HKCEE Past Paper Questions: Part 9 Rate of Reactions Part A: Multiple Choices 1. HKCEE 1996 II Q11 In an experiment, 1.6 g of sulphur are burnt completely in air to form sulphur dioxide. What volume of sulphur

dioxide, measured at room temperature and pressure, is formed? (Relative atomic mass : S = 32.0; Molar volume of gas at room temperature and pressure = 24.0 dm3)

A. 0.6 dm3 B. 1.2 dm3 C. 2.4 dm3 D. 12.0 dm3

Ans:B 2. HKCEE 1996 II Q19 Under certain conditions, 60 cm3 of a gaseous compound, NxOy, decompose completely to give 60 cm3 of

nitrogen gas and 30 cm3 of oxygen gas. (All gas volumes are measured at the same temperature and pressure.) Which of the following combinations is correct? x y A. 1 1 B. 1 2 C. 2 1 D. 2 3

Ans:C 3. HKCEE 1996 II Q32

Which of the following statements concerning one mole of nitrogen gas is/are correct? (1) It has a mass of 14.0 g. (2) It occupies the same volume as 4.0 g of helium gas at room temperature and pressure.

(3) It contains 6.02×1023 atoms of nitrogen. (Relative atomic masses : He = 4.0, N = 14.0; Avogadro constant = 6.02×1023 mol–1) A. (1) only B. (2) only C. (1) and (3) only D. (2) and (3) only

Ans:B 4. HKCEE 1997 II Q17 Which of the following gases occupies the largest volume at room temperature and pressure? (Relative atomic masses : H = 1.0, C = 12.0, N = 14.0, O = 16.0; molar volume of gas at room temperature and pressure = 24.0 dm3)

A. 1.0 g of ammonia B. 2.0 g of nitrogen C. 3.0 g of oxygen D. 4.0 g of carbon dioxide

Ans:C

1

5. HKCEE 1997 II Q34 One mole of sulphur atoms has a mass twice that of one mole of oxygen atoms. Which of the following

statement is/are correct? (1) 2 g of sulphur and 1 g of oxygen each occupy the same volume at room temperature and pressure. (2) 2 g of sulphur and 1 g of oxygen each contain the same number of atoms. (3) The number of atoms contained in one mole of sulphur is twice that contained in one mole of oxygen. A. (1) only B. (2) only C. (1) and (3) only D. (2) and (3) only

Ans:B 6. HKCEE 1999 II Q16 At room temperature and pressure, 8.0 g of oxygen and 20.0 g of gas X occupy the same volume. What is the

molar mass of X? (Relative atomic mass : O = 16.0; molar volume of gas at room temperature and pressure = 24.0 dm3)

A. 20.0 g B. 40.0 g C. 60.0 g D. 80.0 g

Ans:D 7. HKCEE 2001 II Q10 Consider the reaction; 4H2(g) + Fe3O4(s) 3Fe(s) + 4H2O(l)

What mass of iron would be obtained if 96.0 cm3 of hydrogen, measured at room temperature and pressure, is consumed in the reaction?

(Relative atomic mass : Fe = 56.0; molar volume of gas at room temperature and pressure = 24 dm3) A. 0.056 g B. 0.084 g C. 0.168 g D. 0.224 g

Ans:C 8. HKCEE 2001 II Q27 Suppose that the Avogadro number is L. How many atoms does 600 cm3 of oxygen at room temperature and

pressure contain? (Molar volume of gas at room temperature and pressure = 24 dm3)

A. 401

L B. 201

L

C. 25L D. 50L

Ans:B

2

9. HKCEE 2002 II Q16 Gases X and Y react to give a gaseous product Z. The reaction can be represented by the equation : X(g) + 3Y(g) 2Z(g)

In an experiment, 40 cm3 of X and 60 cm3 of Y are mixed and are allowed to react in a closed vessel. What is the volume of the resultant gaseous mixture?

