. When it melts, it breaks the circuit it is fitted to and stops the current flowing.
Copper is NOT used to make fuses nowadays.
Option C — Lithium ion cells are commonly used in mobile phones and other portable electronic
devices.
Option D — Zinc CANNOT be used to make cans for canned food as zinc ions are poisonous.
26 B Option B — Magnesium is used in pyrotechnic and incendiary devices.
It is lighter than aluminium, and its alloys are used for aircraft, car engine casings and missile
construction.
145
27 A Option B — Silver is NOT very strong.
–3
Option C — The density of silver is 10.5 g cm .
Option D — The abundance of silver in the Earth’s crust is very low, 0.075 ppm (parts per million).
28 D Pure gold is quite soft.
29 D
Option
Ore
Main metallic compound in the ore
A
calcite
calcium carbonate
B
cinnabar
mercury(II) sulphide
C
gypsum
calcium sulphate
D
haematite
iron(III) oxide
30 B Option B — The main metallic compound in galena is lead(II) sulphide.
31 A
32 A Mercury can be extracted from cinnabar (containing mercury(II) sulphide) by heating the ore in air.
heat in air
mercury(II) sulphide + oxygen
33 B
mercury(II) sulphide + oxygen
mercury + sulphur dioxide
heat in air
mercury + sulphur dioxide
34 B Options A, C and D — Positions of calcium, magnesium and potassium in the reactivity series are high.
Their oxides are stable and difficult to reduce. Hence they CANNOT be reduced
by hydrogen.
Option B — The position of copper in the reactivity series is low. Its oxide is easy to reduce.
35 A
36 D During the extraction, oxygen gas bubbles off from the positive electrodes. Reaction occurs between the
graphite electrodes and oxygen.
37 D Option D — Silver is extracted by displacement from solution or mechanical separation.
38 C
39 B
146
Option
Metal
Year of discovery
A
gold
about 8000 B.C.
B
magnesium
1808
C
mercury
about 8000 B.C.
D
silver
about 8000 B.C.
40 B Metal resources are limited.
41 D
42 A (3) Zinc exists as compounds in its ores. For example, the main metallic compound in zinc blende is zinc
sulphide.
43 B (1) Aluminium is the most abundant metal in the Earth’s crust. Oxygen is the most abundant element in
the Earth’s crust.
(3) Stainless steel is an alloy of iron, chromium and nickel.
44 D (3) Copper pipes are more expensive than iron pipes. Copper pipes rather than iron pipes are used for
carrying hot water because iron will corrode slowly in hot water but copper will not.
45 B (2) Iron is the second most abundant metal in the Earth’s crust.
46 A (1) Solder is an alloy of lead and tin.
(2) The main metallic compound in galena is lead(II) sulphide.
(3) Lead is extracted from galena (containing lead(II) sulphide) in two steps:
roast in air
sulphide
oxide, then
heat with carbon
oxide
metal
47 A (3) Titanium is light, but its density (4.5 g cm–3) is NOT lower than that of aluminium (2.7 g cm–3).
48 A (3) Oxygen gas bubbles off from the positive electrodes.
49 C
Option
Ore
Main metallic compound in the ore
(1)
Bauxite
hydrated aluminium oxide
(2)
Cinnabar
mercury(II) sulphide
(3)
Copper pyrite
copper(II) iron(II) sulphide
50 A
Option
Metal
Common ore
Main metallic
compound in the ore
(1)
Copper
copper pyrite
copper(II) iron(II) sulphide
Extraction method
controlled
sulphide
metal
heat in air
(2)
Mercury
cinnabar
mercury(II) sulphide
heat in air
sulphide
metal
roast in air
(3)
Zinc
zinc blende
zinc sulphide
sulphide
oxide, then
heat with carbon
oxide
metal
147
51 A
Option
Metal
Common ore
Main metallic
compound in the ore
(1)
Iron
haematite
iron(III) oxide
Extraction method
heat with carbon
oxide
metal
roast in air
(2)
Lead
galena
lead(II) sulphide
sulphide
oxide, then
heat with carbon
oxide
(3)
Magnesium
magnesite
magnesium carbonate
metal
electrolysis of molten ore
52 B (2) During the extraction, a mixture of iron ore, coke and limestone is added to the furnace.
53 B (1) Heating mercury(II) sulphide in air gives mercury and sulphur dioxide.
mercury(II) sulphide + oxygen
heat
mercury + sulphur dioxide
(3) Heating lead(II) oxide with carbon gives lead and carbon dioxide.
Carbon removes the oxygen from lead(II) oxide.
lead(II) oxide + carbon
heat
lead + carbon dioxide
54 D
55 D
56 A
57 C Electric wires are made of pure copper, but NOT of brass.
Brass is an alloy of copper and zinc. The presence of zinc atoms in the copper hinders the movement of
mobile electrons. Hence the electrical conductivity of brass is lower than that of copper.
58 C Brass is harder and more corrosion resistant than pure copper. It has a golden appearance. Hence brass
is commonly used to make ornaments.
59 C Historically, the sequence of discovery of various metals relates closely with the ease of extracting the
metals from their ores.
Aluminium is difficult to extract. It was discovered after the invention of electrolysis in 1800. Therefore
aluminium was used later than iron although it is more abundant than iron in the Earth’s crust.
60 D 24-carat gold is pure gold.
148
Unit 11 Reactivity of metals
Fill in the blanks
1
lilac; white
potassium oxide
2
brick-red; white
calcium oxide
3
very bright light; white
magnesium oxide
4
yellow; white
zinc oxide
5
black
iron(II, III) oxide
6
golden yellow
sodium hydroxide; hydrogen
7
magnesium oxide; hydrogen
8
salts; hydrogen
iron(II) sulphate; hydrogen
iron(II) chloride; hydrogen
9
a) electrolysis
b) carbon
c) heating
10 zinc sulphate; copper
True or false
11 T
12 F
Silver shows NO observable change when heated in air.
13 F
Iron burns with sparks.
14 T Aluminium gives aluminium oxide when burnt in air.
aluminium + oxygen
15 F
aluminium oxide
When lead is heated in air, a powder (orange when hot but yellow when cold) forms on the surface.
lead + oxygen
lead(II) oxide
149
16 F
There is NO need to keep magnesium in paraffin oil.
17 T
18 F
The density of calcium is higher than that of water. Hence it sinks in water.
19 T Sodium reacts with water to give sodium hydroxide solution (an alkaline solution) and hydrogen.
sodium + water
20 F
sodium hydroxide + hydrogen
Magnesium shows little reaction with cold water.
21 T
22 F
Iron(II) sulphate is formed when iron reacts with dilute sulphuric acid.
iron + dilute sulphuric acid
iron(II) sulphate + hydrogen
23 T Zinc is more reactive than copper. Zinc atoms lose outermost shell electrons to form cations more easily
than copper atoms.
24 F
Metals at the top of the reactivity series (potassium, sodium, magnesium and aluminium) are extracted
by electrolysis.
25 T When a mixture of iron(III) oxide powder and aluminium powder is ignited, a very vigorous reaction
occurs.
iron(III) oxide + aluminium
iron + aluminium oxide
Aluminium is more reactive than iron. It can remove oxygen from iron(III) oxide.
Multiple choice questions
26 B Magnesium burns with a very bright light. A white powder forms.
magnesium + oxygen
magnesium oxide
27 A
28 A sodium + water
sodium hydroxide + hydrogen
29 B Option A — Aluminium shows little reaction with cold water.
Option B — The density of calcium is higher than that of water. Hence it sinks in water.
Calcium reacts with water to give calcium hydroxide and hydrogen.
calcium + water
calcium hydroxide + hydrogen
Option C — Lead shows NO reaction with cold water.
Option D — The density of lithium is lower than that of water. Hence it floats on water.
150
30 C
Option
Metal
Reaction with water
Extracted from its oxide
by carbon reduction?
A
iron
little reaction with water
yes
B
magnesium
little reaction with water
no
C
potassium
D
zinc
potassium + water
potassium hydroxide + hydrogen
little reaction with water
no
yes
∴ X could be potassium.
31 B Option A — The reactivity of Group II elements increases down the group.
CFSZMMJVN
NBHOFTJVN
DBMDJVN
SFBDUJWJUZ
JODSFBTJOHEPXO
UIFHSPVQ
TUSPOUJVN
CBSJVN
Hence barium is more reactive than calcium. Calcium reacts readily with water. Barium
probably reacts vigorously with water.
Option B — Calcium reacts with water to give calcium hydroxide and hydrogen.
calcium + water
calcium hydroxide + hydrogen
Hence barium reacts with water to give barium hydroxide and hydrogen as well.
Option C — The densities of Group II elements are higher than that of water. Hence barium sinks in
water.
Option D — The melting point of Group II elements decreases down the group (except magnesium).
Element
Melting point (°C)
Beryllium
1 280
Magnesium
650
Calcium
838
Strontium
769
Barium
725
The melting point of barium is lower than that of calcium.
32 C aluminium + steam
aluminium oxide + hydrogen
151
33 D Options A, B and C — Copper, gold and lead do NOT react with steam.
Option D — Magnesium reacts readily with steam.
magnesium oxide + hydrogen
magnesium + steam
34 A Option A — A copper container is suitable for holding hot water as copper does not corrode in hot
water.
Option B — Lead is poisonous and should not be used to make a hot water container.
Option C — Magnesium is reactive and tarnishes when exposed to air.
Option D — Silver is too expensive.
35 A zinc + dilute sulphuric acid
36 D
zinc sulphate + hydrogen
Option
Pair of chemicals
Reaction occurs
when mixed?
Word equation
A
magnesium and
dilute sulphuric acid
yes
magnesium + dilute sulphuric acid
magnesium sulphate + hydrogen
B
calcium and water
yes
calcium + water
C
iron and steam
yes
iron + steam
D
silver and dilute
hydrochloric acid
no
(silver is very
unreactive)
calcium hydroxide + hydrogen
iron(II,III) oxide + hydrogen
—
37 C Iron reacts with dilute hydrochloric acid to give iron(II) chloride and hydrogen. Heat is also released.
iron + dilute hydrochloric acid
iron(II) chloride + hydrogen
38 D
39 D Only Y has no observable change when heated in air. Therefore Y is the least reactive.
X reacts with water while Z does not. Therefore X is more reactive than Z.
40 C Option A — Both iron and lead exist as compounds in the Earth’s crust. Hence X probably exists as
compounds in the Earth’s crust.
Option B — Both iron and lead form oxide when heated in air. Hence X probably forms an oxide when
heated in air.
Option C — Both iron and lead can be obtained by reduction of their oxides with carbon. Hence X
probably can be obtained from its oxide in the same way.
Option D — Both iron and lead show little / no reaction with cold water. Hence X probably shows NO
reaction with cold water as well.
152
41 D Follow the steps below when writing a balanced chemical equation for the reaction of aluminium with
dilute hydrochloric acid.
Write down the chemical formulae of the reactants and
products.
Al + HCl
To balance the chlorine atom, put the coefficient ‘3’ before
HCl.
Al + 3HCl
AlCl3 + H2
3
To balance the hydrogen atom, put the coefficient ‘ ’
2
before H2.
Al + 3HCl
AlCl3 +
To get a whole number coefficient for each substance,
multiply all coefficients by 2. Now the chemical equation
is balanced.
2Al + 6HCl
AlCl3 + H2
3
H2
2
2AlCl3 + 3H2 (Balanced)
42 C Follow the steps below when writing a balanced chemical equation for the reaction of magnesium with
ammonia.
Write down the chemical formulae of the reactants and
products.
Mg + NH3
To balance the magnesium atom, put the coefficient ‘3’
before Mg.
3Mg + NH3
To balance the nitrogen atom, put the coefficient ‘2’ before
NH3.
3Mg + 2NH3
Mg3N2 + H2
To balanced the hydrogen atom, put the coefficient ‘3’ before
H2. Now the chemical equation is balanced.
3Mg + 2NH3
Mg3N2 + 3H2 (Balanced)
Mg3N2 + H2
Mg3N2 + H2
43 C Follow the steps below when writing a balanced chemical equation for the reaction of chlorine with hot
concentrated potassium hydroxide solution.
Write down the chemical formulae of the reactants and
products.
Cl2 + KOH
To balance the chlorine atom, put the coefficient ‘5’ before
KCl.
3Cl2 + KOH
To balance the potassium atom, put the coefficient ‘6’
before KOH.
3Cl2 + 6KOH
5KCl + KClO3 + H2O
3Cl2 + 6KOH
5KCl + KClO3 + 3H2O
(Balanced)
To balanced the hydrogen atom, put the coefficient ‘3’ before
H2O. Now the chemical equation is balanced.
KCl + KClO3 + H2O
5KCl + KClO3 + H2O
44 B Follow the steps below when writing a balanced chemical equation for the reaction between zinc and
dilute nitric acid.
Write down the chemical formulae of the
reactants and products.
Zn + HNO3
To balance the nitrogen atom, put the
coefficient ‘6 + y’ before HNO3.
3Zn + (6 + y)HNO3
3Zn(NO3)2 + yNO + H2O
To balance the hydrogen atom, put the
6 + y
coefficient ‘
’ before H2O.
2
3Zn + (6 + y)HNO3
3Zn(NO3)2 + yNO +
Zn(NO3)2 + NO + H2O
( 6 2+ y )H O
2
153
Number of oxygen atoms on the reactant side = 3 x (6 + y)
= number of oxygen atoms on the product side
= 3 x 3 x 2 + y + 6 + y
2
6
+
y
∴ 3 x (6 + y) = 18 + y +
2
36
+
2y
+
6
+ y
18 + 3y =
2
(
(
)
)
y = 2
45 C Follow the steps below when writing a balanced chemical equation for the action of dilute sulphuric acid
on iron(III) hydroxide.
Write down the chemical formulae of the
reactants and products.
Fe(OH)3 + H2SO4
To balance the iron atom, put the coefficient
‘2’ before Fe.
2Fe(OH)3 + H2SO4
To balance the polyatomic ion SO42–, put the
coefficient ‘3’ before H2SO4.
2Fe(OH)3 + 3H2SO4
Fe2(SO4)3 + H2O
To balance the hydrogen atom, put the
coefficient ‘6’ before H2O. Now the chemical
equation is balanced.
2Fe(OH)3 + 3H2SO4
Fe2(SO4)3 + 6H2O (Balanced)
Fe2(SO4)3 + H2O
Fe2(SO4)3 + H2O
∴z = 6
46 D CaCO3 is a solid. It reacts with rainwater containing dissolved carbon dioxide to give calcium hydrogencarbonate
solution.
CaCO3(s) + CO2(g) + H2O(l)
Ca(HCO3)2(aq)
47 A ZnO is a solid. It reacts with dilute sulphuric acid to give zinc sulphate solution and water.
ZnO(s) + H2SO4(aq)
ZnSO4(aq) + H2O(l)
48 A Pb3O4 is a solid. It is reduced by hydrogen gas to give lead (a solid) and water.
Pb3O4(s) + 4H2(g)
3Pb(s) + 4H2O(l)
49 A Option A — Magnesium displaces copper from copper(II) sulphate solution.
50 D
Tube
Chemicals involved
Observation
I
copper + silver nitrate solution
grey coating on copper
II
copper + iron(II) nitrate solution
no observable change
III
copper + zinc nitrate solution
no observable change
IV
copper + magnesium nitrate solution
no observable change
∴ option D is correct.
154
51 C R displaces all other metals from solutions of their nitrates. Therefore R is the most reactive.
Q displaces P from a solution of the nitrate of P. Therefore Q is more reactive than P.
∴ the descending order of reactivity of the metals is R > Q > P.
52 B A and C are more reactive than B as both of them can displace copper from copper(II) nitrate solution
but B cannot.
C is more reactive than A as it gives a colourless gas (hydrogen) with the water in the copper(II) nitrate
solution.
∴ the ascending order of reactivity of the metals is B < A < C.
53 B Option A — There is no relationship between the charge on the ion of a metal and the reactivity of the
metal.
Option B — A metal will always displace a less reactive metal from a solution of the compound of the
less reactive metal, but not vice versa.
X can displace Y from an aqueous solution of a salt of Y. This shows that X is more reactive
than Y.
Option C — The oxide of a less reactive metal can be reduced by heating it with a more reactive
metal.
When the oxide of X is heated with Y, X is formed. This shows that Y is more reactive than
X.
Option D — The oxide of a metal lower in the reactivity series has a lower stability. Thus the reduction
of the oxide of the metal is easier.
The oxide of X undergoes decomposition upon strong heating but the oxide of Y does not.
This shows that Y is more reactive than X.
54 B In the reaction between copper and silver nitrate solution, copper displaces silver from silver nitrate solution
to form copper(II) nitate solution and silver deposits.
Cu(s) + 2AgNO3(aq)
Cu(NO3)2(aq) + 2Ag(s)
In the reaction, each copper atom loses two electrons to form a copper(II) ion:
Cu(s)
Cu2+(aq) + 2e–
Each silver ion gains one electron to form a silver atom:
Ag+(aq) + e–
Ag(s)
Nitrate ions do not take part in the reaction. We can delete them from the equation. The equation for
the reaction thus becomes:
Cu(s) + 2Ag+(aq)
Cu2+(aq) + 2Ag(s)
55 C Only Y is found as a free element in the Earth’s crust. Therefore Y is the least reactive.
The oxide of a metal lower in the reactivity series has a lower stability. Thus the reduction of the oxide
of the metal is easier.
