A student tried to extract iron metal from haematite by strong heating. However, he failed. He could not
explain the failure. He thought that silver, also a kind of metal, could be obtained by heating its ore, so
iron could also be extracted by the same method. 
a. Explain the extraction failure and the misunderstanding of the student.
b. Suggest a method to extract iron from haematite.
c. Write a balanced chemical equation for the reaction involved in your method stated in (b).
d. The student said, “The bonds in iron and haematite are not the same, but they are both formed from
the electrostatic force between oppositely charged particles”. Explain what he said.
e. The student at last extracted iron metal from haematite successfully. He found that the colour of
iron metal is very different from that of haematite. Why?
a. Metals that situated above silver in the reactivity series cannot be extracted by heating its ores. Iron
is more reactive than silver, therefore it cannot be extracted by strong heating alone. The extraction
method stated above failed. (1)
b. Heat haematite with carbon or carbon monoxide at about 1000°C / high temperature. (1)
c. 2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g) /
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) (1)
d. The bond in iron metal is metallic bond (1). In metals, the outermost electrons of atoms form a sea
of delocalized electrons (1), leaving the atoms as cations (1). The electrostatic attraction force
between cations and the electron sea holds the particles together. (1). The bond in haematite,
iron(III) oxide, is ionic bond (1). The electrostatic forces between the positively charged iron(III)
ions and the negatively charged oxide ions help to hold the particles together. (1)
e. The colour of haematite is due to the presence of iron(III) ions, not iron. (1)
Answer the following questions with the information given in the table. 
Aluminium Iron Gold
% abundance by mass
8.1 5.0 0.0000004
in the Earth’s crust
World production per
15000 716000 1.4
year (thousand tonnes)
Approximate price per
10500 900 82000000
a. Why the price of aluminium is higher than that of iron, although aluminium is more abundant than
b. Account for the high price of gold.
c. Suggest a reason why, aluminium, instead of iron, is usually used in making window frames.
d. A man would like to use iron to make window frames as iron is much cheaper. Suggest a method to
prevent the frames from rusting.
e. Recycling helps saving metal reserve in the Earth’s crust, but people sometimes would like to
extract metals rather than recycling. Why?
a. Aluminium is extracted by electrolysis of its molten ore, while iron is extracted by using a blast
furnace. (1) The cost of operating electrolysis is much higher than that of using a blast furnace. (1)
b. Gold is a rare metal (1), and the demand for it is high as it is an unreactive metal and can be used
for making ornaments, jewellries and protective coatings. (1). With low supply but high demand,
the price for gold is high.
c. Aluminium is more corrosion-resistant. (1) Aluminium reacts with oxygen in air to form a
nonporous oxide layer (1), protecting aluminium from corrosion; iron alone is subjected to the
attack of oxygen and water. It forms porous rust and therefore it goes further rusting easily. (1)
d. Painting (1). It can keep the iron away from oxygen and moisture.
e. The cost of recycling is sometimes higher than the extraction cost. (1)
A student added excess zinc powder into a beaker of copper(II) sulphate solution. He found that there was
a colour change in the solution. 
a. Describe and explain the colour change in the solution.
b. Write a balanced equation for the reaction.
c. What would happen to the observation of colour change if a zinc bar was used instead of zinc
powder in the experiment? Explain your answer.
d. What would happen if silver powder is added to the product solution of the above experiment?
a. The colour of the solution changed from blue to colourless gradually. (1) As zinc is more reactive
than copper, a displacement reaction took place (1): blue copper(II) ions were reduced to copper
metal gradually (1); while zinc gradually dissolved to form colourless zinc ions. (1)
b. Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) (1)
c. The colour change would not be observed (1). As copper formed in the displacement reaction would
deposit on the surface of the zinc bar (1), the reaction would then stop. (1)
OR The blue colour would disappear very slowly (1)
As the surface area for the reaction to take place would be greatly reduced if zinc powder
were replaced with zinc bar (1), the reaction would be very slow (1), leading to slow colour
d. There is no reaction (1) because silver is less reactive than zinc, it cannot displace zinc. (1)
The table summarizes some information about three metals, X, Y and Z. 