(All volumes are measured at room temperature and pressure.) A. 40 cm3 B. 60 cm3 C. 80 cm3 D. 100 cm3

Ans:B 10. HKCEE 2003 II Q6 Sodium azide, NaN3, is used in air bags in cars. When there is a serious collision, the azide will decompose to

give nitrogen. The decomposition can be represented by the equation: 2NaN3(s) 2Na(s) + 3N2(g)

What is the mass of sodium azide required to produce 72 dm3 of nitrogen at room temperature and pressure? (Relative atomic masses: N = 14.0, Na = 23.0; molar volume of gas at room temperature and pressure = 24 dm3)

A. 65.0 g B. 130.0 g C. 195.0 g D. 292.5 g

Ans:B 11. HKCEE 2004 II Q3 The relative atomic masses of hydrogen and oxygen are 1.0 and 16.0 respectively. Which of the following

statements concerning 36.0 g of water is correct?

(Molar volume of gas at room temperature and pressure = 24 dm3; Avogadro constant = 6.02 × 1023 mol–1) A. It consists 4 mol of hydrogen atoms.

B. It contains 3 × 6.02 × 1023 atoms. C. It contains 6 × 6.02 × 1023 molecules. D. It has a volume of 48 dm3 at room temperature and pressure.

Ans:A 12. HKCEE 2004 II Q6 Decomposition of KClO3(s) at elevated temperatures gives KCl(s) and O2(g) as the only products. What is the

volume of O2(g) produced, measured at room temperature and pressure, when 61.3 g of KClO3(s) undergoes complete decomposition?

(Relative atomic masses : O = 16.0, Cl = 35.5, K = 39.1; molar volume of gas at room temperature and pressure = 24 dm3) A. 8 dm3

B. 12 dm3

C. 18 dm3

D. 36 dm3 Ans:C

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13. HKCEE 2004 II Q21 A gaseous mixture consists of methane and ethane in a mole ratio of 1 : 1. It has a volume of 200 cm3 at room

temperature and pressure. What is the volume of oxygen required, measured at room temperature and pressure, for the complete combustion of the mixture? A. 400 cm3

B. 550 cm3 C. 700 cm3 D. 1100 cm3

Ans:B 14. HKCEE 2005 II Q35

NaHCO3 decomposes upon heating to form Na2CO3, CO2, H2O. What is the volume of CO2 formed at room temperature and pressure if 336 g of NaHCO3 undergoes complete decomposition? (Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0; Molar volume of gases at room temperature and pressure = 24 dm3) A. 12 dm3

B. 24 dm3

C. 48 dm3

D. 96 dm3

15. HKCEE 2005 II Q44

Which of the following statements concerning 1 mole of aluminium is/are correct? (Avogadro’s constant = 6.02 × 1023; Molar volume of gas at temperature and pressure = 24 dm3) (1) It can form 1 mole of Al3+ ions. (2) It can form 3 × 6.02 × 1023 Al3+ ions. (3) It occupies 24 dm3 at room temperature and pressure. A. (1) only

B. (2) only C. (1) and (3) only D. (2) and (3) only

Ans:A

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16. HKCEE 2006 II Q20 The Contact Process involves the following reaction:

2 SO2(g) + O2(g) 2 SO3(g) △H < 0

V2O5(s)

Which of the following statements can be correctly deduced from the above information? A. The formation of sulphur trioxide is endothermic. B. The reaction should be carried out at high temperatures. C. Vanadium(V) oxide is one of the reactants. D. Sulphur trioxide will decompose into sulphur trioxide and oxygen.