Carbon can remove oxygen from the oxide of X but not from that of Z. This shows that Z is more reactive
than X.
∴ the reactivity series of these metals in decreasing reactivity is Z > X > Y.
155
56 D Metals at the top of the reactivity series are extracted by electrolysis. Metals in the middle are extracted by
reduction of their oxides with carbon. Metals at the bottom are extracted by physical methods. Therefore
Z is the most reactive.
57 D Option A — Both zinc and iron can displace copper from copper(II) sulphate solution. Hence chromium
can also displace copper from copper(II) sulphate solution.
Option B — Both zinc and iron react with dilute hydrochloric acid to liberate hydrogen. Hence chromium
also reacts with dilute hydrochloric acid to liberate hydrogen.
Option C — Magnesium is more reactive than zinc and iron. Hence magnesium is more reactive than
chromium.
Option D — Both zinc and iron are usually extracted by reduction with carbon. Hence chromium should
also be obtained in the same way.
58 A Y is the most reactive as it gives a gas (hydrogen) with the water in the zinc sulphate solution.
The oxide of a metal lower in the reactivity series has a lower stability. Thus the reduction of the oxide
of the metal is easier.
The oxide of X can be reduced by heating while the oxide of Z cannot. This shows that X is less reactive
than Z.
∴ the order of increasing reactivity of the three metals is X < Z < Y.
59 B (1) Y is a reactive metal, e.g. potassium, sodium or calcium.
It is more reactive than zinc. When the oxide of Y is heated with zinc, zinc CANNOT remove the
oxygen from the oxide of Y.
(2) Metals at the top of the reactivity series (e.g. potassium, sodium) are extracted by electrolysis.
(3) Y was probably discovered after the invention of electrolysis in 1800.
60 C Option A — Silver is less reactive than copper. There is NO reaction between silver and copper(II) sulphate
solution.
Option B — NO reaction occurs when copper(II) oxide and carbon are mixed without heating.
Option C — Hydrogen can reduce hot copper(II) oxide to copper.
copper(II) oxide + hydrogen
copper + water
Option D — Solid copper(II) sulphide does NOT conduct electricity.
61 A
Option
Metal
Oxide formed when heated in air
Colour of oxide
(1)
Aluminium
aluminium oxide
white
(2)
Iron
iron(II, III) oxide
black
(3)
Mercury
mercury(II) oxide
red
∴ only aluminium gives a white powder when heated in air.
156
62 B (1) Lithium burns in air to form lithium oxide.
lithium oxide
lithium + oxygen
(2) The reactivity of Group I elements increases down the group.
MJUIJVN
SFBDUJWJUZ
JODSFBTJOH
TPEJVN
EPXO
UIF
HSPVQ
QPUBTTJVN
Lithium is the least reactive Group I element. Hence lithium atoms lose electrons LEAST readily.
(3) Lithium ion rechargeable dry cells are commonly used in mobile phones and other portable electronic
devices.
63 A (1) When copper is heated strongly, a black powder (copper(II) oxide) forms on the surface.
copper + oxygen
copper(II) oxide
Copper shows NO reaction with dilute sulphuric acid.
(2) Iron forms a black powder when it is burnt in air. However, it reacts with dilute sulphuric acid.
(3) Lead forms a yellow powder (orange when hot) when it is burnt in air.
64 B Calcium reacts with water readily. A steady stream of bubbles forms.
calcium + water
calcium hydroxide + hydrogen
(1) NO explosion occurs in the reaction.
(3) Calcium sinks in water.
65 D (1) The reactivity of alkali metals increases down the group.
MJUIJVN
SFBDUJWJUZ
TPEJVN
QPUBTTJVN
SVCJEJVN
JODSFBTJOH
EPXO
UIF
HSPVQ
DBFTJVN
Caesium is more reactive than potassium. Hence it loses electrons more readily than potassium
does.
157
(2) Caesium tarnishes rapidly in air because it reacts with oxygen in the air to form an oxide layer on
the surface.
(3) Caesium is an alkali metal. It reacts with water to give caesium hydroxide solution (an alkaline solution)
and hydrogen.
caesium + water
66 C
caesium hydroxide + hydrogen
Option
Process
Product(s)
(1)
Adding calcium and water
calcium hydroxide and hydrogen
(2)
Heating magnesium in air
magnesium oxide
(3)
Passing steam over heated aluminium
aluminium oxide and hydrogen
67 B Calcium reacts with water to give calcium hydroxide and hydrogen.
calcium + water
calcium hydroxide + hydrogen
The clear solution obtained is limewater. A white precipitate (calcium carbonate) forms when carbon
dioxide is bubbled into limewater.
carbon dioxide + calcium hydroxide
calcium carbonate + water
68 B (1) and (3) Gold and silver have NO reaction with dilute sulphuric acid.
(2) Magnesium reacts with dilute sulphuric acid to give magnesium sulphate and hydrogen.
magnesium + dilute sulphuric acid
69 D
magnesium sulphate + hydrogen
Option
Process
Products
(1)
Adding zinc granules to dilute sulphuric acid
zinc sulphate and hydrogen
(2)
Passing steam over heated iron powder
iron(II,III) oxide and hydrogen
(3)
Electrolysis of acidified water
hydrogen and oxygen
70 C (2) Sodium reacts explosively with dilute sulphuric acid.
(3) Potassium is very reactive. It reacts with oxygen in the air.
71 D (1) Pure gold is very soft while an alloy of copper and zinc is quite strong.
(2) The density of gold (19.3 g cm–3) is much higher than that of copper and zinc (both densities lower
than 10 g cm–3).
(3) The zinc in the alloy reacts with dilute sulphuric acid while gold does not.
72 B (1) There is no relationship between the charge on the ion of a metal and the reactivity of the metal.
(2) The oxide of a metal lower in the reactivity series has a lower stability. Thus the reduction of the
oxide of the metal is easier.
Carbon can remove oxygen from the oxide of X but not from that of Y. This shows that Y is more
reactive than X.
(3) X reacts with dilute hydrochloric acid but Y does not. This shows that X is more reactive than Y.
158
73 A Copper reacts with silver nitrate solution to form copper(II) nitrate solution and silver deposits.
Cu(s) + 2AgNO3(aq)
Cu(NO3)2(aq) + 2Ag(s)
(1) A grey solid (silver) formed.
(2) The solution became blue in colour because it contained copper(II) ions.
74 D (2) and (3) Iron can displace copper and silver from copper(II) sulphate solution and silver nitrate solution
respectively. Hence iron gradually dissolves in both cases.
75 B X reacts with the solution of nitrate of Y according to the following equation:
X(s) + Y2+(aq)
X2+(aq) + Y(s)
X displaces Y from the solution of nitrate of Y. This shows that X is more reactive than Y.
(1) X and Y may NOT be able to react with water. Hence (1) is incorrect.
(2) X displaces Y from the solution of nitrate of Y. This shows that X is more reactive than Y. Hence (2)
is correct.
(3) Although X is more reactive than Y, we cannot deduce that the oxide of Y undergoes decomposition
upon strong heating but the oxide of X does not.
For example, magnesium is more reactive than zinc but the oxide of zinc does not undergo decomposition
upon strong heating.
76 A
Tube
Experimental result
Deduction
1
grey coating on X
X is more reactive than zinc.
2
no observable change
magnesium is more reactive than X.
3
mud-like deposit on X
X is more reactive than Y.
(3) The experimental results indicate that X is more reactive than both zinc and Y. However, it is impossible
to deduce the relative reactivity of Y and zinc from the experimental results.
77 A (1) Zinc displaces lead from dilute lead(II) nitrate solution.
Zn(s) + Pb2+(aq)
Zn2+(aq) + Pb(s)
(2) Lead(II) oxide CANNOT be reduced by heating alone.
(3) NO reaction occurs when lead(II) oxide and carbon are mixed without heating.
78 B
Option
Experiment
Products
(1)
Heating silver oxide
silver and oxygen
(2)
Heating iron pyrite (FeS2)
iron(III) oxide and sulphur dioxide
(3)
Heating copper(II) oxide with magnesium
copper and magnesium oxide
79 A X displaces iron from iron(II) nitrate solution. This shows that X is more reactive than iron.
(1) As iron reacts with steam, X can probably react with steam.
159
(2) X is more reactive than iron. Hence X is also more reactive than copper. It displaces copper from
copper(II) sulphate solution.
(3) X is more reactive than iron. Its oxide CANNOT be reduced by heating with iron powder.
80 A
Option
Process
Remark
(1)
Adding calcium to dilute
hydrochloric acid
calcium + dilute hydrochloric acid
(2)
Adding zinc to magnesium
nitrate solution
zinc is less reactive than magnesium; there is NO reaction between
zinc and magnesium nitrate solution
(3)
Mixing copper(II) oxide with
charcoal powder
NO reaction occurs when copper(II) oxide and charcoal powder are
mixed without heating
calcium chloride + hydrogen
81 A Sodium is very reactive. It appears dull when exposed to air. This is because it reacts with oxygen in the
air to form an oxide layer on the surface.
When it is freshly cut or scratched, it appears shiny.
82 C Unreactive metals (e.g. copper) do not react with dilute sulphuric acid.
83 D Zinc is more reactive than iron. It reacts with dilute hydrochloric acid more vigorously than iron.
84 A When magnesium is added to dilute sulphuric acid, hydrogen gas is formed.
Mg(s) + H2SO4(aq)
MgSO4(aq) + H2(g)
In the reaction, each magnesium atom loses two electrons to form a magnesium ion:
Mg(s)
Mg2+(aq) + 2e–
Two hydrogen ions gain two electrons to form a hydrogen molecule:
+
–
2H (aq) + 2e
H2(g)
Sulphate ions do not take part in the reaction. We can delete them from the equation. The equation for
the reaction thus becomes:
Mg(s) + 2H+(aq)
2+
Mg (aq) + H2(g)
85 A The reactivity of Group I elements increases down the group.
MJUIJVN
SFBDUJWJUZ
JODSFBTJOH
TPEJVN
EPXO
UIF
HSPVQ
QPUBTTJVN
Hence potassium is more reactive than sodium.
160
Atoms of more reactive metals lose outermost shell electrons to form cations more readily. Hence potassium
atoms lose electrons more readily than sodium atoms.
86 D Lead is less reactive than iron. It CANNOT displace iron from iron(II) nitrate solution.
87 C Sodium is very reactive. It reacts with the water in zinc chloride solution to give hydrogen gas. Sodium
dissolves in the solution.
88 C Magnesium is above copper in the reactivity series. NO reaction occurs when magnesium oxide is heated
with copper powder. Copper CANNOT remove the oxygen from magnesium oxide.
Unit 12 Reacting masses
Fill in the blanks
1
mole
2
molar mass; gram per mole
3
crystallization
4
anhydrous
5
empirical
6
molecular
True or false
7
F
The mole is the amount of a substance that contains the same number of particles as there are atoms
in 12.0 g of carbon-12.
12.0 g of carbon-12 contains 6.02 x 1023 atoms (the Avogadro constant).
8
T
9
T One nitrogen molecule contains two nitrogen atoms. One mole of nitrogen molecules (i.e. 6.02 x 1023
molecules) contains two moles of nitrogen atoms (i.e. 2 x 6.02 x 1023 atoms).
10 F
The chemical formula of ammonium dichromate is (NH4)2Cr2O7. One formula unit of (NH4)2Cr2O7 contains
two NH4+ ions and one Cr2O72– ion. Hence one mole of (NH4)2Cr2O7 contains three moles of ions.
11 F
One aluminium atom forms one aluminium ion by losing three electrons.
Al
Al3+ + 3e–
Hence one mole of aluminium atoms can form one mole of aluminium ions, i.e. 6.02 x 1023 Al3+ ions.
12 T One mole of sulphur dioxide (SO2) molecules contains two moles of oxygen atoms. Hence two moles of
sulphur dioxide molecules contain four moles of oxygen atoms.
13 T 1 mole of O2(g) contains 2 moles of oxygen atoms while 1 mole of O3(g) contains 3 moles of oxygen
atoms.
161
14 T
15 F
–1
The units of molar mass are g mol .
16 F
The empirical formula of a compound gives the simplest whole number ratio of atoms or ions present in
the compound.
17 F
The chemical formula of hydrated copper(II) sulphate is CuSO4•5H2O.
Anhydrous copper(II) sulphate contains NO water of crystallization.
18 F
The empirical and molecular formulae of a compound are NOT necessarily the same. For example, the
molecular formula of hydrogen peroxide is H2O2. However, its empirical formula is HO.
19 T Consider the balanced chemical equation for a reaction:
aA + bB
cC + dD
The equation tells us that ‘a’ moles of A react with ‘b’ moles of B to give ‘c’ moles of C and ‘d’ moles
of D.
20 T The actual yield of the reaction is often less than the theoretical yield because of incomplete reaction or
the loss of some products during purification.
Multiple choice questions
21 D
Option
Substance
Molar mass of
substance
No. of moles of
substance present
No. of atoms /
molecules present
A
1 g of hydrogen
molar mass of H2
= 2 x 1.0 g mol–1
0.5 mole of H2
0.5 x 6.02 x 1023
molecules
B
16 g of sulphur
molar mass of S
= 32.1 g mol–1
0.5 mole of S
C
24 g of carbon
molar mass of C
= 12.0 g mol–1
2 moles of C
2 x 6.02 x 1023 atoms
D
32 g of oxygen
molar mass of O2
= 2 x 16.0 g mol–1
1 mole of O2
6.02 x 1023 molecules
23
0.5 x 6.02 x 10
atoms
∴ the Avogadro constant (6.02 x 1023) is the same as the number of molecules in 32 g of oxygen.
22 C Number of magnesium atoms = number of moles of magnesium atoms x L
= 4.50 mol x 6.02 x 1023 mol–1
= 2.71 x 1024
number of sodium ions
L
23
9.03 x 10
=
6.02 x 1023 mol–1
23 B Number of moles of sodium ions =
= 1.50 mol
162
24 D Number of carbon dioxide molecules present = number of moles of carbon dioxide molecules x L
= 6.00 mol x 6.02 x 1023 mol–1
= 3.61 x 1024
Each carbon dioxide molecule contains one carbon atom and two oxygen atoms.
Number of atoms present = 3 x number of carbon dioxide molecules
= 3 x 3.61 x 1024
= 1.08 x 1025
25 B Number of formula units of Na2S present = number of moles of Na2S x L
= 0.350 mol x 6.02 x 1023 mol–1
= 2.107 x 1023
One formula unit of sodium sulphide contains two sodium ions and one sulphide ion.
Number of ions present = 3 x number of formula units of Na2S
= 3 x 2.107 x 1023
= 6.32 x 1023
26 D The chemical formula of calcium phosphate is Ca3(PO4)2. Hence 1 mole of calcium phosphate contains 5
moles of ions.
27 B The chemical formula of magnesium chloride is MgCl2. Hence 1 mole of magnesium chloride contains 1
mole of magnesium ions and 2 moles of chloride ions, i.e. 6.02 x 1023 magnesium ions and 2 x 6.02 x
1023 chloride ions.
28 A
No. of moles of
ions in 1 mole of
substance
No. of moles of
ions present
Option
Substance
No. of moles of
substance present
A
aluminium sulphate Al2(SO4)3
2
5
10
B
calcium chloride CaCl2
2
3
6
C
zinc oxide ZnO
3
2
6
D
potassium nitrate KNO3
4
2
8
∴ 2 moles of aluminium sulphate contain the largest number of moles of ions.
29 B
Option
Substance
No. of moles of
substance present
No. of moles of
ions in 1 mole of
substance
No. of moles of
ions present
A
aluminium fluoride AlF3
5
4
20
B
iron(III) chloride FeCl3
6
4
24
C
copper(II) nitrate Cu(NO3)2
7
3
21
D
potassium permanganate KMnO4
8
2
16
∴ 6 moles of iron(III) chloride have the greatest number of moles of ions, i.e. the greatest number of
ions.
163
30 C 1 mole of silane contains 4 moles of hydrogen atom, i.e. 4L hydrogen atoms.
1
mol
4L
y
number of moles of silane containing y hydrogen atoms =
mol
4L
∴ number of moles of silane containing 1 hydrogen atom =
31 B 1 mole of XO contains 2 moles of atoms, i.e. n atoms.
∴ 2 moles of atoms = n atoms
1 mole of atoms =
n
atoms
2
1 mole of X2O3 contains 5 moles of atoms, i.e.
5
n atoms.
2
32 B
Option
Substance
A
CO2
B
Na2CO3
C
CH3COOH
D
NH3
Number of moles
of substance
present
Mass of substance
present
(12.0 + 2 x 16.0) g mol–1
= 44.0 g mol–1
0.500 mol
0.500 mol x 44.0 g mol–1
= 22.0 g
(2 x 23.0 + 12.0 + 3 x 16.0) g mol–1
= 106.0 g mol–1
1.00 mol
1.00 mol x 106.0 g mol–1
= 106 g
(2 x 12.0 + 4 x 1.0 + 2 x 16.0) g mol–1
= 60.0 g mol–1
1.50 mol
1.50 mol x 60.0 g mol–1
= 90.0 g
2.00 mol
2.00 mol x 17.0 g mol–1
= 34.0 g
Molar mass of substance
–1
(14.0 + 3 x 1.0) g mol
–1
= 17.0 g mol
∴ 1.00 mole of Na2CO3 has the greatest mass.