X Y Z
A white solid A white solid formed A black solid
Reaction with oxygen
Reaction with cold water Gas evolved Gas evolved No reaction
Reaction with dilute hydrochloric acid Gas evolved Explosion Gas evolved
a. Arrange the three metals in descending order of reactivity.
b. How Y is stored in laboratory?
c. Y is a Group I metal. Write a balanced equation for the reaction of metal Y and water.
d. State the observation when metal Y reacts with water.
e. What would metal X be? Write a balanced equation for the reaction between metal X and dilute
hydrochloric acid, using the metal name you stated.
f. A piece of metal X is burnt in Bunsen flame. What is the colour of the flame?
g. Suggest what metal Z would be. By what method is metal Z extracted from its ore? Write an
equation for the reaction involved in the extraction.
a. Y > X > Z (1)
b. Y is stored under paraffin oil in laboratory. (1)
c. 2Y(s) + 2H2O(l) 2YOH(aq) + H2(g) (1)
d. The metal moves quickly around the water surface (1) and burns with coloured flames. (1) It melts
and dissolves in water gradually. (1)
e. Calcium; (1) Ca(s) + 2HCl(aq) CaCl2(aq) + H2(g) (1)
f. Brick red. (1)
g. Iron. (1) Iron is extracted by carbon reduction of its ore. (1)
2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g) (1)
Copper can be extracted from copper pyrites, an ore containing copper(II) sulphide, by carbon reduction.
a. Suggest a solid reducing agent used in the extraction.
b. Write balanced equation(s) to illustrate the reaction(s) in the extraction process.
c. Explain why lead extraction requires a higher temperature than copper extraction does.
d. Suggest a method to extract copper other than by carbon reduction.
e. How many grams of reducing agent suggested in part (a) is required to extract all copper from
8.604 g of copper(II) sulphide? (Relative atomic masses: C= 12.0; O= 16.0; S= 32.1; Cu= 63.5)
a. Carbon / coke (1)
b. 2CuS(s) + 3O2(g) 2CuO(s) + 2SO2(g) (1)
2CuO(s) + C(s) 2Cu(s) + CO2(g) (1)
c. Lead is more reactive than copper, (1) thus lead compounds are more stable than copper compounds.
(1) As a result, more energy, and hence higher temperature, is needed to convert lead compounds
into lead. (1)
d. Add zinc powder into copper(II) sulphate solution. (1)
Or any correct displacement reactions.
e. Consider the following equations:
2CuS(s) + 3O2(g) 2CuO(s) + 2SO2(g) …………..(I)
2CuO(s) + C(s) 2Cu(s) + CO2(g) ….……………(II)
Number of moles of CuS(s) present = 8.604/(63.5+32.1) = 0.09 mol (1)
From equation (I), mole ratio of CuS to CuO = 1:1
number of moles of CuO formed = 0.09 mol (1)
From equation (II), mole ratio of CuO to C = 2:1
number of moles of C needed = 0.045 mol (1)
Mass of C needed = 0.045 x 12.0 = 0.54 g (1)
In an experiment, a sample of 24.08 g iron burns in oxygen to form an oxide. 
a. What is the formula of the oxide if the oxide contains 70% of iron by mass?
b. Find the formula mass of the oxide.
c. What is the mass of oxygen required for a complete combustion of the iron sample? Write down all
steps in your calculation.
d. The oxide is then reduced back to iron by carbon. Calculate the amount of carbon needed to obtain
all iron in the product formed from the reaction stated in part (c).
(The relative atomic masses: C= 12.0; O= 16.0; Fe= 56.0)
a. Let there be 100 g of the iron oxide.
Mole ratio of Fe : O = 70/56 : (100-70)/16 (1)
= 1.25 : 1.875
The formula of the oxide is Fe2O3. (1)
b. The formula mass of Fe2O3
= 2 56 + 3 16
= 160 (1)
c. The equation for the reaction is
4Fe(s) + 3O2(g) 2Fe2O3(s) (1)
Mole ratio 4 : 3
No. of moles 24.08/ 56 0.43 3/4
= 0.43 (1) = 0.3225 (1)
The mass of oxygen required for complete combustion
= 0.3225 (16 2)
= 10.32 g (1)
d. Refer to the equation in part (c), the amount of Fe2O3 produced is 0.43/ 2 = 0.215 mol.