Ans:D 17. HKCEE 2007 II Q33 50 cm3 of carbon monoxide burns completely in 50 cm3 of oxygen. Assuming that all volumes are measured

at room temperature and pressure, what is the final gaseous volume at the end of the combustion? (Molar volume of gas at room temperature and pressure = 24 dm3)

A. 50 cm3 B. 75 cm3 C. 100 cm3 D. 150 cm3

Ans:B 18. HKCEE 2007 II Q39 CO2(g), SO3(g), O2(g) are composed of atoms of different elements. At room temperature and pressure,

what is the ratio of the number of atoms involved in 100 cm3 of CO2(g), 100 cm3 of SO3(g) and 200 cm3 of O2(g)? (Molar volume of gas at room temperature and pressure = 24 dm3)

A. 3 : 4 : 4 B. 3 : 4 : 2 C. 2 : 3 : 4 D. 1 : 1 : 2

Ans:A

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19. HKCEE 2007 II Q40 When 10 g of PURE calcium carbonate (molar mass = 100.1 g) reacted with excess hydrochloric acid, 2.40

dm3 of carbon dioxide was obtained at room temperature and pressure. However, in a similar experiment using 10 g of IMPURE calcium carbonate, 2.50 dm3 of carbon dioxide was obtained. Assuming that the impurity is a metallic carbonate, what would this impurity be?

(Molar masses : MgCO3 = 84.3 g, ZnCO3 = 125.4 g, FeCO3 = 115.8 g, CuCO3 = 123.5 g; molar volume of gas at room temperature and pressure = 24 dm3)

A. MgCO3 B. ZnCO3 C. FeCO3 D. CuCO3

Ans:A 20. HKCEE 2008 II Q32 Nitroglycerin ( )is an explosive and can explode according to the following equation : 9353 ONHC

(g)O(g)6NO(g)10H(g)12CO)(ONH4C 22229353 +++→l

0.1 mole of nitroglycerin undergoes explosion and the products are allowed to cool to room temperature.

What is the total volume of gases left behind at room temperature and pressure?

(Molar volume of gas at room temperature and pressure ) 3dm 24=

A. 3dm 4.11 B. 3dm 4.17 C. 3dm 6.45 D. 3dm 6.69

Ans:A 21. HKCEE 2009 II Q33 An oxide of metal M reacts completely with carbon to give 12.6 g of metal M and 2.38 dm3 of carbon dioxide

measured at room temperature and pressure. What is the chemical formula of the oxide?

(Relative atomic masses: M=63.5, O= 16.0; Molar volume of gas at room temperature and pressure = 24 dm3)

A. MO B. MO2

C. M2O D. M2O3

Ans:A

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22. HKCEE 2009 II Q40 Assuming that air contains 20% of oxygen by volume, how much air is required to burn completely 100 cm3 of

ethane? (All volumes are measured at the same temperature and pressure.) A. 350 cm3

B. 1000 cm3

C. 1750 cm3

D. 3500 cm3

Ans:C 23. HKCEE 2010 II Q37 What is the theoretical volume of carbon dioxide gas, measured at room temperature and pressure, that can be

obtained by adding 100 cm3 of 2.0 M HCl(aq) to 0.80 g of Na2CO3(s) ? (Relative atomic masses : H = 1.0, C = 12.0,O = 16.0, Na = 23.0, Cl = 35.5; molar volume of gas at room temperature and pressure = 24 dm3) A. 90 cm3

B. 180 cm33

C. 240 cm3

D. 480 cm3

Ans:B

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Part B: Long Questions 1. HKCEE 1979 Q3a A small quantity of powdered manganese(IV) oxide (MnO2) was added to 50.0 cm3 of hydrogen peroxide solution. The

volume of the gas liberated at room temperature and 1 atmospheric pressure was plotted against time as shown in the following graph:

(i) Using the graph, determine: 1. the total volume of gas liberated at the end of the experiment; 2. the time required for the liberation of the first half of the total volume of gas liberated (to the nearest 0.1 minute);

and 3. the time required for three-quarters of the reactant to be used up (to the nearest 0.1 minute). (ii) At which point on the graph, A, B, C or D was the rate of the reaction fastest? (iii) Explain the shape of the graph by considering changes in the rates of reaction. (iv) Write an equation for the reaction. Calculate the concentration of the hydrogen peroxide solution in mol dm-3. (Assume the

molar volume of gas at room temperature and 1 atmospheric pressure = 24 dm3).