33 A Number of moles of sulphur atoms =
=
mass of sulphur
molar mass of sulphur
4.80 g
32.1 g mol–1
= 0.150 mol
34 A Molar mass of CaSO4•2H2O = [40.1 + 32.1 + 4 x 16.0 + 2 x (2 x 1.0 + 16.0)] g mol–1
= 172.2 g mol–1
mass of CaSO4•2H2O
molar mass of CaSO4•2H2O
60.2 g
=
172.2 g mol–1
Number of moles of formula units in 60.2 g of CaSO4•2H2O =
= 0.350 mol
164
35 C Molar mass of NH3 = (14.0 + 3 x 1.0) g mol–1
= 17.0 g mol–1
mass of NH3
molar mass of NH3
51.0 g
=
–1
17.0 g mol
Number of moles of NH3 =
= 3.00 mol
Number of NH3 molecules present = number of moles of NH3 x L
= 3.00 mol x 6.02 x 1023 mol–1
= 3 x 6.02 x 1023
36 D Molar mass of Mg3N2 = (3 x 24.3 + 2 x 14.0) g mol–1
= 100.9 g mol–1
mass of Mg3N2
molar mass of Mg3N2
202 g
=
–1
100.9 g mol
Number of moles of Mg3N2 =
= 2.00 mol
1 mole of magnesium nitride contains 3 moles of magnesium ions and 2 moles of nitride ions, i.e. a total
of 5 moles of ions.
Number of moles of ions present = 5 x 2.00 mol
= 10.0 mol
Number of ions present = 10.0 mol x 6.02 x 1023 mol–1
= 1.00 x 6.02 x 1024
37 D Number of moles of aluminium =
=
mass of aluminium
molar mass of aluminium
6.75 g
–1
27.0 g mol
= 0.25 mol
Number of aluminium atoms = number of moles of aluminium x L
= 0.25 mol x 6.02 x 1023 mol–1
= x
mass of carbon
molar mass of carbon
24.0 g
=
–1
12.0 g mol
Number of moles of carbon =
= 2.00 mol
Number of carbon atoms = number of moles of carbon x L
= 2.00 mol × 6.02 x 1023 mol–1
= y
y
2.00 x 6.02 x 1023
=
x
0.25 x 6.02 x 1023
y = 8x
165
38 C 1 mole of oxygen contains 2 moles of atoms, i.e. x atoms.
∴ 2 moles of atoms = x atoms
x
atoms
2
1 mole of ozone contains 3 moles of atoms.
1 mole of atoms =
2 moles of ozone contain 2 x 3 x
x
atoms; i.e. 3x atoms.
2
39 A Molar mass of HCl = (1.0 + 35.5) g mol–1
= 36.5 g mol–1
Number of moles of HCl =
=
mass of HCl
molar mass of HCl
21.9 g
–1
36.5 g mol
= 0.600 mol
0.600 mole of CO2 has the same number of molecules as 21.9 g of HCl.
Molar mass of CO2 = (12.0 + 2 x 16.0) g mol–1
= 44.0 g mol–1
Mass of 0.600 mole of CO2 = number of moles of CO2 x molar mass of CO2
= 0.600 mol x 44.0 g mol–1
= 26.4 g
40 C
Relative atomic mass Mass of substance
of substance
present
Option
Substance
Number of moles of atoms present
—
oxygen
16.0
8.00 g
8.00 g
= 0.500 mol
–1
16.0 g mol
A
hydrogen
1.00
1.00 g
1.00 g
= 1.00 mol
–1
1.00 g mol
B
carbon
12.0
18.0 g
18.0 g
= 1.50 mol
–1
12.0 g mol
C
neon
20.2
10.1 g
10.1 g
= 0.500 mol
20.2 g mol–1
D
silicon
28.1
28.1 g
28.1 g
= 1.00 mol
28.1 g mol–1
The number of moles of atoms in 8.00 g of oxygen is the same as that in 10.1 g of neon. Hence they
contain the same number of atoms.
166
41 B
Mass of gas present
Number of moles of molecules
present
(14.0 + 2 x 16.0) g mol
= 46.0 g mol–1
10 g
10 g
= 0.217 mol
–1
46.0 g mol
SO2
(32.1 + 2 x 16.0) g mol–1
= 64.1 g mol–1
10 g
10 g
= 0.156 mol
–1
64.1 g mol
C
CO
(12.0 + 16.0) g mol
= 28.0 g mol–1
10 g
10 g
= 0.357 mol
28.0 g mol–1
D
CH4
(12.0 + 4 x 1.0) g mol–1
= 16.0 g mol–1
10 g
10 g
= 0.625 mol
16.0 g mol–1
Option
Gas
A
NO2
B
Molar mass of gas
–1
–1
10 g of SO2 contain the smallest number of moles of molecules, i.e. the smallest number of molecules.
42 B
No. of moles
No. of moles
of ions in
of ions
one mole of
present
substance
Mass of
Molar mass of substance substance
present
No. of moles
of substance
present
(40.1 + 2 x 35.5) g mol–1
= 111.1 g mol–1
12 g
12 g
111.1 g mol–1
= 0.11 mol
3
3 x 0.11 mol
= 0.33 mol
(24.3 + 16.0) g mol–1
= 40.3 g mol–1
14 g
14 g
40.3 g mol–1
= 0.35 mol
2
2 x 0.35 mol
= 0.70 mol
Option
Substance
A
calcium
chloride
CaCl2
B
magnesium
oxide
MgO
C
iron(III)
chloride
FeCl3
(55.8 + 3 x 35.5) g mol–1
= 162.3 g mol–1
16 g
16 g
162.3 g mol–1
= 0.10 mol
4
4 x 0.10 mol
= 0.40 mol
D
sodium
bromide
NaBr
(23.0 + 79.9) g mol–1
= 102.9 g mol–1
18 g
18 g
102.9 g mol–1
= 0.17 mol
2
2 x 0.17 mol
= 0.34 mol
14 g of magnesium oxide contain the greatest number of moles of ions, i.e. the greatest number of
ions.
43 A Number of moles of X2 =
mass of X2
molar mass of X2
mass of X2
number of moles of X2
70.0 g
=
2.50 mol
Molar mass of X2 =
–1
= 28.0 g mol
∴ the relative atomic mass of X is 14.0.
167
44 A Option A — Molar mass of oxygen gas (O2) = 2 x 16.0 g mol–1 = 32.0 g mol–1
Number of moles of molecules in 32.0 g of oxygen gas = 1.00 mol
–1
Molar mass of nitrogen gas (N2) = 2 x 14.0 g mol
–1
= 28.0 g mol
Number of moles of molecules in 28.0 g of nitrogen gas = 1.00 mol
∴ 32.0 g of oxygen gas and 28.0 g of nitrogen gas contain the same number of mole of
molecules, i.e. they contain the same number of molecules.
Option B — Molar mass of ammonia gas (NH3) = (14.0 + 3 x 1.0) g mol–1 = 17.0 g mol–1
Number of moles of molecules in 34.0 g of ammonia gas =
34.0 g
–1
17.0 g mol
= 2.00 mol
∴ 32.0 g of oxygen gas and 34.0 g of ammonia gas contain different number of moles of
molecules, i.e. they contain DIFFERENT number of molecules.
Option C — Each oxygen molecule contains two oxygen atoms.
Number of moles of atoms in 32.0 g (1 mole) of oxygen gas (O2) = 2.00 mol
∴ number of oxygen atoms present = 2.00 mol x 6.02 x 1023 mol–1
23
= 2.00 x 6.02 x 10
Option D — Molar mass of carbon = 12.0 g mol–1
Number of moles of atoms in 12.0 g of carbon = 1.00 mol
Number of moles of atoms in 32.0 g (1 mole) of oxygen gas = 2.00 mol
∴ 32.0 g of oxygen gas and 12.0 g of carbon contain different number of moles of atoms,
i.e. they contain DIFFERENT number of atoms.
45 D Every ammonia (NH3) molecule contains one nitrogen atom and three hydrogen atoms.
Option A — 1 mole of ammonia contains 1 mole of nitrogen atoms. Hence 2 moles of ammonia contain
2 moles of nitrogen atoms.
Option B — 2 moles of ammonia contain 6 moles of hydrogen atoms, i.e. 6 x 6.02 x 1023 hydrogen
atoms.
Option C — 2 moles of ammonia contain 2 moles of nitrogen atoms and 6 moles of hydrogen atoms,
i.e. a total of 8 moles of atoms.
1 mole of hydrogen chloride contains 1 mole of hydrogen atoms and 1 mole of chlorine
atoms. Hence 4 moles of hydrogen chloride contain 4 moles of hydrogen atoms and 4 moles
of chlorine atoms, i.e. a total of 8 moles of atoms.
∴ 2 moles of ammonia and 4 moles of hydrogen chloride contain the same number of
moles of atoms, i.e. the same number of atoms.
Option D — number of moles of CO2 in 44.0 g of the substance
mass of CO2
=
molar mass of CO2
44.0 g
=
–1
(12.0 + 2 x 16.0) g mol
= 1.00 mol
168
1 mole of carbon dioxide contains 1 mole of carbon atoms and 2 moles of oxygen atoms,
i.e. a total of 3 moles of atoms.
On the other hand, 2 moles of ammonia contain 8 moles of atoms.
∴ 2 moles of ammonia and 44.0 g of carbon dioxide contain DIFFERENT number of
atoms.
46 C Formula mass of K2CO3 = 2 x 39.1 + 12.0 + 3 x 16.0 = 138.2
Percentage by mass of K in K2CO3
number of atoms of K in formula x relative atomic mass of K
x 100%
formula mass of K2CO3
2 x 39.1
=
x 100%
138.2
=
= 56.6%
47 A Formula mass of CuSO4•5H2O = 63.5 + 32.1 + 4 x 16.0 + 5 x (2 x 1.0 +16.0)
= 249.6
5 x (2 x 1.0 + 16.0)
x 100%
249.6
= 36.1%
Percentage by mass of water in CuSO4•5H2O =
48 B Let m be the relative atomic mass of X.
Formula mass of X2CrO4 = 2 x m + 52.0 + 4 x 16.0 = 2m + 116.0
Percentage by mass of X in X2CrO4
number of atoms of X in formula x relative atomic mass of X
x 100%
formula mass of X2CrO4
2m
=
x 100%
2m + 116.0
=
2m
x 100% = 65.1%
2m + 116.0
m = 108
∴ the relative atomic mass of X is 108.
49 C Formula mass of Na2CO3•nH2O = 2 x 23.0 + 12.0 + 3 x 16.0 + n(2 x 1.0 + 16.0)
= 106.0 + 18n
Percentage by mass of water in Na2CO3•nH2O =
18n
x 100%
106.0 + 18n
18n
x 100% = 60.4%
106.0 + 18n
n= 9
169
50 D Mass of oxygen in the oxide = (54.8 – 49.7) g = 5.1 g
X
Oxygen
Mass of element in the compound
49.7 g
5.1 g
Relative atomic mass
207.2
16.0
49.7 g
= 0.24 mol
207.2 g mol–1
5.1 g
= 0.32 mol
16.0 g mol–1
Number of moles of atoms that
combine
0.24 mol
0.24 mol
Mole ratio of atoms
Simplest whole number ratio of
atoms
= 1.3
1 x 3 = 3
1.3 x 3 = 4
Cu
O
25.4 g
3.20 g
63.5
16.0
25.4 g
= 0.400 mol
63.5 g mol–1
3.20 g
= 0.200 mol
16.0 g mol–1
0.400 mol
= 2
0.200 mol
0.200 mol
= 1
0.200 mol
51 A
Mass of element in the compound
Relative atomic mass
Number of moles of atoms that
combine
0.32 mol
0.24 mol
= 1
Mole ratio of atoms
∴ the empirical formula of the compound is Cu2O.
52 B For 100 g of the alcohol, there are 60.0 g of carbon, 13.3 g of hydrogen and 26.7 g of oxygen.
Carbon
Hydrogen
Oxygen
60.0 g
13.3 g
26.7 g
Relative atomic mass
12.0
1.0
16.0
Number of moles of
atoms that combine
60.0 g
= 5.00 mol
12.0 g mol–1
Mole ratio of atoms
5.00 mol
= 3
1.67 mol
Mass of element in the
compound
13.3 g
1.0 g mol–1
= 13.3 mol
13.3 mol
= 8
1.67 mol
∴ the empirical formula of the alcohol is C3H8O.
53 C Formula mass of ZnSO4•xH2O = (65.4 + 32.1 + 4 x 16.0) + x(2 x 1.0 + 16.0)
= 161.5 + 18x
1 mole of ZnSO4•xH2O contains 1 mole of ZnSO4.
i.e. (161.5 + 18x) g of ZnSO4•xH2O contain 161.5 g of ZnSO4.
5.75 g of ZnSO4•xH2O contain 3.23 g of ZnSO4.
161.5 g
(161.5 + 18x) g
170
=
x = 7
3.23 g
5.75 g
26.7 g
= 1.67 mol
16.0 g mol–1
1.67 mol
= 1
1.67 mol
54 B Mass of MgSO4•7H2O heated = (27.855 – 20.465) g
= 7.39 g
Mass of residue = (25.157 – 20.465) g
= 4.692 g
Molar mass of MgSO4•7H2O = [24.3 + 32.1 + 4 x 16.0 + 7 x (2 x 1.0 + 16.0)] g mol–1
= 246.4 g mol–1
mass of MgSO4•7H2O
molar mass of MgSO4•7H2O
7.39 g
=
–1
246.4 g mol
Number of moles of MgSO4•7H2O heated =
= 0.0300 mol
Let MgSO4•xH2O be the chemical formula of the residue.
Molar mass of MgSO4•xH2O = [24.3 + 32.1 + 4 x 16.0 + x(2 x 1.0 + 16.0)] g mol–1
= (120.4 + 18x) g mol–1
1 mole of MgSO4•7H2O gives 1 mole of MgSO4•xH2O upon heating.
∴
4.692 g
= 0.0300 mol
–1
(120.4 + 18x) g mol
x = 2
∴ the chemical formula of the residue is MgSO4•2H2O.
55 B For 100 g of compound X, there are 82.8 g of carbon and 17.2 g of hydrogen.
Mass of element in the compound
Carbon
Hydrogen
82.8 g
17.2 g
12.0
1.0
Relative atomic mass
Number of moles of atoms that
combine
Mole ratio of atoms
Simplest whole number ratio of
atoms
82.8 g
= 6.90 mol
12.0 g mol–1
17.2 g
1.0 g mol–1
= 17.2 mol
6.90 mol
= 1
6.90 mol
17.2 mol
= 2.49
6.90 mol
1 x 2 = 2
2.49 x 2 = 5
∴ the empirical formula of compound X is C2H5.
Let (C2H5)n be the molecular formula of compound X.
Relative molecular mass of X =
=
∴ 29n =
n =
n(2 x 12.0 + 5 x 1.0)
29n
58.0
2
∴ the molecular formula of compound X is C4H10.
171
56 C Let m be the relative atomic mass of X.
Number of moles of X in the oxide =
=
Number of moles of O in the oxide =
=
mass of X
molar mass of X
4.68 g
–1
m g mol
mass of O
molar mass of O
2.16 g
–1
16.0 g mol
The chemical formula of the oxide is X2O3,
∴
number of moles of X
number of moles of O
=
2
=
3
4.68 g
–1
m g mol
2.16 g
–1
16.0 g mol
m = 52.0
∴ the relative atomic mass of X is 52.0.
57 C The balanced equation of the reaction is:
4Fe(OH)2 + O2 + 2H2O
4Fe(OH)3
Hence 4 moles of Fe(OH)3 can be obtained when 1 mole of O2 reacts.
58 B TiCl4(g) + 2Mg(l)
114 g
? g
Ti(s) + 2MgCl2(l)
Molar mass of TiCl4 = (47.9 + 4 x 35.5) g mol–1
= 189.9 g mol–1
mass of TiCl4
molar mass of TiCl4
114 g
=
189.9 g mol–1
Number of moles of TiCl4 =
= 0.600 mol
According to the equation, 1 mole of TiCl4 requires 2 moles of Mg for complete reaction.
∴ number of moles of Mg required = 2 x 0.600 mol
= 1.20 mol
Molar mass of Mg = 24.3 g mol–1
Mass of Mg required = number of moles of Mg x molar mass of Mg
= 1.20 mol x 24.3 g mol–1
= 29.2 g
172
59 C CaCO3(s)
7.51 g
CaO(s) + CO2(g)
? g
Molar mass of CaCO3 = (40.1 + 12.0 + 3 x 16.0) g mol–1
= 100.1 g mol–1
mass of CaCO3
molar mass of CaCO3
7.51 g
=
–1
100.1 g mol
Number of moles of CaCO3 =
= 0.0750 mol
According to the equation, 1 mole of CaCO3 produces 1 mole of CaO upon decomposition.
∴ number of moles of CaO = 0.0750 mol
Molar mass of CaO = (40.1 + 16.0) g mol–1 = 56.1 g mol–1
Mass of CaO obtained = number of moles of CaO x molar mass of CaO
= 0.0750 mol x 56.1 g mol–1
= 4.21 g
60 B Fe2O3(s) + 3CO(g)
? g
2Fe(s) + 3CO2(g)
83.7 g
Molar mass of Fe = 55.8 g mol–1
Number of moles of Fe =
=
mass of Fe
molar mass of Fe
83.7 g
–1
55.8 g mol
= 1.50 mol
According to the equation, 1 mole of Fe2O3 produces 2 moles of Fe.