The equation for the reduction is
2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g) (1)
Mole ratio 2 : 3
No. of mole 0.215 0.215 3/2
= 0.3225 (1)
The mass of carbon required for complete reduction
= 0.3225 12.0
= 3.87 g (1)
A student lost the labels of three bottles, and each of which contains one kind of metal. In order to find
out what the metals (they are temporarily called P, Q and R respectively) are, he placed these metals in
different solutions to see if displacement reactions occur. The following table summarizes the results of
the experiments. In the table, a (+) sign indicates that a displacement reaction occurred; while a () sign
indicates that no reaction occurred. 
Metal MgCl2(aq) FeCl3(aq) CuCl2(aq)
P + + +
R + +
a. What metals would P, Q and R possibly be? Explain your answer.
b. The experiments conducted do not provide sufficient information to identify each metal. Suggest
experiments to identify them based on your answer in part (a).
c. In part (a), positions of metals in the reactivity series can be predicted approximately by those
displacement reactions. Can the method of placing those metals in dilute hydrochloric acid be used
in laboratory for the same purpose?
a. P is more reactive than magnesium. (1) It would be potassium, sodium or calcium; (1) Q is more
reactive than copper but less reactive than iron. (1) It would be lead; (1) R is more reactive than iron
but less reactive than magnesium. (1) It would be aluminium or zinc. (1)
b. P may be potassium, sodium or calcium. Place a piece of P in cold water. (1) Potassium and sodium
will move around on the water surface quickly. Potassium will give a lilac flame, while sodium will
give a golden yellow flame; (2) calcium will sink in water and no flame will be produced. (1)
Or Carry out a flame test to test P. (1) Potassium will give a lilac flame; sodium will give a
golden yellow flame; calcium will give a brick red flame.(3) \
R may be aluminium or zinc. Pass steam through a sample of R. (1) If gas is evolved, it may be zinc
as zinc reacts with steam to produce hydrogen gas; if no gas is evolved, it may be aluminium as
aluminium has a protective oxide layer on its surface to prevent it from reacting with steam. (1)
c. No. It is very dangerous (1) to put unknown metals in dilute hydrochloric acid, as some metals,
such as potassium and sodium, can cause explosion when reacting with acid. (1)
Barium (Ba), with atomic number 56, is in the same group with calcium and magnesium in the Periodic
a. Which metal, calcium or barium, is more reactive? Explain your answer.
b. Write a balanced equation for the reaction between barium and dilute hydrochloric acid.
c. By what method is barium extracted from its ore?
d. Can barium be used to prevent iron from rusting? Why?
a. Barium is more reactive. (1)
Barium has a lower position than calcium in the Periodic Table. (1) The lower a metal element
situates in the group, the more reactive it is. (1)
b. Ba(s) + 2HCl(aq) BaCl2(aq) + H2(g) (1)
c. Barium is extracted by electrolysis of its molten ore. (1)
d. Yes. (1) Barium should be more reactive than iron because it is more reactive than calcium, which
is at a higher position than iron in the metal reactivity series. (1) Barium can prevent iron rusting by
sacrificial protection. (1)
XXX is a hot city located near the coast. Its air quality is poor due to overpopulation and intense
industrial development. Her citizens found that articles made of iron rust much faster than those in other
a. Explain this phenomenon.
b. Suggest three ways to slow down iron rusting in the city.
c. A man bought a new aluminium cooking pan. A few days later, he found that there was a thin white
layer covering the pan surface. He scratched the layer off but it appeared again after a few days. He
then scratched off the layer once it formed. After a period of time, the pan broke. Explain the
phenomenon with appropriate chemical equation(s).
d. Anodization is a well-grown industry in the city.
i) State the main function of anodization.
ii) Describe how anodization can be done in laboratory.
a. Air near the coast contains many electrolytes, such as sodium chloride, which can accelerate
corrosion by providing a better conducting medium (for reaction of iron with water and oxygen).