Solution

(i) 1. 60 cm3

2. 0.9 min

3. 1.7 min

(ii) The greatest rate is at the point D.

(iii) At the beginning the slope is the steepest, the rate of reaction is faster. It is because the concentration of the reactant,

hydrogen peroxide is higher. At the intermediate stage, the slope is less steep than the beginning, the rate of

reaction is slower. It is because some of the hydrogen peroxide is used, the concentration of H2O2 is lower.

At the end, the slope is equal to zero, it means that the rate is equal to zero. It is because the H2O2 is used up

completely.

(iv) 2H2O2 → 2H2O + O2

No. of moles of O2 liberated = 0.06/24 = 2.5×10-3 mol

No. of moles of H2O2 = (2.5×10-3)(2) = 5×10-3 mol

Concentration of H2O2 = (5×10-3 mol ) / (0.05 dm3) = 0.1 M

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2. HKCEE 1980 Q4a

A student poured 4 cm3 of hydrogen peroxide solution into a test tube. After shaking the test-tube, he observed a few tiny

bubbles of a colourless gas G rising to the surface of the liquid. He then added 1 spatula of a black solid to act as a catalyst

for the reaction. A steady stream of bubbles of G was then evolved.

(i) What is the function of the catalyst in this reaction?

(ii) Two minutes after adding the catalyst, the student wished the gas G to be evolved at a faster rate. State how this could be

achieved.

(iii) Sketch a graph, showing the volume of G evolved against time, starting from the moment of the catalyst was added.

(iv) Draw a labeled diagram of the apparatus you would use to prepare G from the hydrogen peroxide solution and collect the

gas in a dry state.

(vi) The volume of G collected in this experiment was 100 cm3. Calculate the mass of hydrogen peroxide contained in the 4

cm3 of solution.

(Molar volume of a gas at the temperature and pressure of experiment = 25 dm3.)

(vii) Using the result in (vi), deduce the molarity of the hydrogen peroxide solution used in this experiment.

Solution

(i) To increase the rate of the reaction.

(ii) Warm the mixture.

(iii)

(iv)

(v) 2H2O2 → 2H2O + O2

(vi) No. of moles of O2 liberated = 0.1 /25 = 4×10-3 mol

No. of moles of H2O2 in 4 cm3 = (4×10-3)(2) = 8×10-3 mol

Mass of H2O2 = (8×10-3) (1.0×2 + 16.0×2 ) = 0.272 g

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3. HKCEE 1982 Q4a A student experimentally determined the volume of carbon dioxide evolved when 1.0 g of powdered calcium carbonate

was allowed to react with 40 cm3 of 1.0 M hydrochloric acid at room temperature and atmospheric pressure. The results can be graphically represented by curve X:

(i) Calculate whether the calcium carbonate or the hydrochloric acid was present in excess. Hence deduce the theoretical volume of carbon dioxide that would be evolved when the reaction was completed. (ii) Compare the volume of carbon dioxide calculated in (i) with the actual volume obtained as shown by curve X. Explain

any difference between the two volumes. (iii) Describe briefly what you would do in order to make sure that the volume of carbon dioxide collected is as close to the

theoretical volume as possible. (iv) Using the same mass of calcium carbonate, but by varying the experimental conditions, curve Y could be obtained.

Suggest two different conditions (other than changing the temperature and pressure) necessary to obtain curve Y. (v) Curve Z was obtained when 1.0 g of calcium carbonate lumps were put in 40 cm3 of 0.5M sulphuric acid, and the volume

of CO2 evolved was measured at room temperature and atmospheric pressure. Explain why a smaller final volume of carbon dioxide was obtained.

(Relative atomic masses: H = 1.0, C=12.0, O = 16.0, Cl = 35.5, Ca = 40.0; Molar volume of a gas at room temperature and atmospheric pressure = 24 dm3)

Solution

(i) CaCO3 + 2HCl → CaCl2 + H2O + CO2

No. of moles of CaCO3 = 1 ÷ (40.0 + 12.0 + 16.0×3) = 0.01 mol

No. of moles of HCl = 1.0 × 0.04 = 0.04 mol

Since 1 mole of CaCO3 reacts with 2 moles of HCl,

Therefore, HCl is in excess.