1.50
mol
2
= 0.750 mol
∴ number of moles of Fe2O3 consumed =
Molar mass of Fe2O3 = (2 x 55.8 + 3 x 16.0) g mol–1
= 159.6 g mol–1
Mass of Fe2O3 consumed = number of moles of Fe2O3 x molar mass of Fe2O3
= 0.750 mol x 159.6 g mol–1
= 120 g
173
61 C CaCO3(s) + SiO2(s)
? tonne(s) 1.00 tonne
CaSiO3(l) + CO2(g)
Molar mass of CaCO3 = (40.1 + 12.0 + 3 x 16.0) g mol–1
= 100.1 g mol–1
Molar mass of SiO2 = (28.1 + 2 x 16.0) g mol–1
= 60.1 g mol–1
According to the equation, 1 mole of CaCO3 is needed to remove 1 mole of SiO2.
∴ 100.1 g of CaCO3 are needed to remove 60.1 g of SiO2.
CaCO3(s) + SiO2(s)
100.1 g
60.1 g
? tonne(s) 1.00 tonne
CaSiO3(l) + CO2(g)
Mass of CaCO3 needed = 1 tonne x
100.1 g
60.1 g
= 1.67 tonnes
62 D Pb3O4(s) + 4H2(g)
178 g
3Pb(s) + 4H2O(l)
? g
Molar mass of Pb3O4 = (3 x 207.2 + 4 x 16.0) g mol–1 = 685.6 g mol–1
mass of Pb3O4
molar mass of Pb3O4
178 g
=
–1
685.6 g mol
Number of moles of Pb3O4 =
= 0.260 mol
According to the equation, 1 mole of Pb3O4 gives 3 moles of Pb in the reaction.
∴ number of moles of Pb obtained = 3 x 0.260 mol
= 0.780 mol
Mass of Pb obtained = number of moles of Pb x molar mass of Pb
= 0.780 mol x 207.2 g mol–1
= 162 g
63 C KClO3 decomposes to give KCl and O2 according to the following equation:
2KClO3(s)
24.5 g
2KCl(s) + 3O2(g)
? g
Molar mass of KClO3 = (39.1 + 35.5 + 3 x 16.0) g mol–1
= 122.6 g mol–1
mass of KClO3
molar mass of KClO3
24.5 g
=
–1
122.6 g mol
Number of moles of KClO3 =
= 0.200 mol
174
According to the equation, 2 moles of KClO3 gives 3 moles of O2 upon decomposition.
3
x 0.200 mol
2
= 0.300 mol
∴ number of moles of O2 produced =
Molar mass of O2 = 2 x 16.0 g mol–1
= 32.0 g mol–1
Mass of O2 produced = number of moles of O2 x molar mass of O2
= 0.300 mol x 32.0 g mol–1
= 9.60 g
64 A The total mass of the reactants equals that of the products.
∴ y = a + b – x
65 A 2XHCO3(s)
1.50 g
X2CO3(s) + H2O(l) + CO2(g)
0.330 g
Molar mass of CO2 = (12.0 + 2 x 16.0) g mol–1
= 44.0 g mol–1
Number of moles of CO2 produced =
=
mass of CO2
molar mass of CO2
0.330 g
–1
44.0 g mol
= 0.00750 mol
According to the equation, 2 moles of XHCO3 give 1 mole of CO2 upon heating.
∴ number of moles of XHCO3 heated = 2 x 0.00750 mol
= 0.0150 mol
Mass of XHCO3 heated = number of moles of XHCO3 x molar mass of XHCO3
mass of XHCO3 heated
number of moles of XHCO3
1.50 g
=
0.0150 mol
= 100 g mol–1
Molar mass of XHCO3 =
∴ the formula mass of XHCO3 is 100.
66 C The oxide of M is reduced by hydrogen according to the following equation:
M2O(s) + H2(g)
23.2 g
2M(s) + H2O(l)
1.80 g
Let m be the relative atomic mass of M.
Molar mass of H2O = (2 x 1.0 + 16.0) g mol–1
–1
= 18.0 g mol
175
mass of H2O
molar mass H2O
1.80 g
=
–1
18.0 g mol
= 0.100 mol
Number of moles of H2O =
According to the equation, 1 mole of M2O produces 1 mole of H2O upon reduction.
∴ number of moles of M2O reduced = 0.100 mol
Mass of M2O = number of moles of M2O x molar mass of M2O
mass of M2O
number of moles of M2O
23.2 g
=
0.100 mol
–1
= 232 g mol
Molar mass of M2O =
= (2m + 16.0) g mol–1
∴ m = 108
67 D Copper and silver nitrate solution react according to the following equation:
Cu(s) + 2AgNO3(aq)
5 g
Cu(NO3)2(aq) + 2Ag(s)
? g
According to the equation, 1 mole of Cu gives 2 moles of Ag in the reaction. As the relative atomic
mass of Ag is greater than that of Cu, hence the mass of Ag obtained is more than double of that of
Cu, i.e. more than 10 g.
68 D CaCO3(s) + 2HCl(aq)
? g
CaCl2(aq) + H2O(l) + CO2(g)
0.380 g
Molar mass of CO2 = (12.0 + 2 x 16.0) g mol–1
= 44.0 g mol–1
Number of moles of CO2 =
=
mass of CO2
molar mass of CO2
0.380 g
–1
44.0 g mol
= 0.00864 mol
According to the equation, 1 mole of CaCO3 gives 1 mole of CO2 upon reaction with dilute hydrochloric
acid.
∴ number of moles of CaCO3 reacted = 0.00864 mol
Molar mass of CaCO3 = (40.1 + 12.0 + 3 x 16.0) g mol–1
= 100.1 g mol–1
Mass of CaCO3 reacted = number of moles of CaCO3 x molar mass of CaCO3
= 0.00864 mol x 100.1 g mol–1
= 0.865 g
0.865 g
1.00 g
= 86.5%
Percentage by mass of CaCO3 in the sample =
176
x 100%
69 D 4FeS2(s) + 11O2(g)
? tonnes
2Fe2O3(s) + 8SO2(g)
36.7 tonnes
Molar mass of Fe2O3 = (2 x 55.8 + 3 x 16.0) g mol–1
= 159.6 g mol–1
mass of Fe2O3
molar mass of Fe2O3
6
36.7 x 10 g
=
–1
159.6 g mol
Number of moles of Fe2O3 =
= 0.230 x 106 mol
According to the equation, 4 moles of FeS2 give 2 moles of Fe2O3 when roasted.
∴ number of moles of FeS2 reacted = 2 x 0.230 x 106 mol
= 0.460 x 106 mol
Molar mass of FeS2 = (55.8 + 2 x 32.1) g mol–1
= 120.0 g mol–1
Mass of FeS2 reacted = number of moles of FeS2 x molar mass of FeS2
= 0.460 x 106 mol x 120.0 g mol–1
= 55.2 x 106 g
6
55.2 x 10 g
x 100%
6
60.0 x 10 g
= 92.0%
Percentage by mass of FeS2 in the ore =
70 D Suppose 10 g of Mg and Fe react with excess hydrochloric acid separately.
Mg(s) + 2HCl(aq)
Fe(s) + 2HCl(aq)
MgCl2(aq) + H2(g)
FeCl2(aq) + H2(g)
Number of moles of Mg in 10 g =
=
mass of Mg
molar mass of Mg
10 g
24.3 g mol–1
= 0.412 mol
Number of moles of Fe in 10 g =
=
mass of Fe
molar mass of Fe
10 g
55.8 g mol–1
= 0.179 mol
According to the equations, 1 mole of Mg / Fe gives 1 mole of hydrogen with excess hydrochloric
acid.
10 g of Mg give 0.412 mole of hydrogen while 10 g of Fe give 0.179 mole of hydrogen. Hence Mg
gives more hydrogen than iron does because the relative atomic mass of Mg is smaller than that of Fe.
177
71 A
S(s) + O2(g)
48.2 g
64.0 g
SO2(g)
? g
mass of S
molar mass of S
48.2 g
–1
32.1 g mol
Number of moles of S present =
=
= 1.50 mol
Number of moles of O2 present =
=
mass of O2
molar mass of O2
64.0 g
32.1 g mol–1
= 1.99 mol
According to the equation, 1 mole of S reacts with 1 mole of O2 to produce 1 mole of SO2. During the
reaction, 1.50 moles of S react with 1.50 moles of O2. Therefore O2 is in excess. The amount of S limits
the amount of SO2 produced.
Number of moles of SO2 produced = 1.50 mol
Molar mass of SO2 = (32.1 + 2 x 16.0) g mol–1
= 64.1 g mol–1
Mass of SO2 produced = number of moles of SO2 x molar mass of SO2
= 1.50 mol x 64.1 g mol–1
= 96.2 g
72 C N2H4(l) + O2(g)
2.40 g 3.40 g
N2(g) + 2H2O(l)
? g
mass of N2H4
molar mass of N2H4
2.40 g
=
–1
32.0 g mol
Number of moles of N2H4 present =
= 0.0750 mol
Number of moles of O2 present =
=
mass of O2
molar mass of O2
3.40 g
32.0 g mol–1
= 0.106 mol
According to the equation, 1 mole of N2H4 reacts with 1 mole of O2 to produce 2 moles of H2O. During
the reaction, 0.0750 mole of N2H4 reacts with 0.0750 mole of O2. Therefore O2 is in excess. The amount
of N2H4 limits the amount of water produced.
Number of moles of H2O produced = 2 x 0.0750 mol
= 0.150 mol
178
Molar mass of H2O = (2 x 1.0 + 16.0) g mol–1
= 18.0 g mol–1
Mass of H2O produced = number of moles of H2O x molar mass of H2O
= 0.150 mol x 18.0 g mol–1
= 2.70 g
73 A N2(g) + 3H2(g)
60 g
2NH3(g)
80 g
mass of H2
molar mass of H2
60 g
=
–1
2.0 g mol
Number of moles of H2 =
= 30 mol
According to the equation, 3 moles of hydrogen react with excess nitrogen to produce 2 moles of
ammonia.
2
x 30 mol
3
= 20 mol
∴ number of moles of NH3 produced =
Molar mass of NH3 = (14.0 + 3 x 1.0) g mol–1
= 17.0 g mol–1
Theoretical yield of NH3 = number of moles of NH3 x molar mass of NH3
= 20 mol x 17.0 g mol–1
= 340 g
80 g
x 100%
340 g
80
=
x 100%
340
Percentage yield of NH3 =
74 B Percentage yield =
actual yield
theoretical yield
x 100%
16.0 g
75.0%
= 21.3 g
Theoretical yield of CO2 =
75 D 2Cu(s) + S(s)
2.40 g
Cu2S(s)
2.85 g
mass of Cu
molar mass of Cu
2.40 g
=
–1
63.5 g mol
Number of moles of Cu =
= 0.0378 mol
According to the equation, 2 moles of Cu react with excess S to produce 1 mole of Cu2S.
179
0.0378
mol
2
= 0.0189 mol
∴ number of moles of Cu2S produced =
Molar mass of Cu2S = (2 x 63.5 + 32.1) g mol–1
= 159.1 g mol–1
Theoretical yield of Cu2S = number of moles of Cu2S x molar mass of Cu2S
= 0.0189 mol x 159.1 g mol–1
= 3.01 g
2.85 g
x 100%
3.01 g
= 94.7%
Percentage yield of Cu2S =
76 A (1) One ozone molecule contains 3 oxygen atoms. Hence 1 mole of ozone contains 3 moles of atoms,
i.e. 3 x 6.02 x 1023 atoms.
(2) One mole of ozone contains 1 mole of molecules.
(3) One oxygen molecule contains 2 oxygen atoms. Hence 1 mole of oxygen contains 2 moles of atoms,
i.e. 2 x 6.02 x 1023 atoms.
∴ 1 mole of ozone and 1 mole of oxygen contain DIFFERENT number of atoms.
77 A (1) One magnesium atom can form 1 Mg2+ ion. Hence 1 mole of magnesium can form 1 mole of Mg2+
ions.
(2) One mole of magnesium can form 1 mole of Mg2+ ions, i.e. 6.02 x 1023 Mg2+ ions.
(3) The molar mass of Mg2+ ions is 24.3 g mol–1.
78 A Molar mass of CO2 = (12.0 + 2 x 16.0) g mol–1
= 44.0 g mol–1
88.0 g of CO2 contain 2 moles of CO2.
(1) One carbon dioxide molecule contains 2 oxygen atoms. Hence 88.0 g of CO2 (i.e. 2 moles of CO2)
contain 4 moles of oxygen atoms.
(2) One carbon dioxide molecule contains 3 atoms. Hence 88.0 g of CO2 (i.e. 2 moles of CO2) contain
6 moles of atoms, i.e. 6 x 6.02 x 1023 atoms.
(3) 88.0 g of CO2 (i.e. 2 moles of CO2) contain 2 moles of molecules, i.e. 2 x 6.02 x 1023 molecules.
79 B (1) One mole of sulphur atoms and one mole of oxygen atoms contain the same number of atoms.
(2) Suppose 2 g of sulphur and 1 g of oxygen contain x atoms.
2
g
x
1
Mass of 1 oxygen atom =
g
x
∴ the mass of 1 sulphur atom is twice that of 1 oxygen atom. Hence the mass of 1 mole of sulphur
atoms is twice that of 1 mole of oxygen atoms.
Mass of 1 sulphur atom =
180
80 D
Option
Mass of
Substance Molar mass of substance substance
present
No. of moles
of substance
present
No. of moles
of atoms in
one mole of
substance
No. of moles
of atoms
present
—
ammonia
NH3
(14.0 + 3 x 1.0) g mol–1
–1
= 17.0 g mol
5.10 g
5.10 g
17.0 g mol–1
= 0.300 mol
4
4 x 0.300 mol
= 1.20 mol
(1)
carbon
dioxide
CO2
(12.0 + 2 x 16.0) g mol–1
= 44.0 g mol–1
17.6 g
17.6 g
44.0 g mol–1
= 0.400 mol
3
3 x 0.400 mol
= 1.20 mol
(2)
nitrogen
monoxide
NO
(14.0 + 16.0) g mol–1
= 30.0 g mol–1
18.0 g
18.0 g
30.0 g mol–1
= 0.600 mol
2
2 x 0.600 mol
= 1.20 mol
(3)
chlorine
Cl2
(2 x 35.5) g mol–1
–1
= 71.0 g mol
42.6 g
42.6 g
71.0 g mol–1
= 0.600 mol
2
2 x 0.600 mol
= 1.20 mol
5.10 g of ammonia, 17.6 g of carbon dioxide, 18.0 g of nitrogen monoxide and 42.6 g of chlorine all
contain the same number of moles of atoms, i.e. the same number of atoms.
Unit 13 Corrosion of metals and their protection
Fill in the blanks
1
corrosion
2
rusting; iron(III) oxide
3
oxygen; water
4
galvanized
5
electroplating
6
sacrificial
7
magnesium; zinc
8
cathodic protection
9
chromium; nickel
10 anodization
True or false
11 F
The chemical name of rust is hydrated iron(III) oxide.
12 T
181
13 T An industrial environment speeds up the rusting process. This is due to the emission of acidic gases from
factories. These gases form acids with moisture in the air, thus speeding up the rusting process.
14 T
15 F
Rusting occurs more quickly where the iron surface is scratched.
16 T A layer of paint prevents both oxygen and water from reaching the iron beneath. Hence painting can
protect iron from rusting.
17 F
Tin is less reactive than iron. It CANNOT protect iron from rusting by sacrificial protection.
18 T Food cans are usually made of mild steel coated with a thin layer of tin. The tin protects the steel from
both oxygen and water. Furthermore, tin ions are non-poisonous.
19 F
When part of the zinc plated on an iron object is scratched, zinc still protects the iron from rusting. This
is because zinc is more reactive than iron. Hence zinc corrodes instead of iron. Zinc ‘sacrifices’ itself to
prevent iron from rusting.
20 T Magnesium is more reactive than iron. When a piece of magnesium is joined to an undergound pipeline,
magnesium corrodes instead of iron. Magnesium provides sacrificial protection.
21 F
Iron coated with zinc is NOT used for making cans for drinks because zinc ions are poisonous.
22 F
We can protect iron from rusting by connecting it to the negative terminal of a battery while a conductor
such as graphite is connected to the positive terminal. This supplies electrons to the iron and prevents
the formation of iron(II) ions. This method is called impressed current cathodic protection.
m
QSPUFDUFENFUBM
DBUIPEF
QPXFS
TVQQMZ
BDPOEVDUPS
BOPEF
Steel structures of a pier can be protected from rusting by this method.
23 F
182
The rust tends to fall off as it is formed. A fresh iron surface is then exposed and so the rusting goes
on.
24 F
Aluminium anodization can increase the corrosion resistance of aluminium, but NOT the strength of
aluminium.
25 F
An alloy is a mixture of a metal with one or more other elements.
During aluminium anodization, the thickness of the aluminium oxide layer on an aluminium object is
increased. Anodized aluminium is NOT an alloy.
Multiple choice questions
26 A
27 B Option B — The iron nail is in contact with water only. Therefore it will NOT rust.
28 C
29 D Option A — The iron nail will NOT rust as it is in contact with water only.
Option B — The iron nail will NOT rust as it is in contact with dry air only.
Option C — The iron nail will rust.