High temperature causes high reaction rate, making rusting goes faster. (1)
Overpopulation and intense industrial development produces many pollutants in the air. Common
air pollutants are sulphur dioxide, nitrogen oxides and carbon monoxide. These air pollutants
dissolve in rainwater to form acid, which provides hydrogen ions, and then corrodes iron. (1)
b. Painting (1)
Sacrificial protection (1) or any other correct methods
c. When the new aluminium cooking pan was in use, the oxygen in air reacted with aluminium to
form a white aluminium oxide layer covering the pan surface. (1)
4Al(s) + 3O2(g) 2Al2O3(s) (1)
After the layer was scratched off, the clean aluminium continued to react with oxygen and form
oxide layer again. (1) The repeated scratching caused more aluminium reacting with oxygen and
finally all aluminium had reacted and the pan broke. (1)
d. (i) Anodization is to thicken the oxide layer on the surface of aluminium (1), and hence to
enhance its resistance to corrosion. (1)
(ii) In a setup of electrolysis, the aluminium object to be anodized is placed at the positive
electrode (1), and the negative electrode (1) is an aluminium sheet, using dilute sulphuric acid
as the electrolyte. (1) The oxygen evolved at the positive electrode reacts with aluminium to
increase the thickness of the oxide layer. (1)
Five metals (P, Q, R, S, and T) were tested in the following experiments. 
(i) Small pieces of the five metals were added to dilute hydrochloric acid and the volumes of gas
collected in the first two minutes were recorded.
Metal P Q R S T
Volume/cm3 53 90 0 32 75
(ii) Steam was allowed to pass through small pieces of the five metals and the volumes of gas collected
in the first two minutes were recorded.
Metal P Q R S T
Volume/cm3 49 103 0 0 0
a. Arrange the five metals in descending order of reactivity. Explain your answer.
b. Suggest and explain what T is.
c. Q is a Group II element. Write the equations for the reactions of Q with steam and with dilute
d. Describe a test for the gas formed in experiment (i).
a. Q > T > P > S > R (2)
Q is the most reactive as the largest volume of gas was evolved in the two reactions. (1) R is the
least reactive as no reaction occurred in both experiments. (1) T is more reactive than P because T
gave more gases in experiment (i). (1) P is more reactive than S because P gave more gases in
experiment (i) and S had no reaction in experiment (ii). (1)
b. T is aluminium. (1) P reacts with both dilute hydrochloric acid and steam, so P is above lead in the
reactivity series. (1) From part (a), T is more reactive than P, so T is above lead in the series. (1)
That T did not react with steam was not due to its low position in the series but the nonporous
protective oxide layer of aluminium prevents the attack from steam. Among the reactive metals,
only aluminum has a protective oxide layer on its surface. (1)
c. Q(s) + H2O(g) QO(s) + H2(g) (1)
Q(s) + 2HCl(aq) QCl2(aq) + H2(g) (1)
d. Place a burning splint near the mouth of the test tube containing the gas. (1) The gas is hydrogen if
the splint burns with a “pop” sound. (1)
Silver is formed if zinc powder or magnesium powder is added to silver nitrate solution. 
a. Adding which powder, zinc or magnesium, would the process of silver formation be faster?
b. Explain why magnesium powder helps obtaining silver faster than magnesium ribbon does.
c. How many grams of magnesium powder should be added into silver nitrate solution to obtain 5 g of
d. If 0.8 g of zinc reacts with excess silver nitrate solution to form silver. What mass of silver can be
(Relative atomic masses: N=14.0; O=16.0; Mg=24.3; Zn=65.4; Ag = 107.9;)
a. Magnesium powder (1). As magnesium is more reactive than zinc (1), magnesium undergoes
displacement reaction faster.
b. Magnesium powder provides greater surface area for the reaction to occur. (1)
c. The ionic equation for the reaction is
Mg(s) + 2Ag+(aq) Mg2+(aq) + 2Ag(s) (1)
Mole ratio 1 : 2
No. of moles 0.0463/2 5/107.9
= 0.0232 = 0.0463 (1)
Mass of magnesium required to obtain 5 g of silver
= 0.0232 24.3
= 0.5634g (1)
d. The equation for the reaction is
Zn(s) + 2AgNO3(aq) Zn(NO3)2(aq) + 2Ag(s) (1)
Mole ratio : 1 : 2
No. of moles 0.8/65.4 0.0122 2
= 0.0122 = 0.0245 (1)
The mass of silver obtained
= 0.0245 107.9
= 2.64 g (1)
X, Y and Z are three different metal solids, having +2 charge when in compounds. The table below shows
the results of two experiments about the metals. 