Volume of CO2 evolved = 0.01 × 24000 = 240 cm3

(ii) The actual volume of CO2 obtained is 210 cm3, which is smaller than the theoretical value. It is because the CO2

may dissolve in the solution / the CaCO3 is not pure / the CO2 is trapped in the flask.

(iii) Shake the contents frequently to get rid of the dissolved or trapped CO2 /

Use pure calcium carbonate

(iv) a. use lumps of CaCO3 instead of CaCO3 powder

b. use diluted acid

(v) When CaCO3 reacts with sulphuric acid, a coating of CaSO4 will form which protect the lumps of carbonate in

contact with acid, therefore the reaction will stop before the carbonate reacted completely.

10

4. HKCEE 1984 Q1a In an experiment to study the rate of reaction of zinc with an excess of hydrochloric acid, a student carried out the

following tests at room temperature and pressure:

Test Reactants mixed 1 1 g zinc granules + 50 cm3 2 M HCl 2 1 g zinc granules + 100 cm3 2M HCl 3 1 g zinc granules + 50 cm3 4M HCl

For test 1, a plot of the volume of gas evolved against time was obtained, as shown in the graph below:

(Relative atomic masses: H=1.0, Cl=35.5, Zn=65.0 Molar volume of a gas at room temperature and pressure = 24.0 dm3) (i) Write an ionic equation for the reaction involved. (ii) Draw a labeled diagram to show how test 1 can be carried out in the laboratory. (iii) What would be the final volume of gas collected in test 1? (iv) Compare the initial rates of reaction in (1) test 1 with test 2, and (2) test 1 with test 3. Give a brief explanation for each case. (v) Copy the graph in your answer book and draw on the same graph the expected curve if the reactants in test 1 were allowed

to react at a temperature of 50oC.

Solution

(i) Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

(ii)

(iii) As HCl is in excess,

No. of moles of Zn = 1/65.0 = 0.0154 mol

Volume of gas collected in test 1 = 0.0154 × 24.0 = 0.370 dm3 or 370 cm3

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(iv)

(1) As the concentration of HCl is the same in both cases, the initial rates in test 1 and 2 would be almost identical.

(2) For higher concentration of HCl, the initial rate in test 3 would be faster than that in test 1, thus the rate would be

increased.

(v)

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5. HKCEE 1985 Q5c

Two pieces of magnesium ribbon of the same mass were separately added to 50 cm3 of 1 M hydrochloric acid and 50 cm3

of 1M ethanoic acid. In each case, the acid was in excess. The time required to complete the reactions were as follows:

Experiment Reactants Time required

I Magnesium + 1M HCl 150s

II Magnesium + 1M ethanoic acid 240s

(i) How could you know when reactions were completed?

(ii) Explain why it took a longer time to complete the reaction in experiment II.

(iii) If another piece of magnesium ribbon of the same mass were added to 50 cm3 of 1M sulphuric acid, how would the time

required for complete reaction differ from that recorded in experiment I? Suggest an explanation.

Solution

(i) All the magnesium has been dissolved.

(ii) Ethanoic acid is a weak acid, the concentration of H+(aq) is lower than that of the 1M HCl. The reaction rate is

proportional to H+(aq) concentration, therefore, the reaction between Mg and ethanoic acid is slower.

(iii) The time would be shorter than 150 s.

H2SO4 is a dibasic and strong acid, therefore, the concentration of H+(aq) is higher than that of the HCl and hence

the reaction rate is faster.