Option D — The iron nail will rust most as the ionic substances in the sea water speed up the rusting
process.
30 A Option A — Dilute hydrochloric acid speeds up the rusting process.
31 B Option A — Sodium sulphate is an ionic substance. The rusting process speeds up if the water present
contains an ionic substance.
Option B — Glucose is a covalent substance. Adding it to the water would NOT speed up the rusting
process.
Option C — Attaching copper (a less reactive metal) to the iron nail speeds up the rusting process.
Option D — Carbonic acid is formed when carbon dioxide is passed through water. The acid speeds up
the rusting process.
32 C A rust indicator contains potassium hexacyanoferrate(III). Potassium hexacyanoferrate(III) gives a blue colour
in the presence of iron(II) ions.
Nail X is connected to tin, a metal less reactive than iron. The rusting process of the iron nail is speeded
up.
During rusting, iron atoms first lose electrons to form iron(II) ions.
Fe(s)
Fe2+(aq) + 2e–
Hence a blue colour will appear at X after a short time.
Nail Y is connected to zinc, a metal more reactive than iron. Zinc corrodes instead of iron. Hence NO
blue colour appears at Y after a short time.
183
33 B Potassium hexacyanoferrate(III) gives a blue colour in the presence of iron(II) ions. Phenolphthalein gives
a pink colour in the presence of excess hydroxide ions.
In dish 1, wrapping with a copper strip speeds up the rusting process of the iron nail.
During rusting, iron atoms first lose electrons to form iron(II) ions.
Fe(s)
Fe2+(aq) + 2e–
The electrons then combine with oxygen and water to form hydroxide ions.
1
O2(g) + H2O(l) + 2e–
2OH–(aq)
2
Hence a blue colour would develop around the iron nail and a pink colour would develop around the
copper strip.
34 C In dish 1 and dish 3, copper and silver are less reactive than iron. Both of them speed up the rusting
process of the iron nail.
In dish 2 and dish 4, magnesium and zinc are more reactive than iron. Both of them corrode instead of
iron. They protect iron from rusting via sacrificial protection.
35 C Option B — Tin is less reactive than iron.
36 D
37 C Using stainless steel is the most expensive way to prevent rusting. However, this method is the best
because no maintenance is needed.
Medical instruments require very good protection from rusting. It is worthwhile to use stainless steel to
make these instruments.
38 A Option A — Grease can prevent oxygen and water from reaching the iron. It is used to protect the chain
of a bicycle which cannot be painted. Grease serves as a lubricant also.
Option C — A layer of paint prevents both oxygen and water from reaching the iron beneath. However,
as soon as the paint is scratched, the metal is exposed to air, and rusting starts.
TDSBUDIFE
QBJOU
JSPO
QBJOU
PYZHFOBOEXBUFS
DBOOPUSFBDIJSPO
OPSVTUJOHPDDVST
PYZHFOBOEXBUFS
SFBDIJSPOSVTUJOH
PDDVST
Therefore painting is NOT suitable for protecting the moving parts of machines, e.g. the
chain of a bicycle.
Option D — It is NOT feasible to attach a more reactive metal to the chain of a bicycle for rust
prevention.
184
39 D Option A — Coat hanger is NOT protected by greasing as grease on the hanger would be messy.
Option B — Galvanized iron is NOT used for making food cans because zinc ions are poisonous.
Option C — Lead is less reactive than iron. It CANNOT provide sacrificial protection.
Option D — A car body is protected from rusting by connecting it to the negative terminal of the car
battery. The method is called impressed current cathodic protection.
m
QSPUFDUFENFUBM
DBUIPEF
QPXFS
TVQQMZ
BDPOEVDUPS
BOPEF
40 A Option A — Copper is less reactive than iron. It CANNOT provide sacrificial protection.
41 D When aluminium reacts with oxygen in the air, an even coating of aluminium oxide forms. This oxide
layer sticks to the surface of the metal and is impermeable to oxygen and water. It protects the metal
beneath from further attack.
42 D Option A — Anodization CANNOT increase the strength of aluminium.
Option D — During anodization, oxygen produced at the positive electrode reacts with the aluminium
object to increase the thickness of the oxide layer on the object.
Experimental set-up for aluminium anodization:
7
BMVNJOJVNTIFFU
BTUIFOFHBUJWFFMFDUSPEF
EJMVUFTVMQIVSJDBDJE
BMVNJOJVNPCKFDUUPCFBOPEJ[FE
BTUIFQPTJUJWFFMFDUSPEF
43 B During the anodization process, oxygen produced at the positive electrode reacts with the aluminium
object to increase the thickness of the oxide layer on the object.
185
44 B (1) The chemical formula of rust is Fe2O3•xH2O.
(3) The rust is not firmly attached to the iron surface. It tends to fall off as it forms. A fresh iron surface is
then exposed and so the rusting goes on. This eventually causes structural weakness and disintegration
of the metal.
45 A (1) The iron nail is in contact with both water and oxygen (dissolved in the sea water). Therefore it will
rust.
(2) Calcium sulphate is NOT a drying agent. The iron nail is in contact with both oxygen and moisture.
Therefore it will rust. Furthermore, the acidic car exhaust gas speeds up the rusting process.
(3) The iron nail is in contact with water only. Therefore it will NOT rust.
46 D (3) Attaching iron to tin (a less reactive metal) speeds up the rusting process.
47 C (1) Anodization is a process used to increase the corrosion resistance of aluminium.
48 B (2) Anodization CANNOT improve the strength of aluminium.
49 B (1) Iron is stronger than aluminium.
(3) Iron is cheaper than aluminium.
50 C An alloy is a mixture of a metal with one or more other elements.
(1) During aluminium anodization, the thickness of the aluminium oxide layer on an aluminium object is
increased. Anodized aluminium is NOT an alloy.
(2) Bronze is an alloy of copper and tin.
(3) 18-carat gold contains 18 parts by mass of gold in 24 parts. Carat gold is an alloy of gold, silver and
copper.
51 A At a higher temperature, chemical reaction becomes faster. The rusting process becomes faster too.
52 D Iron coated with zinc is called galvanized iron. Galvanized iron is NOT used to make food cans because
zinc ions are poisonous.
Tin is less reactive than iron. It CANNOT provide sacrificial protection.
53 D Steel structures of a pier may be protected from rusting by connecting to the negative terminal of a
battery. This supplies electrons to the iron and prevents the formation of iron(II) ions. This method is called
impressed current cathodic protection.
54 B Stainless steel is seldom used to make large structures because it is expensive.
55 C Anodization CANNOT increase the strength of aluminium.
186
Part B
Topic-based exercise
Multiple choice questions
1
A
2
A A fuse is a safety device that protects electric circuits from the effects of excessive electric currents. It
commonly consists of a current-conducting wire of low melting point. If we try to pass a current higher
than the rated value through the wire, it will heat up so much that it melts. When it melts, it breaks the
circuit it is fitted to and stops the current flowing.
3
C Option A — Duralumin is an alloy of aluminium containing copper, magnesium and manganese.
Option B — Solder is an alloy of lead and tin.
Option D — The casing of a zinc-carbon dry cell is made of zinc.
4
C The removal of oxygen from a metal oxide is called reduction.
When the oxides of copper and zinc are heated with carbon, carbon can remove oxygen from the oxides.
The oxides are reduced.
5
A In pure copper, the layers of atoms can slide over each other easily.
Mixing copper with zinc causes a distortion to the regular arrangement of the copper atoms. So, it is
difficult for the layers of atoms to slide over each other.
6
A
7
B Aluminium is difficult to extract. It was discovered after the invention of electrolysis in 1800.
8
C Lead does NOT react with steam.
9
D Option D — Zinc oxide forms when zinc is heated in air.
zinc + oxygen
zinc oxide
10 D Options A and C — The water tank could NOT be made of calcium and potassium because both of them
react with water.
Option B — The water tank could NOT be made of copper because copper does not react with dilute
sulphuric acid.
11 A Only Z reacts with water. Therefore Z is the most reactive.
Metals near to the bottom of the reactivity series can be extracted from their ores by heating with carbon.
Only X can be extracted by heating its oxide with carbon powder. Therefore X is the least reactive.
∴ the order of increasing reactivity of the metals is X < Y < Z.
12 C X and Y react with cold water but only Y is stored in paraffin oil. Therefore Y is the most reactive.
Z does not react with water. Therefore Z is the least reactive.
∴ the order of decreasing reactivity of the metals is Y > X > Z.
187
13 B Option A — Aluminium does NOT react with cold water.
Option B — Calcium reacts with cold water. This fits with the description of X.
Option C — Copper does NOT react with cold water.
Option D — Zinc does NOT react with cold water.
14 D
Option
Initial rate of
hydrogen formation
A
Mg and H2O
B
C
D
Initial rate of
hydrogen formation
Remark
<
Ba and HCl
The reactivity of Group II elements
increases down the group. Hence barium
(Ba) is more reactive than magnesium (Mg).
Fe and HCl
<
K and H2O
Potassium (K) reacts with water explosively
while iron (Fe) reacts with dilute
hydrochloric acid readily.
K and H2O
<
Rb and H2O
The reactivity of Group I elements increases
down the group. Hence rubidium (Rb) is
more reactive than potassium (K).
Mg and HCl
The reactivity of Group I elements increases
down the group. Hence rubidium (Rb)
is more reactive than sodium. Sodium
is more reactive than magnesium (Mg).
Thus rubidium (Rb) is more reactive than
magnesium (Mg). Thus the comparison for
the initial rate of hydrogen formation is
correct.
Rb and HCl
>
15 D Option A — X is stored in paraffin oil while Y is not. This shows that X is more reactive than Y.
Option B — X reacts with dilute hydrochloric acid but Y does not. This shows that X is more reactive
than Y.
Option C — X is a Group I element while Y is a Group II element. Group I elements are more reactive
than Group II elements.
Option D — The oxide of a metal lower in the reactivity series has a lower stability. Thus the reduction
of the oxide of the metal is easier.
The oxide of X gives X when heated with carbon while the oxide of Y does not. This shows
that Y is more reactive than X.
16 C Follow the steps below when writing the balanced chemical equation:
188
Write down the chemical formulae of the reactants and products.
H2 + Fe3O4
Fe + H2O
To balance the iron atom, put the coefficient ‘3’ before Fe.
H2 + Fe3O4
3Fe + H2O
To balance the oxygen atom, put the coefficient ‘4’ before H2O.
H2 + Fe3O4
3Fe + 4H2O
To balance the hydrogen atom, put the coefficient ‘4’ before H2.
Now the chemical equation is balanced.
4H2 + Fe3O4
3Fe + 4H2O
(Balanced)
17 C Follow the steps below when writing a balanced chemical equation for the reaction between aqua-regia
and gold.
Write down the chemical formulae of the reactants
and products.
Au + HNO3 + HCl
To balance the chlorine atom, put the coefficient ‘4’
before HCl.
Au + HNO3 + 4HCl
HAuCl4 + H2O + NO2
HAuCl4 + H2O + NO2
Consider the following equation:
Au + wHNO3 + 4HCl
HAuCl4 + yH2O + zNO2
Number of hydrogen atoms on the reactant side = w + 4
= number of hydrogen atoms on the product side
= 1 + 2y
∴ y =
w + 3
2
Number of oxygen atoms on the reactant side = 3w
= number of oxygen atoms on the product side
= y + 2z
∴
w + 3
2
5w – 3 = 4z
3w =
+ 2z
The answers in option C (i.e. w = 3, y = 3 and z = 3) fit this equation.
18 B S is a precipitate. Hence (x) is (s).
H2O is water. Hence (y) is (l).
19 C Options A, B and D — Iron, magnesium and zinc are more reactive than copper. Copper CANNOT displace
them from solutions of their compounds.
20 B The reaction in the test tube can be represented by the following chemical equation:
Cu(s) + 2AgNO3(aq)
Cu(NO3)2(aq) + 2Ag(s)
Silver nitrate solution is colourless while copper(II) nitrate solution is blue. Therefore the solution in the
test tube changes from colourless to greenish blue.
21 D Option D — Magnesium displaces copper from copper(II) nitrate solution.
22 B Option B — Zinc displaces copper from copper(II) sulphate solution. Hence copper(II) sulphate solution
CANNOT be stored in a zinc-plated watering can.
23 B Magnesium displaces zinc from zinc nitrate solution. Magnesium CANNOT displace calcium from calcium
nitrate solution.
189
24 D X reacts with the solution of sulphate of Z according to the following equation:
2+
X(s) + Z (aq)
2+
X (aq) + Z(s)
X displaces Z from the solution of sulphate of Z. Hence X is more reactive than Z.
Option A — X and Z may NOT be able to react with dilute hydrochloric acid.
Option B — X and Z may NOT be Group II elements.
Option C — X is more reactive than Z.
Option D — X is more reactive than Z. Hence atoms of X lose electrons to form cations more readily.
25 D The more reactive a metal, the more difficult to extract and the more recent the year of discovery.
26 B Y is the most reactive as only it gives a gas (hydrogen) with the water in the silver nitrate solution.
X is more reactive than Z as X displaces silver from silver nitrate solution but Z does not.
∴ the descending order of reactivity of the metals is Y > X > Z.
27 D The chemical formula of chromium(III) sulphate is Cr2(SO4)3. One formula unit of Cr2(SO4)3 contains 5 ions.
Hence 1 mole of Cr2(SO4)3 contains 5 moles of ions.
28 D The chemical formula of sodium chloride is NaCl while that of sodium sulphate is Na2SO4.
Number of moles of Cl– ions in the mixture = 4 mol
∴ number of moles of Na+ ions in NaCl = 4 mol
Number of moles of SO42– ions in the mixture = 3 mol
∴ number of moles of Na+ ions in Na2SO4 = 2 x 3 mol
= 6 mol
∴ total number of Na+ ions in the mixture = (4 + 6) mol
= 10 mol
29 B
Option
Substance
No. of moles of
substance present
No. of moles of
ions in 1 mole of
substance
No. of moles of
ions present
A
iron(III) sulphate Fe2(SO4)3
4
5
20
B
calcium phosphate Ca3(PO4)2
5
5
25
C
barium hydrogencarbonate
Ba(HCO3)2
6
3
18
D
magnesium hydroxide Mg(OH)2
7
3
21
∴ 5 moles of calcium phosphate have the greatest number of moles of ions, i.e. the greatest number
of ions.
190
30 A Molar mass of C60 = (60 x 12.0) g mol–1
= 720 g mol–1
mass of C60
molar mass of C60
12 g
=
720 g mol–1
Number of moles of C60 =
= 0.017 mol
Number of C60 molecules = number of moles of C60 x L
= 0.017 mol x 6.02 x 1023 mol–1
= 1.0 x 1022
31 C 1 mole of glucose contains 6 moles of oxygen atoms, i.e. 6L oxygen atoms.
∴ number of moles of glucose containing 1 oxygen atom =
Number of moles of glucose containing x oxygen atoms =
1
mol
6L
x
mol
6L
32 D 1 mole of Cl2O contains 3 moles of atoms, i.e. n atoms.
∴ 3 moles of atoms = n atoms
n
atoms
3
1 mole of Cl2O7 contains 9 moles of atoms.
1 mole of atoms =
∴ 2 moles of Cl2O7 contains 2 x 9 x
n
atoms, i.e. 6n atoms.
3
33 C Molar mass of H2O = (2 x 1.0 + 16.0) g mol–1
= 18.0 g mol–1
54.0 g of H2O contain 3 moles of H2O.
Option A — 3 moles of water contain 3 moles of molecules.
Option B — 1 water molecule contains 3 atoms. Hence 3 moles of water molecules contain 9 moles of
atoms, i.e. 9 x 6.02 x 1023 atoms.
Option C — 1 water molecule contains 2 hydrogen atoms. Hence 3 moles of water molecules contain
6 moles of hydrogen atoms.
Option D — 1 water moleccule contains 1 oxygen atom. Hence 3 moles of water molecules contain 3
moles of oxygen atoms, i.e. 3 x 6.02 x 1023 oxygen atoms.
number of carbon atoms
L
2 x 1022
=
6.02 x 1023 mol–1
34 C Number of moles of carbon atoms =
= 0.033 mol
Mass of carbon atoms = number of moles of carbon atoms x molar mass of carbon
= 0.033 mol x 12.0 g mol–1
= 0.40 g
191
35 D
Mass of
Substance Molar mass of substance substance
present
Option
No. of moles
of substance
present
No. of moles
of atoms in
1 mole of
substance
No. of moles
of atoms
present
—
sulphur
dioxide
SO2
(32.1 + 2 x 16.0) g mol–1
= 64.1 g mol–1
6.41 g
6.41 g
64.1 g mol–1
= 0.100 mol
3
3 x 0.100 mol
= 0.300 mol
A
ammonia
NH3
(14.0 + 3 x 1.0) g mol–1
= 17.0 g mol–1
1.70 g
17.0 g
17.0 g mol–1
= 0.100 mol
4
4 x 0.100 mol
= 0.400 mol
B
carbon
dioxide
CO2
(12.0 + 2 x 16.0) g mol–1
= 44.0 g mol–1
2.20 g
2.20 g
44.0 g mol–1
= 0.0500 mol
3
3 x 0.0500 mol
= 0.150 mol
C
hydrogen
chloride
HCl
(1.0 + 35.5) g mol–1
= 36.5 g mol–1
3.65 g
3.65 g
36.5 g mol–1
= 0.100 mol
2
2 x 0.100 mol
= 0.200 mol
D
nitrogen
dioxide
NO2
(14.0 + 2 x 16.0) g mol–1
= 46.0 g mol–1
4.60 g
4.60 g
46.0 g mol–1
= 0.100 mol
3
3 x 0.100 mol
= 0.300 mol
∴ 4.60 g of nitrogen dioxide and 6.41 g of sulphur dioxide contain the same number of moles of atoms,
i.e. they contain the same number of atoms.