Experiment Metal X Metal Y Metal Z
Adding metal to dilute Gas evolved No reaction Gas evolved
Adding metal to cold water No reaction No reaction Gas evolved
a. Arrange the three metals in ascending order of reactivity.
b. Which metal forms the most stable oxide? Explain your answer.
c. Suggest and explain what metal Y would be.
d. When 0.42 g of molten oxide of metal Y undergoes electrolysis, it decomposes completely to give
0.12 g of oxygen. Calculate the formula mass of the oxide and the relative atomic mass of Y.
a. Y < X < Z (2)
b. Z. (1) Z is the most reactive one and hence its oxide is most stable. (1)
c. Copper. (1) As Y does not react with both cold water and dilute hydrochloric acid, it must be lower
than lead in the reactivity series. (1) The metals lower than lead are copper, mercury, silver and gold.
(1) Mercury is liquid (1), the charge of silver in compound is usually +1 (1), and gold usually does
not form any oxide at all (1). Copper is a solid and its charge in compound can be +2 (1). Therefore,
it is reasonable to believe that Y is copper.
d. No. of mole of oxygen atoms in YO = 0.12/16 = 0.0075 (1)
No. of mole of YO = no. of mole of oxygen atoms = 0.007 (1)
Formula mass of YO = 0.42/0.0075
= 56 (1)
Relative atomic mass of Y = 56 –16 = 40 (1)
A mixture of zinc sulphide and zinc oxide is heated. After strong heating for 30 minutes, 0.52 g of a
gaseous product A is formed. The residue is then heated strongly with excess carbon. 0.70 g of another
gaseous product B is formed after the reaction is complete. 
a. Identify the two gaseous products in the reactions involved.
b. Write balanced equations for the reactions as described above.
c. Calculate the percentage by mass of zinc oxide in the mixture.
a. A is sulphur dioxide. B is carbon dioxide. (2)
b. 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g) (1)
2ZnO(s) + C(s) 2Zn(s) + CO2(g) (1)
c. The first reaction is
2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g)
Mole ratio 1 : 1 : 1
No. of moles 0.52/(32.1 + 16 2) (1)
0.0081 0.0081 = 0.0081 (1)
Mass 0.0081 (65.4 + 32.1) 0.0081 (65.4 + 16)
= 0.79 g = 0.66 g (2)
The second reaction is
2ZnO(s) + C(s) 2Zn(s) + CO2(g)
Mole ratio 2 : 1
No. of moles 0.016 2 0.7/(12 + 16 + 16) (1)
= 0.032 = 0.016
Mass 0.032 (65.4 + 16)
= 2.6048 g (1)
The total mass of ZnO reacted is 2.6048 g. This includes the ZnO initially present in the mixture and the
product from ZnS. Therefore,
mass of ZnO initially present in the mixture = 2.6048 – 0.66 = 1.9448g (1)
The total mass of the mixture = mass of ZnS + mass of ZnO
= 0.79 + 1.9448
= 2.7348 g (1)
The percentage by mass of ZnO in the mixture
= (1.9448/2.7348) 100%
= 71.11 % (1)
Five iron nails are placed at different conditions. The setup is shown below. 
calcium chloride oil layer
fresh distilled water
C. D. E.
sodium chloride tap water
tap water solution
iron nail iron nail iron nail
a. What factors are essential for the rusting process of iron?
b. What is the purpose of placing anhydrous calcium chloride in tube A?
c. Why oil and fresh distilled water were used in tube B?
d. If the air in tube C is replaced with car exhaust gas, would the iron nail in it rust faster? Explain
e. Would the nail in tube B rust? Explain your answer.
f. If the magnesium ribbon is replaced with magnesium powder, would the iron nail start rusting at an
earlier time? Why?
a. The presence of both oxygen (1) and water (1).
b. It absorbs moisture in air (1), removing one essential factor, water, to prevent or slow down rusting.
c. Oil prevents oxygen dissolving into the water below; (1) fresh distilled water ensures no ions / no
air present. (1)
d. Yes. (1) Car exhaust gas contains many acidic pollutants (1), such as sulphur dioxide, carbon
dioxide and nitrogen oxides. When they dissolve in water, acid forms and the ions accelerate the
rusting process. (1)
e. Yes (1), as there is some oxygen in the unboiled distilled water. (1)
f. Yes. (1) As magnesium powder provides greater surface area for its reaction with air, (1)
magnesium would run out earlier and therefore the nail would start rusting earlier. (1)