36 C Molar mass of X2 = (2 x 19.0) g mol–1
–1
= 38.0 g mol
mass of X2
molar mass of X2
76.0 g
=
–1
38.0 g mol
Number of moles of molecules =
= 2.00 mol
Number of molecules present = number of moles of molecules x L
= 2.00 mol x L mol–1
= 2L
37 A
Option
Gas
Molar mass of gas
A
fluorine F2
(2 x 19.0) g mol–1 = 38.0 g mol–1
B
nitrogen N2
(2 x 14.0) g mol–1 = 28.0 g mol–1
C
oxygen O2
(2 x 16.0) g mol–1 = 32.0 g mol–1
D
hydrogen H2
–1
(2 x 1.0) g mol
–1
= 2.0 g mol
The molar mass of F2 is the greatest. Hence 100 g of F2 contain the smallest number of moles of molecules,
i.e. the smallest number of molecules.
192
38 C Formula mass of FeSO4•xH2O = (55.8 + 32.1 + 4 x 16.0) + x(2 x 1.0 + 16.0)
= 151.9 + 18x
1 mole of FeSO4•xH2O contain x moles of H2O.
i.e. (151.9 + 18x) g of FeSO4•xH2O contain 18x g of H2O.
18.1 g of FeSO4•xH2O contain 8.20 g of H2O.
8.20 g
18x g
=
18.1 g
(151.9 + 18x) g
x = 7
39 C For 100 g of the oxide, there are 81.6 g of X and 18.4 g of oxygen.
X
Oxygen
81.6 g
18.4 g
35.5
16.0
81.6 g
= 2.30 mol
35.5 g mol–1
18.4 g
= 1.15 mol
16.0 g mol–1
Mass of element in the compound
Relative atomic mass
Number of moles of atoms that
combine
2.30 mol
1.15 mol
Mole ratio of atoms
= 2
1.15 mol
1.15 mol
= 1
40 B Relative molecular mass of CO2 = (12.0 + 2 x 16.0)
= 44.0
2 x 16.0
x 100%
44.0
= 72.7%
Percentage by mass of oxygen in CO2 =
Mass of oxygen in 3.29 g of CO2 = 3.29 g x 72.7%
= 2.39 g
Mass of M in 11.9 g of the oxide = (11.9 – 2.39) g
= 9.5 g
M
Oxygen
Mass of element in the compound
9.5 g
2.39 g
Relative atomic mass
63.5
16.0
9.5 g
= 0.15 mol
–1
63.5 g mol
2.39 g
= 0.149 mol
16.0 g mol–1
0.15 mol
= 1
0.149 mol
0.149 mol
= 1
0.149 mol
Number of moles of atoms that
combine
Mole ratio of atoms
∴ the empirical formula of the oxide is MO.
193
41 A Let y% be the percentage by mass of MnO2 in the sample.
For 100 g of the sample, there are (100 – y) g of MnSiO3 and y g of MnO2.
Mass of Mn in the sample = 49.2 g
Mass of Mn in (100 – y) g of MnSiO3 = (100 – y) x
Mass of Mn in y g of MnO2 =
∴ (100 – y) x
54.9
+
131.0
54.9
86.9
54.9
86.9
54.9
131.0
y
y = 49.2
y = 34.3
42 B Mass of oxygen in the oxide = (6.10 – 4.53) g
= 1.57 g
X
Oxygen
4.53 g
1.57 g
69.7
16.0
4.53 g
= 0.0650 mol
–1
69.7 g mol
1.57 g
= 0.0981 mol
16.0 g mol–1
0.0650 mol
= 1
0.0650 mol
0.0981 mol
= 1.51
0.0650 mol
2 x 1 = 2
2 x 1.51 = 3
Mass of element in the compound
Relative atomic mass
Number of moles of atoms that
combine
Mole ratio of atoms
Simplest whole number ratio of
atoms
∴ the empirical formula of the oxide is X2O3.
43 C 2NaN3(s)
2Na(s) + 3N2(g)
According to the above equation, 1.00 mole of NaN3 decomposes to give 1.00 mole of Na and 1.50
moles of N2.
10Na(s) + 2KNO3(s)
N2(g) + 5Na2O(s) + K2O(s)
According to the above equation, 1.00 mole of Na reacts to give 0.10 mole of N2.
∴ total number of N2 produced by 1.00 mole of NaN3 = (1.50 + 0.10) mol
= 1.60 mol
44 A The total mass of the reactants equals that of the products.
∴ s = p + q – r
194
45 A 2Ca(NO3)2(s)
82.0 g
2CaO(s) + 4NO2(g) + O2(g)
? g
–1
Molar mass of Ca(NO3)2 = [40.1 + 2 x (14.0 + 3 x 16.0)] g mol
= 164.1 g mol–1
mass of Ca(NO3)2
molar mass of Ca(NO3)2
82.0 g
=
–1
164.1 g mol
Number of moles of Ca(NO3)2 =
= 0.500 mol
According to the equation, 2 moles of Ca(NO3)2 give 1 mole of O2 upon heating.
0.500
mol
2
= 0.250 mol
∴ number of moles of O2 obtained =
Molar mass of O2 = (2 x 16.0) g mol–1
= 32.0 g mol–1
Mass of O2 obtained = number of moles of O2 x molar mass of O2
= 0.250 mol x 32.0 g mol–1
= 8.00 g
46 B 2LiOH(s) + CO2(g)
? g
11.0 g
Li2CO3(s) + H2O(l)
Molar mass of CO2 = (12.0 + 2 x 16.0) g mol–1
= 44.0 g mol–1
mass of CO2
molar mass of CO2
11.0 g
=
–1
44.0 g mol
Number of moles of CO2 =
= 0.250 mol
According to the equation, 2 moles of LiOH can absorb 1 mole of CO2.
∴ number of moles of LiOH needed = 2 x 0.250 mol
= 0.500 mol
Molar mass of LiOH = (6.9 + 16.0 + 1.0) g mol–1
= 23.9 g mol–1
Mass of LiOH needed = number of moles of LiOH x molar mass of LiOH
= 0.500 mol x 23.9 g mol–1
= 12.0 g
195
47 C Upon heating, NaHCO3 decomposes according to the following equation:
2NaHCO3(s)
33.6 g
Na2CO3(s) + H2O(l) + CO2(g)
? g
Molar mass of NaHCO3 = (23.0 + 1.0 + 12.0 + 3 x 16.0) g mol–1
= 84.0 g mol–1
mass of NaHCO3
molar mass of NaHCO3
33.6 g
=
–1
84.0 g mol
= 0.400 mol
Number of moles of NaHCO3 =
According to the equation, 2 moles of NaHCO3 decompose to form 1 mole of Na2CO3.
0.400
mol
2
= 0.200 mol
∴ number of moles of Na2CO3 formed =
Molar mass of Na2CO3 = (2 x 23.0 + 12.0 + 3 x 16.0) g mol–1
= 106.0 g mol–1
Mass of Na2CO3 formed = number of moles of Na2CO3 x molar mass of Na2CO3
= 0.200 mol x 106.0 g mol–1
= 21.2 g
48 D The chemical formula of the oxide of M is MO. The oxide is reduced by hydrogen according to the
following equation:
MO(s) + H2(g)
13.4 g
M(s) + H2O(l)
1.08 g
Let m be the relative atomic mass of M.
Molar mass of H2O = (2 x 1.0 + 16.0) g mol–1
= 18.0 g mol–1
mass of H2O
molar mass of H2O
1.08 g
=
–1
18.0 g mol
= 0.0600 mol
Number of moles of H2O =
According to the equation, 1 mole of MO produces 1 mole of H2O upon reduction.
∴ number of moles of MO reduced = 0.0600 mol
Mass of MO = number of moles of MO x molar mass of MO
mass of MO
number of moles of MO
13.4 g
=
0.0600 mol
= 223 g mol–1
Molar mass of MO =
= (m + 16.0) g mol–1
196
∴ m = 207
49 C CO2(g) + 3H2(g)
6.60 g 1.20 g
CH3OH(g) + H2O(g)
Number of moles of CO2 present =
=
mass of CO2
molar mass of CO2
6.60 g
–1
44.0 g mol
= 0.150 mol
mass of H2
molar mass of H2
1.20 g
–1
2.0 g mol
Number of moles of H2 present =
=
= 0.600 mol
According to the equation, 1 mole of CO2 reacts with 3 moles of H2 to produce 1 mole of CH3OH. During
the reaction, 0.150 mole of CO2 reacts with 0.450 mole of H2. Therefore H2 is in excess. The amount of
CO2 limits the amount of CH3OH obtained.
Number of moles of CH3OH obtained = 0.150 mol
Molar mass of CH3OH = (12.0 + 4 x 1.0 + 16.0) g mol–1
–1
= 32.0 g mol
Mass of CH3OH obtained = number of moles of CH3OH x molar mass of CH3OH
= 0.150 mol x 32.0 g mol–1
= 4.80 g
50 B 2NH3(g) + 3CuO(s)
6.80 g
35.8 g
N2(g) + 3H2O(l) + 3Cu(s)
Number of moles of NH3 present =
=
mass of NH3
molar mass of NH3
6.80 g
–1
17.0 g mol
= 0.400 mol
mass of CuO
molar mass of CuO
35.8 g
=
79.5 g mol–1
Number of moles of CuO present =
= 0.450 mol
According to the equation, 2 moles of NH3 react with 3 moles of CuO to produce 1 mole of N2. During
the reaction, 0.450 mole of CuO reacts with 0.300 mole of NH3. Therefore NH3 is in excess. The amount
of CuO limits the amount of N2 obtained.
0.450
mol
3
= 0.150 mol
Number of moles of N2 obtained =
197
Molar mass of N2 = (2 x 14.0) g mol–1
= 28.0 g mol–1
Mass of N2 obtained = number of moles of N2 x molar mass of N2
= 0.150 mol x 28.0 g mol–1
= 4.20 g
51 D 2Al(s) + 3Cl2(g)
8.10 g
2AlCl3(s)
38.2 g
mass of Al
molar mass of Al
8.10 g
–1
27.0 g mol
Number of moles of Al =
=
= 0.300 mol
According to the equation, 2 moles of Al react with Cl2 to produce 2 moles of AlCl3.
∴ number of moles of AlCl3 obtained = 0.300 mol
Molar mass of AlCl3 = (27.0 + 3 x 35.5) g mol–1
= 133.5 g mol–1
Theoretical yield of AlCl3 = number of moles of AlCl3 x molar mass of AlCl3
= 0.300 mol x 133.5 g mol–1
= 40.1 g
38.2 g
x 100%
40.1 g
= 95.3%
Percentage yield of AlCl3 =
52 D Silver oxide decomposes according to the following equation when heated:
2Ag2O(s)
8.90 g
4Ag(s) + O2(g)
7.75 g
mass of Ag
molar mass of Ag
7.75 g
=
–1
107.9 g mol
Number of moles of Ag =
= 0.0718 mol
According to the equation, 2 moles of Ag2O give 4 moles of Ag when heated.
0.0718
mol
2
= 0.0359 mol
∴ number of moles of Ag2O =
Molar mass of Ag2O = (2 x 107.9 + 16.0) g mol–1
= 231.8 g mol–1
198
Mass of Ag2O = number of moles of Ag2O x molar mass of Ag2O
= 0.0359 mol x 231.8 g mol–1
= 8.32 g
8.32 g
x 100%
8.90 g
= 93.5%
Percentage by mass of Ag2O in the sample =
53 D CaCO3(s) + 2HCl(aq)
Na2CO3(s) + 2HCl(aq)
CaCl2(aq) + CO2(g) + H2O(l)
2NaCl(aq) + CO2(g) + H2O(l)
1.0 mole of CaCO3 requires 2.0 moles of HCl to liberate all the CO2.
1.0 mole of Na2CO3 requires 2.0 moles of HCl to liberate all the CO2.
∴ number of moles of HCl required = (2.0 + 2.0) mol
= 4.0 mol
54 D Let m be the relative atomic mass of X.
Suppose 100 g of X and Zn react with excess dilute sulphuric acid separately.
X(s) + 2H+(aq)
X2+(aq) + H2(g)
Zn(s) + 2H+(aq)
Zn2+(aq) + H2(g)
Number of moles of X in 100 g =
=
mass of X
molar mass of X
100 g
–1
m g mol
Number of moles of Zn in 100 g =
=
mass of Zn
molar mass of Zn
100 g
–1
65.4 g mol
According to the equations, 1 mole of X and Zn gives 1 mole of H2 respectively when reacted with excess
dilute sulphuric acid.
Number of moles of H2 given by X =
Number of moles of H2 given by Zn =
100 g
m g mol–1
100 g
–1
65.4 g mol
X gives more hydrogen than Zn does.
i.e.
100 g
m g mol–1
>
100 g
–1
65.4 g mol
It can be deduced that m < 65.4
i.e. the relative atomic mass of X is smaller than that of Zn.
199
55 A CaCO3(s) + 2HCl(aq)
CaCl2(aq) + CO2(g) + H2O(l)
According to the equation, 1 mole of pure CaCO3 (i.e. 100.1 g of pure CaCO3) reacts with excess HCl
to give 1 mole of CO2 (i.e. 44 g of CO2).
1 mole of impure CaCO3 gives 46 g of CO2, i.e. more than 1 mole of CO2.
It can be concluded that 100 g of impure sample contain more than 1 mole of carbonate. Hence the
formula mass of the impurity should be smaller than that of CaCO3. Only MgCO3 (option A) fits this.
56 C Option C — Acidic gas from a factory speeds up the rusting process.
57 B Option B — Connecting to copper (a metal less reactive than iron) speeds up the rusting process of the
iron nail.
58 C Option C — Apply grease to a door hinge to prevent it from rusting.
59 D Option D — The oxide layer on the aluminium foil prevents any further reaction.
60 D (1) Nickel-cadmium rechargeable cells are used to power professional cameras.
(2) Stainless steel is an alloy of iron containing chromium and nickel.
(3) Titanium is used to make supersonic aircraft because it is light but very strong, resists corrosion, and
has a very high melting point.
61 C (1) Titanium is used to make replacement hip joints because it is light and strong. Its density is
4.5 g cm–3.
62 D (3) Brass is an alloy of copper containing zinc.
In pure copper, the layers of atoms can slide over each other easily.
Adding zinc to copper causes a distortion to the regular arrangement of the copper atoms. It is difficult
for the layers of atoms to slide over each other. Hence brass is stronger than pure copper.
63 A (2) Sodium reacts with water to give sodium hydroxide.
2Na(s) + 2H2O(l)
2NaOH(aq) + H2(g)
The universal indicator turns blue / purple (alkaline colour).
(3) Sodium burns with a golden yellow colour.
64 A (1) Heating magnesium carbonate strongly gives magnesium oxide and carbon dioxide gas.
magnesium carbonate
heat
magnesium oxide + carbon dioxide
(2) Adding iron to dilute sulphuric acid gives hydrogen gas.
iron + dilute sulphuric acid
iron(II) sulphate + hydrogen
(3) Copper is unreactive. It does NOT react with dilute hydrochloric acid.
200
65 D (1) Iron pyrite is quite hard while gold is very soft.
(2) Iron pyrite is not malleable while gold is malleable.
(3) When iron pyrite (FeS2) is heated, iron(III) oxide is produced.
4FeS2(s) + 11O2(g)
2Fe2O3(s) + 8SO2(g)
Gold does NOT undergo any chemical change when heated.
66 A (2) Carbon CANNOT remove oxygen from magnesium oxide.
(3) Zinc is less reactive than magnesium. It CANNOT remove oxygen from magnesium oxide.
67 B (1) Iron corrodes after exposed to air. As X is slightly more reactive than iron, it probably also corrodes
after exposed to air.
(2) Iron reacts readily with dilute hydrochloric acid. As X is slightly more reactive than iron, it probably
reacts with dilute hydrochloric acid as well.
(3) At the temperature of Bunsen flame, carbon cannot remove oxygen from an oxide of iron. Therefore
carbon probably CANNOT remove oxygen from the oxide of X at this temperature.
68 B (1) Silver is less reactive than copper. There is NO reaction between silver and copper(II) sulphate
solution.
(2) Magnesium is more reactive than copper. It can remove oxygen from copper(II) oxide.
copper(II) oxide + magnesium
copper + magnesium oxide
(3) NO reaction occurs when copper(II) oxide and carbon are mixed without heating.
69 A (1) and (2) One aluminium atom forms one aluminium ion by losing three electrons.
Al
Al3+ + 3e–
Hence one mole of aluminium atoms can form one mole of Al3+ ions, i.e. 6.02 x 1023 Al3+
ions.
(3) One mole of aluminium contains the same number of atoms as 1 mole of sodium.
70 B Molar mass of H2O = (2 x 1.0 + 16.0) g mol–1
= 18.0 g mol–1
36.0 g of water contain 2 moles of H2O.
23
(1) 2 moles of H2O contain 2 x 6.02 x 10 molecules.
(2) One H2O molecule contains 3 atoms. 2 moles of H2O contain 6 moles of atoms, i.e. 6 x 6.02 x 1023
atoms.
(3) One H2O molecule contains 2 hydrogen atoms. 2 moles of H2O contain 4 moles of hydrogen
atoms.
71 B (1) The magnesium strip provides sacrificial protection. Therefore the iron nail will NOT rust.
(2) Silver is less reactive than iron. Attaching to the silver strip speeds up the rusting process.
(3) The copper layer prevents both oxygen and water from reaching the iron nail. Therefore the iron nail
will NOT rust.
201
72 C (1) It is NOT feasible to prevent a car body from rusting by applying a layer of grease. This is because
grease on the car body would be messy.
(2) The car body can be protected by connecting it to the negative terminal of the car battery. This
supplies electrons to the car body and prevents the formation of iron(II) ions.
(3) A layer of paint prevents both oxygen and water from reaching the car body. This can prevent the
car body from rusting.
73 A
74 C Copper was the first metal extracted from its ore. It was used easlier than iron in the history.
Iron is the second most abundant metal in the Earth’s crust, more abundant than copper.
75 B Suppose 10 g of Ca and Zn react with excess dilute HCl separately.
Ca(s) + 2HCl(aq)
CaCl2(aq) + H2(g)
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
Number of moles of Ca in 10 g =
=
mass of Ca
molar mass of Ca
10 g
–1
40.1 g mol
= 0.249 mol
Number of moles of Zn in 10 g =
=
mass of Zn
molar mass of Zn
10 g
–1
65.4 g mol
= 0.153 mol
According to the equations, 1 mole of Ca / Zn produces 1 mole of H2 with excess dilute HCl.
10 g of Ca produce 0.249 mole of hydrogen while 10 g of Zn produce 0.153 mole of hydrogen. Hence
Ca produces a greater amount of hydrogen than Zn does because the relative atomic mass of Ca is
smaller than that of Zn.
76 C Suppose 10 g of Na and Li are added to water separately.
2Na(s) + 2H2O(l)
2Li(s) + 2H2O(l)
2NaOH(aq) + H2(g)
2LiOH(aq) + H2(g)
Number of moles of Na in 10 g =
=
mass of Na
molar mass of Na
10 g
–1
23.0 g mol
= 0.435 mol
202
Number of moles of Li in 10 g =
=
mass of Li
molar mass of Li
10 g
–1
6.9 g mol
= 1.45 mol
According to the equations, 2 moles of Na / Li produce 1 mole of H2 with water.
0.435
1.45
mole of hydrogen while 10 g of Li produce
mole of hydrogen. Hence
2
2
Li produces a greater amount of hydrogen than Na does because the relative atomic mass of Li is smaller
10 g of Na produce
than that of Na.
The reactivity of Group I elements increases down the group. Hence Na is more reactive than Li.
77 B Copper can displace silver from silver nitrate solution because it is more reactive than silver.
There is no relationship between the fact ‘copper can form an ion with a charge of +2 while silver can
form an ion with a charge of +1’ and the reactivity of the metals.
78 C 1 mole of O3(g) contains the same number of molecules as 1 mole of O2(g).
1 molecule of O3(g) contains 3 atoms while 1 molecule of O2(g) contains 2 atoms. Hence the number of
atoms in 1 mole of O3(g) is greater than that in 1 mole of O2(g).
79 A The rusting process speeds up if the water present contains ionic substances such as sodium chloride.
Hence a seaside environment speeds up the rusting process since the thin water layer on the iron surface
contains dissolved sodium chloride.
80 C A layer of paint prevents both oxygen and water from reaching the iron beneath. However, as soon as
the paint is scratched, the metal is exposed to air, and rusting starts. Therefore painting is NOT suitable
for protecting the moving parts of machines.
Short questions
81 a) B
(1)
Any two of the following:
High corrosion resistance (1) / high mechanical strength (1) / high density (1)
b) C
(1)
Any two of the following:
Low cost (1) / high mechanical strength (1) / medium density (1)
c) D
(1)
Medium price
(1)
Good conductivity of heat
(1)
203
82 a) 2Mg(s) + O2(g)
2MgO(s)
(1)
b) 4Al(s) + 3O2(g)
2Al2O3(s)
(1)
c) 2Na(s) + 2H2O(l)
2NaOH(aq) + H2(g)
d) Zn(s) + H2O(g)
ZnO(s) + H2(g)
e) Ca(s) + 2HCl(aq)
f) 2Ag2O(s)
CaCl2(aq) + H2(g)
4Ag(s) + O2(g)
g) 2CuO(s) + C(s)
2Cu(s) + CO2(g)
(1)
(1)
(1)
(1)
(1)
h) Fe2O3(s) + 2Al(s)
2Fe(s) + Al2O3(s)
(1)
83 a) Mg(s) + 2H+(aq)
Mg2+(aq) + H2(g)
(1)
b) Zn(s) + 2H+(aq)
Zn2+(aq) + H2(g)
(1)
c) Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
(1)
d) Cu(s) + 2Ag+(aq)
Cu2+(aq) + 2Ag(s)
84 a) Sodium burns vigorously with a golden yellow flame. / A white smoke forms.
4Na(s) + O2(g)
2Na2O(s)
b) Gas bubbles are given off. / Magnesium dissolves in the acid.
Mg(s) + H2SO4(aq)
MgSO4(aq) + H2(g)
(1)
(1)
(1)
(1)
(1)
c) Potassium melts to form a silvery bead. / The bead moves rapidly on the water surface. / Potassium burns
with a lilac flame.
(1)
2K(s) + 2H2O(l)
2KOH(aq) + H2(g)
d) Calcium burns with a brick-red flame. / A white powder forms.
2Ca(s) + O2(g)
2CaO(s)
e) Calcium sinks in water. / A steady steam of bubbles forms. / The water becomes milky.
Ca(s) + 2H2O(l)
Ca(OH)2(s) + H2(g)
(1)
(1)
(1)
(1)
(1)
f) The zinc metal slowly becomes coated with a brown layer. / The blue colour of the solution fades
gradually.
(1)
Zn(s) + CuSO4(aq)
204
ZnSO4(aq) + Cu(s)
(1)
85
Substance
Mass of substance
present (g)
Molar mass of
–1
substance (g mol )
117
18.0
Water
H2O
(1)
Number of moles of
Number of particles
substance present
present
(mol)
(1)
3.91 x 1024
molecules
6.50
24
(1)
159.6
(1)
4.00
(1)
2.41 x 10
units
346
138.2
(1)
2.50
(1)
1.51 x 10 formula
(1)
units
61.6
44.0
(1)
1.40
(1)
8.43 x 1023
molecules
Iron(III) oxide
Fe2O3
638
Potassium carbonate
K2CO3
Carbon dioxide
CO2
(1)
formula
24
(1)
Structured questions
86 a) Rocks from which we obtain metals are called ores.
b) i) Reduction
ii) Fe2O3(s) + 3CO(g)
(1)
(1)
2Fe(s) + 3CO2(g)
(1)
c) In pure iron, all the atoms are of the same size. Hence the layers of atoms are able to slide over each
other easily.
(1)
In the iron alloy, larger metal atoms are present. They cause a distortion to the regular arrangement of
the iron atoms.
(1)
So, it is difficult for the layers of atoms to slide over each other.
d) i) Metal resources can be saved.
(1)
(1)
Fuel can be saved.
(1)
Environmental impact due to metal waste and mining can be reduced.
(1)
ii) Any two of the following:
The cost is high. (1) / It is difficult to get people to save metal waste. (1) / It is difficult to sort the
metal waste. (1)
87 a) i) Rubidium sinks in water while potassium floats on water.
Rubidium reacts more vigorously than potassium.
ii) Rubidium is denser than water and potassium is less dense than water.
(1)
(1)
(1)
The outermost shell electron in a rubidium atom is further away from the nucleus than that in a
potassium atom.
There is less attraction between the nucleus and the outermost shell electron in a rubidium atom.(1)
205
Hence a rubidium atom lose an electron more readily than a potassium atom. Therefore rubidium is
more reactive than potassium.
(1)
iii) 2Rb(s) + 2H2O(l)
2RbOH(aq) + H2(g)
(1)
b) i) This prevents rubidium from reacting with moisture and oxygen in air.
ii) Explosive or flammable
c) i) Ca(s) + 2H2O(l)
(1)
(1)
Ca(OH)2(s) + H2(g)
(1)
ii)
IZESPHFO
XBUFS
JOWFSUFEGVOOFM
DBMDJVN
(1 mark for correct set-up; 1 mark for correct labels; 0 mark if the set-up is not workable)
iii) Potassium floats on water while calcium sinks.
Potassium burns with a lilac flame while calcium does not catch fire.
88 a) Z > Y > X
(2)
(1)
(1)
(1)
Only Z reacts with cold water. Therefore it is the most reactive.
(1)
Y reacts with dilute hydrochloric acid but X shows no reaction with dilute hydrochloric acid. Therefore Y
is more reactive than X.
(1)
b) i) Copper
(1)
ii) X / copper is produced.
2CuO(s) + C(s)
(1)
2Cu(s) + CO2(g)
c) i) Zinc
ii) Zn(s) + 2HCl(aq)
(1)
ZnCl2(aq) + H2(g)
d) i) Calcium
(1)
It gives a ‘pop’ sound when tested with a burning splint.
206
(1)
(1)
ii) Hydrogen
iii) Ca(s) + 2H2O(l)
(1)
Ca(OH)2(s) + H2(g)
(1)
(1)
iv)
IZESPHFO
XBUFS
JOWFSUFEGVOOFM
DBMDJVN;
(1 mark for correct set-up; 1 mark for correct labels; 0 mark if the set-up is not workable)
e) X was discovered earlier.
(1)
It is less reactive than Z and thus easier to extract.
89 a)
(2)
TUFBN
EBNQ
NJOFSBM
XPPM
(1)
NFUBM
EFMJWFSZ
UVCF
IZESPHFO
XBUFS
IFBU
USPVHI
(1 mark for correct set-up; 1 mark for correct collection method for gaseous product; 1 mark for correct
labels; 0 mark if the set-up is not workable)
(3)
b) Z < X < Y
(1)
Only the oxide of Z is decomposed by heating alone. Therefore Z is the least reactive.
(1)
Only Y can react with steam. Therefore it is the most reactive.
(1)
c) i) Silver
ii) 2Ag2O(s)
(1)
4Ag(s) + O2(g) or 2Z2O(s)
4Z(s) + O2(g)
(1)
iii) Yes
Cu(s) + 2Ag+(aq)
(1)
Cu2+(aq) + 2Ag(s) or Cu(s) + 2Z+(aq)
Cu2+(aq) + 2Z(s)
d) Less reactive metals were discovered first and more reactive metals were discovered later.
90 a) Hydrogen
It gives a ‘pop’ sound when tested with a burning splint.
b) Y < Z < X
(1)
(1)
(1)
(1)
(1)
X and Z are more reactive than Y as both X and Z can displace copper from copper(II) nitrate solution
but Y cannot. Therefore Y is the least reactive. (or oxides of X and Z cannot be reduced by carbon but
oxide of Y can.)
(1)
X is more reactive than Z as X can react with cold water but Z cannot.
(1)
207
c) X is a reactive metal.
(1)
It reacts with water in the copper(II) nitrate solution and hydrogen is given off.
d) i) Magnesium
(1)
(1)
ii) Displacement reaction
(1)
iii) The blue colour of the solution fades.
(1)
iv) Mg(s) + Cu2+(aq)
Mg2+(aq) + Cu(s) or Z(s) + Cu2+(aq)
Z2+(aq) + Cu(s)
91 a) C > B > A > D
(1)
(1)
From reactions (v) and (vi), it can be concluded that C is more reactive than B because C reacts vigorously
with steam while B reacts slowly with steam.
(1)
From reactions (i) and (ii), it can be concluded that B is more reactive than A because B reacts with dilute
hydrochloric acid but A does not.
(1)
From reactions (iii) and (iv), it can be concluded that A is more reactive than D because oxide of A is
formed upon heating while there is no reaction for D.
(1)
b) E is more reactive than A because E reacts with dilute hydrochloric acid but A does not.
(1)
E is less reactive than B because B can displace E from a solution of compound of E.
(1)
Thus E should be placed between B and A in the reactivity series.
(1)
c) Let x be the relative atomic mass of A.
The chemical formula of the oxide of A is AO.
AO(s) + H2(g)
A(s) + H2O(l)
Number of moles of H2O produced =
3.24 g
mass of H2O
=
= 0.180 mol
18.0 g mol–1
molar mass of H2O
(1)
According to the equation, 1 mole of AO produces 1 mole of H2O.
∴ number of moles of AO = 0.180 mol
=
mass of AO
molar mass of AO
14.3 g
–1
(x + 16.0) g mol
x = 63.4
=
92 a) 2Mg(l) + TiCl4(g)
Ti(s) + 2MgCl2(l)
b) Magnesium is more reactive than titanium.
208
(1)
(1)
(1)
(1)
Magnesium atoms lose electrons more easily.
(1)
In the reaction between TiCl4 and Mg, magnesium competes for the chlorine and wins.
(1)
c) Number of moles of TiCl4 =
11.4 g
mass of TiCl4
=
= 0.0600 mol
189.9 g mol–1
molar mass of TiCl4
(1)
According to the equation for the reaction between TiCl4 and Mg, 1 mole of TiCl4 produces 1 mole of
Ti.
∴ number of moles of Ti obtained = 0.0600 mol
Theoretical yield of Ti = number of moles of Ti x molar mass of Ti
= 0.0600 mol x 47.9 g mol–1
= 2.87 g
Percentage yield of Ti =
2.54 g
x 100% = 88.5%
2.87 g
(1)
(1)
d) i) Making supersonic aircraft (1) or making tooth implants / replacement hip joints (1)
ii) Any two of the following:
Light (1) / very strong (1) / very high melting point (1) / resists corrosion (1)
OR
Light (1) / very strong (1) / resists corrosion (1) / can be easily shaped (1) / biocompatible (1)
93 a) CaO(s) + 2HCl(aq)
CaCl2(aq) + H2O(l)
(1)
mass of CaO
molar mass of CaO
14.0 g
=
56.1 g mol–1
= 0.250 mol
b) Number of moles of CaO made from the sample =
(1)
∴ number of moles of Ca made from the sample = 0.250 mol
Mass of Ca made from the sample = number of moles of Ca x molar mass of Ca
= 0.250 mol x 40.1 g mol–1
= 10.0 g
Percentage by mass of Ca in the sample =
10.0 g
x 100% = 20.0%
50.0 g
c) Atoms of calcium and strontium have the same number of outermost shell electrons.
(1)
(1)
(1)
d) i)
IZESPHFO
XBUFS
JOWFSUFEGVOOFM
DBMDJVN
(1 mark for correct set-up; 1 mark for correct labels; 0 mark if the set-up is not workable)
(2)
209
ii) Ca(s) + 2H2O(l)
Ca(OH)2(s) + H2(g)
(1)
iii) Calcium is covered by a layer of calcium oxide.
(1)
Reaction between calcium and water starts only when the oxide layer dissolves.
94 a) To prevent the loss of the oxide formed.
(1)
(1)
b) Mass of cobalt heated = (23.13 – 20.49) g = 2.64 g
Mass of oxygen in the oxide = (24.09 – 23.13) g = 0.96 g
Cobalt
Oxygen
Number of moles of atoms
that combine
2.64 g
–1 = 0.0448 mol
58.9 g mol
0.96 g
–1 = 0.060 mol
16.0 g mol
(1)
Mole ratio of atoms
0.0448 mol
= 1
0.0448 mol
0.060 mol
= 1.34
0.0448 mol
(1)
Simplest whole number ratio
of atoms
1 x 3 = 3
1.34 x 3 = 4
(1)
∴ the empirical formula of the cobalt oxide is Co3O4.
c)
60
Co
Number of protons
Number of neutrons
27
33
(0.5)
(0.5)
d) Formula mass of CoCl2•nH2O = 58.9 + 2 x 35.5 + n(2 x 1.0 + 16.0) = 129.9 + 18n
1 mole of CoCl2•nH2O contains n moles of H2O.
i.e. (129.9 + 18n) g of CoCl2•nH2O contain 18n g of H2O.
22.0 g of CoCl2•nH2O contain 9.99 g of H2O.
18n
9.99 g
=
129.9 + 18n
22.0 g
n = 6
(1)
(1)
95 a) i)
/
/
(1)
ii)
'
'
(1)
210
b) i) Suppose there are 100 g of the fluoride, so there are 42.4 g of nitrogen and 57.6 g of fluorine.
Nitrogen
Fluorine
42.4 g
–1 = 3.03 mol
14.0 g mol
Number of moles of
atoms that combine
3.03 mol
3.03 mol
Mole ratio of atoms
57.6 g
–1 = 3.03 mol
19.0 g mol
3.03 mol
3.03 mol
= 1
= 1
(1)
(1)
∴ the empirical formula of the fluoride is NF.
Let (NF)n be the molecular formula of the fluoride.
Relative molecular mass of the fluoride = n(14.0 + 19.0) = 66.0
n = 2
(1)
∴ the molecular formula of the fluoride is N2F2.
ii)
'
/
/
'
(1)
96 a) i)
Common name
Chemical name
Haematite (1)
iron(III) oxide
Limestone
calcium carbonate (1)
Coke (1)
carbon
ii) (hot) air
(1)
iii) B (molten) slag
(1)
C (molten) iron
(1)
b) C(s) + O2(g)
CO2(g)
c) i) CO2(g) + C(s)
(1)
2CO(g)
(1)
ii) Reducing agent / removes oxygen from iron(III) oxide
(1)
iii) 3CO(g) + Fe2O3(s)
(1)
d) Number of moles of Fe2O3 =
3CO2(g) + 2Fe(s)
mass of Fe2O3
399 g
=
= 2.50 mol
molar mass of Fe2O3
159.6 g mol–1
(1)
According to the equation, 1 mole of Fe2O3 produces 2 moles of Fe upon reaction.
∴ number of moles of Fe produced = 2 x 2.50 mol = 5.00 mol
(1)
Mass of Fe produced = number of moles of Fe x molar mass of Fe
= 5.00 mol x 55.8 g mol–1
= 279 g
(1)
211
e) CaCO3(s)
CaO(s) + CO2(g)
(1)
f) Any two of the following:
Loss of animal habitat (1) / noise pollution (1) / dust pollution (1) / damage to the landscape (1) / traffic
density (1)
97 a) i) The glowing splint relights.
ii) 2Ag2O(s)
(1)
4Ag(s) + O2(g)
iii) Number of moles of Ag produced =
(1)
9.17 g
= 0.0850 mol
107.9 g mol–1
(1)
According to the equation, 2 moles of Ag2O undergo decomposition to produce 4 moles of Ag.
∴ number of moles of Ag2O decomposed =
0.0850
mol = 0.0425 mol
2
Mass of Ag2O decomposed = number of moles of Ag2O x molar mass of Ag2O
= 0.0425 mol x 231.8 g mol–1
= 9.85 g
Percentage by mass of Ag2O in the sample =
b) i) Pb3O4(s) + 4H2(g)
9.85 g
x 100% = 84.9%
11.6 g
3Pb(s) + 4H2O(l)
(1)
(1)
(1)
ii) Solids with metallic lustre formed. / Drops of colourless liquid appeared.
(1)
iii) Hydrogen is explosive / flammable.
(1)
c) No. The reactivity of silver and lead can only be compared using the same reaction.
98 a) i) Percentage by mass of H2O in gypsum =
2 x 18.0
x 100%
40.1 + 32.1 + 4 x 16.0 + 2 x 18.0
= 20.9%
(1)
(1)
(1)
ii) Blackboard chalk / white line road marking / plaster board / cement
(1)
iii) Contains water (of crystallization)
(1)
b) i) CaCO3(s)
CaO(s) + CO2(g)
(1)
ii) Carbon
(1)
iii) Carbon dioxide
(1)
iv) Number of moles of CaO =
mass of CaO
157 g
=
= 2.80 mol
molar mass of CaO
56.1 g mol–1
(1)
According to the equation, 1 mole of CaCO3 gives 1 mole of CaO upon decomposition.
∴ number of moles of CaCO3 decomposed = 2.80 mol
Mass of CaCO3 decomposed = number of moles of CaCO3 x molar mass of CaCO3
= 2.80 mol x 100.1 g mol–1
= 280 g
212
(1)
Percentage by mass of CaCO3 in the sample =
280 g
x 100% = 84.8%
330 g
v) Used in making steel from iron / to neutralize acidity in soil / as a drying agent in industry
c) i) CaCO3(s) + 2HCl(aq)
CaCl2(aq) + CO2(g) + H2O(l)
(1)
(1)
(1)
ii) Any two of the following:
Effervescence / gas bubbles are given off. (1) / Calcium carbonate dissolves. (1) / Heat is evolved. (1)
iii) Brick-red
(1)
99 a) PbO(s) + 2HNO3(aq)
Pb(NO3)2(aq) + H2O(l)
(1)
b)
HMBTTSPE
NJYUVSF
GJMUFSQBQFS
SFTJEVF
GJMUFSGVOOFM
GJMUSBUF
(1 mark for correct set-up; 1 mark for labelling filter funnel and filter paper; 1 mark for labelling residue
and filtrate; 0 mark if the set-up is not workable)
(3)
c) i) Pb(NO3)2(aq) + 2NaCl(aq)
PbCl2(s) + 2NaNO3(aq)
ii) To precipitate all the lead(II) ions as lead(II) chloride.
d) To remove water soluble impurities that adhere to the surface of the precipitate.
(1)
(1)
(1)
e) For the overall reaction:
PbO(s)
? g
PbCl2(s) (not balanced)
9.37 g
mass of PbCl2
molar mass of PbCl2
9.37 g
=
278.2 g mol–1
= 0.0337 mol
Number of moles of PbCl2 formed =
(1)
1 mole of PbO produces 1 mole of PbCl2.
∴ number of moles of PbO present = 0.0337 mol
Mass of PbO present = number of moles of PbO x molar mass of PbO
= 0.0337 mol x 223.2 g mol–1
= 7.52 g
(1)
213
7.52 g
x 100%
9.80 g
= 76.7%
Percentage by mass of PbO in the sample =
f) The sample does not contain other ions which form insoluble chloride.
100 a) i) x : s, y : aq, z : l.
(1)
(1)
(1)
ii) Number of moles of calcite added =
mass of CaCO3
2.60 g
=
–1 = 0.0260 mol
molar mass of CaCO3
100.1 g mol
(1)
According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl.
0.0150 mole of HCl reacts with 0.00750 mole of CaCO3.
Number of moles of calcite left = (0.0260 – 0.00750) mol = 0.0185 mol
(1)
Mass of calcite left = number of moles of CaCO3 x molar mass of CaCO3
= 0.0185 mol x 100.1 g mol–1
= 1.85 g
(1)
iii) Separate calcite from the reaction mixture, dry and weigh it.
b) i) 1.10 g (187.6 g – 186.5 g)
(1)
(1)
ii) Number of moles of CO2 liberated =
1.10 g
mass of CO2
=
= 0.0250 mol
44.0 g mol–1
molar mass of CO2
(1)
According to the equation for the reaction between CaCO3 and HCl, 1 mole of CaCO3 produces
1 mole of CO2.
∴ number of moles of CaCO3 in the sample = 0.0250 mol
Mass of CaCO3 in the sample = number of moles of CaCO3 x molar mass of CaCO3
= 0.0250 mol x 100.1 g mol–1
= 2.50 g
Percentage by mass of CaCO3 in the sample =
2.50 g
x 100% = 80.6%
3.10 g
iii) Some carbon dioxide dissolve in the acid.
101 a) Mg3N2(s) + 3H2O(l)
2NH3(g) + 3MgO(s)
b) Number of moles of Mg3N2 present =
Number of moles of H2O present =
7.64 g
mass of Mg3N2
=
= 0.0757 mol
100.9
g mol–1
molar mass of Mg3N2
15.4 g
mass of H2O
=
= 0.856 mol
18.0 g mol–1
molar mass of H2O
(1)
(1)
(1)
(1)
(1)
(1)
According to the equation, 1 mole of Mg3N2 reacts with 3 moles of H2O to produce 3 moles of MgO.
During the reaction, 0.0757 mole of Mg3N2 reacts with 0.227 mole of H2O. Therefore Mg3N2 is the limiting
reactant. It limits the amount of MgO obtained.
(1)
c) Mass of MgO obtained = number of moles of Mg x molar mass of MgO
= 0.227 mol x 40.3 g mol–1
= 9.15 g
214
(1)
102 a)
MJE
DSVDJCMF
QJQFDMBZ
USJBOHMF
IFBU
MFBE**
PYJEFNJYFE
XJUIDBSCPOQPXEFS
USJQPE
(1 mark for correct set-up; 1 mark for correct labels; 0 mark if the set-up is not workable)
b) i) 2PbO(s) + C(s)
2Pb(s) + CO2(g)
ii) Solids with metallic lustre would be observed.
mass of PbO
10.0 g
=
= 0.0448 mol
molar mass of PbO
223.2 g mol–1
mass of C
10.0 g
=
= 0.833 mol
molar mass of C
12.0 g mol–1
c) i) Number of moles of PbO =
Number of moles of C =
(2)
(1)
(1)
(1)
(1)
According to the equation, 2 moles of PbO react with 1 mole of C. Hence 0.0448 mole of PbO would
react with 0.0224 mole of C. Therefore carbon is in excess.
(1)
ii) Maximum mass of Pb obtained = number of moles of Pb x molar mass of Pb
= 0.0448 mol x 207.2 g mol–1
= 9.28 g
d) i) Percentage yield of lead =
7.10 g
x 100% = 76.5%
9.28 g
(1)
(1)
ii) The temperature is not high enough. / Not enough time allowed for the reaction to take place. (1)
e) Let the mole ratio of PbO : PbO2 be x : y.
Number of moles of Pb
Number of moles of O
=
x + y
= 3
x + 2y
4
x = 2, y = 1
(1)
(1)
∴ Pb3O4 is a mixture of PbO and PbO2 in a mole ratio of 2 : 1.
103 a) The reaction is reversible.
b) Number of moles of N2 =
(1)
72.8 g
mass of N2
=
= 2.60 mol
28.0 g mol–1
molar mass of N2
(1)
According to the equation, 1 mole of N2 produces 2 moles of NH3.
∴ number of moles of NH3 produced = 2 x 2.60 mol = 5.20 mol
Theoretical yield of NH3 = number of moles of NH3 x molar mass of NH3
= 5.20 mol x 17.0 g mol–1
= 88.4 g
(1)
c) Mass of NH3 produced = 88.4 g x 15.0% = 13.3 g
(1)
d) i) 2NH3(g) + 3CuO(s)
(1)
N2(g) + 3H2O(l) + 3Cu(s)
215
ii) Number of moles of NH3 =
mass of NH3
20.4 g
=
= 1.20 mol
molar mass of NH3
17.0 g mol–1
(1)
Number of moles of CuO =
mass of CuO
159 g
=
= 2.00 mol
molar mass of CuO
79.5 g mol–1
(1)
According to the equation, 2 moles of NH3 react with 3 moles of CuO. During the reaction, 1.20
moles of NH3 react with 1.8 moles of CuO. Therefore NH3 is the limiting reactant. It limits the amount
of N2 obtained.
(1)
1.20
mol = 0.600 mol
2
Mass of N2 obtained = number of moles of N2 x molar mass of N2
= 0.600 mol x 28.0 g mol–1
= 16.8 g
∴ number of moles of N2 obtained =
(1)
104 a) i) Rusting will not occur in tubes C and D.
(1)
In tube C, the zinc strip provides sacrificial protection.
(1)
In tube D, there is no oxygen.
(1)
ii) Tube A.
(1)
The very dilute hydrochloric acid speeds up the rusting process.
(1)
b) i) Copper is less reactive than iron. Iron corrodes instead of copper.
105
(1)
ii) Sea water contains sodium chloride.
(1)
iii) Copper(II) ions are formed when the copper corrodes.
(1)
Case
a) Iron rod fully plated with zinc
Observation
No observable change
Explanation
(1)
The iron rod does not rust as the
zinc coating prevents oxygen and
water from reaching the iron.
(1)
b) Iron rod fully plated with tin
No observable change
(1)
The iron rod does not rust as the
tin coating prevents oxygen and
water from reaching the iron.
(1)
c) Iron rod partly plated with zinc
No observable change
(1)
The iron rod does not rust as zinc
provides sacrificial protection.
(1)
d) Iron rod partly plated with tin
216
A blue colour appears
(1)
The iron rod rusts rapidly as iron is
in contact with tin, a less reactive
metal.
(1)
106 a) Very small percentage of both metals exists in the Earth’s crust.
(1)
b) The alloy is stronger than pure gold.
(1)
c) Less reactive metals were discovered first and more reactive metals were disscovered later.
(1)
d) i) During the extraction, a mixture of iron ore, coke and limestone is added at the top of the blast
furnace.
(1)
Carbon monoxide is produced in the process. The carbon monoxide reduces the iron ore to produce
iron.
(1)
ii) A large amount of electricity is needed for the extraction of aluminium. This makes aluminium
expensive.
(1)
e) i) The aluminium undergoes anodization to increase the thickness of the oxide layer. This makes the
aluminium much more resistant to corrosion.
(1)
The iron is painted so as to prevent oxygen and water from reaching it. The paint layer protects the
iron beneath from rusting.
(1)
ii) As soon as the paint on the iron is scratched, rusting starts.
(1)
Aluminium reacts with oxygen in the air to give an even coating of oxide on the surface. The oxide
layer is not permeable to oxygen and water. It protects the metal beneath from further attack. (1)
f) i) To protect iron from rusting.
ii) Aluminium is softer than iron so that the ring-pull can be pulled off more easily.
(1)
(1)
iii) Advantage: Aluminium is lighter than iron. / Aluminium can be recycled more easily. / Aluminium is
more corrosion resistant than iron. / Aluminium can be dyed more easily.
(1)
Disadvantage: Aluminium is more expensive than iron. / Aluminium is not as strong as iron.
107 a) Aluminium oxide
b) Electrolysis
(1)
(1)
(1)
c) Any two of the following:
Metal resources can be saved. (1) / Fuel can be saved. (1) / Environmental impact due to metal waste
and mining can be reduced. (1) / Public awareness of conservation can be raised. (1)
d) Any two of the following:
The cost is high. (1) / It is difficult to get people to save metal waste. (1) / It is difficult to sort the metal
waste. (1)
e) Aluminium is more reactive than iron.
The extraction of iron is easier than that of aluminium.
f) Making aeroplanes / window frames / overhead power cables / foils
(1)
(1)
(1)
217
g) i) There is a layer of aluminium oxide attached to the metal surface.
ii) (1) Aluminium anodization
(1)
(1)
(2)
EJMVUFTVMQIVSJDBDJE
BMVNJOJVNTIFFU
BTUIFOFHBUJWF
FMFDUSPEF
BMVNJOJVNPCKFDUUPCFBOPEJ[FE
BTUIFQPTJUJWFFMFDUSPEF
(1 mark for correct set-up; 1 mark for correct labels; 0 mark if the set-up is not workable)
108 a) Aluminium
b) i) 2MgO(s) + 2Cl2(g) + C(s)
(2)
(1)
2MgCl2(s) + CO2(g)
(1)
ii) Displacement reaction
(1)
Potassium is more reactive than magnesium.
(1)
iii)
m
$M
.H
m
$M
(1)
c) Electrolysis
(1)
d) i) Calcium hydroxide
(1)
ii) (1) Mg(OH)2(s) + 2HCl(g)
MgCl2(s) + 2H2O(l)
(2) Number of moles of Mg(OH)2 =
35.0 g
mass of Mg(OH)2
=
= 0.600 mol
58.3
g mol–1
molar mass of Mg(OH)2
(1)
(1)
According to the equation, 1 mole of Mg(OH)2 produces 1 mole of MgCl2.
∴ number of moles of MgCl2 produced = 0.600 mol
Theoretical yield of MgCl2 = number of moles of MgCl2 x molar mass of MgCl2
= 0.600 mol x 95.3 g mol–1
= 57.2 g
(1)
Mass of MgCl2 produced = 57.2 g x 85.0%
= 48.6 g
(1)
e) Duralumin
218
(1)
109 Aluminium is the most abundant metal in the Earth’s crust while iron is the second most abundant one.
(1)
The abundance of copper in the Earth’s crust is less.
(1)
Both copper and iron were discovered in ancient times, while aluminium was discovered about two hundred
years.
(1)
Copper is extracted by controlled heating of its sulphide ore.
(1)
Iron cannot be extracted by simply heating its ore. It is extracted by carbon reduction.
(1)
Aluminium is difficult to extract. It is extracted by electrolyzing its molten ore and discovered after the invention
of electrolysis in 1800.
(1)
Hence the date of discovery of a metal depends more on the ease of the metal, rather than its abundance
in the Earth’s crust.
(3 marks for organization and presentation)
110 To find the empirical formula of magnesium oxide, we need to know the masses of magnesium and oxygen
present in a given mass of the compound.
First weigh a crucible and lid. Record the reading.
(1)
Clean a magnesium ribbon with sandpaper. Put the ribbon into the crucible. Weigh the crucible and lid with
its contents.
Record the reading.
(1)
Heat the magnesium ribbon. Lift the lid carefully with tongs at intervals.
Continue to heat until the magnesium catches fire. When this occurs, remove the burner.
(1)
When the magnesium stops burning, remove the lid and heat the crucible strongly for 5 minutes.
(1)
Allow the crucible to cool. Reweigh the crucible and lid with its contents. Record the reading.
(1)
From the readings, it is possible to obtain the mass of magnesium used and the mass of magnesium oxide
obtained.
The difference between these two masses is the mass of oxygen in the oxide.
(1)
The empirical formula of magnesium oxide can be obtained from these data.
(3 marks for organization and presentation)
219
111 The reaction of a metal with oxygen in the air, moisture or other substances in the environment is called
corrosion.
(1)
When aluminium reacts with oxygen in the air, an even coating of aluminium oxide forms.
(1)
This oxide layer is not permeable to oxygen and water, and thus protects the metal beneath from further
attack.
(1)
The thickness of the oxide layer can be increased by anodization so as to give the surface more
protection.
(1)
When a more reactive metal is attached to iron, the more reactive metal corrodes instead of iron. The metal
provides sacrifical protection.
(1)
Painting on iron prevents both oxygen and water from reaching the iron beneath. Hence iron is protected
from corrosion.
(1)
(3 marks for organization and presentation)
220