13

6. HKCEE 1986 Q5a A student placed equal masses of zinc powder into two identical flasks A and B, and then introduced equal volumes of

hydrochloric acid of different concentrations into the flasks through dropping funnels. The volumes of gas liberated were measured at regular intervals, and the results are shown graphically below:

(Molar volume of a gas at room temperature and atmospheric pressure = 24.0 dm3; Relative atomic mass of zinc = 65.4) (i) Suggest a method for measuring the volume of the gas liberated. (ii) To which flask was a more concentrated sample of the acid added? Explain your answer. (iii) How long did it take for the reaction in flask B to reach completion? (iv) In both of the above experiments the zinc was in excess. Calculate the mass of zinc reacted in flask A. (v) How would you modify the experimental conditions in order to study the effect of temperature on the rate of reaction?

Solution

(i) By using a gas syringe.

(ii) A more concentrated sample of the acid is added to flask B because more gas is liberated.

(iii) The time required is 6 mins

(iv) Zn + 2HCl → ZnCl2 + H2

No. of moles of H2 produced = 48 / 24000 = 0.002 mol

No. of moles of zinc reacted = 0.002 mol

Mass of zinc reacted = 0.002 × 65.4 = 0.13 g

(v) Varying the temperature of the experiment, but keep the other conditions constant.

14

7. HKCEE 1987 Q5a A student carried out an experiment at 20oC to study how the concentration of sodium thiosulphate affects the rate of the

following reaction: S2O3

2-(aq) + 2H+(aq) → H2O(l) + SO2(aq) + S(s)

The student prepared four samples of sodium thiosulphate solution in similar beakers as in the table below. He placed the

beaker containing sample 1 over a black cross mark. He then added 20 cm3 of 2M HCl to the sample, and recorded the time t required for the cross to disappear. He repeated the experiment for the other samples. His results are given below:

Composition of sample Sample

Volume of 0.1 M sodium thiosulphate (cm3)

Volume of water (cm3)t (s) 1/t (s-1)

I 20.0 0.0 37 0.0270II 15.0 5.0 50 0.0200III 10.0 10.0 76 0.0133IV 5.0 15.0 154 0.0065

(i) Explain why

(1) water was added to the sodium thiosulphate solutions in the preparation of samples II, III and IV.

(2) the beaker was placed over the black cross mark.

(3) the black cross disappeared after time t.

(4) the same volume of acid was used in each sample.

(ii) Plot a graph of the volume of 0.1 M sodium thiosulphate used against 1/t.

(iii) From your graph, estimate the time that would be required for the cross to disappear if 20 cm3 of 2M HCl were added to a

solution containing 16 cm3 of 0.1 M sodium thiosulphate and 4 cm3 of water.

(iv) What is the relationship between the concentration of sodium thiosulphate and the rate of reaction? Explain your answer

with reference to the graph which you have plotted.

Solution

(i) (1) Water was added in order to keep the total volume in each sample constant.

(2) The beaker was placed over the black cross mark in order to standardize the time taken for the cross to

disappear from view.

(3) When sufficient sulphur is formed, it will block the sight of the cross from the student.

(4) To ensure that the only variable that changes is the concentration of thiosulphate. The concentration of HCl is

kept constant.

15

(ii)

(iii) From the graph, if volume of thiosulphate is 16 cm3,

1/t = 0.00215 ; t = 46.55 s

(iv) Since volume of thiosulphate is directly proportional to the 1/t and 1/t is directly proportional to the reaction rate,

therefore, the rate of reaction is directly proportional to the concentration of thiosulphate.

16

8. HKCEE 1988 Q3b The industrial preparation of methanol can be represented by the following equation: )()(2)( 32 gOHCHgHgCO catalyst⎯⎯ →←+ The graph below shows the mole percentage of methanol in the reaction mixture against time at the same pressure but at

two different temperatures:

(i) With reference to the graph, answer the following questions: (1) What difference between the two initial rates of reaction can be deduced? Explain your deduction. (2) Why do both curves eventually become horizontal? (3) List the substances present in the reaction mixture during the time interval BC. (4) Explain whether the reaction for the production of methanol is exothermic or endothermic. (ii) At a given temperature and pressure, how would an increase in the amount of carbon monoxide in the system affect the

yield of methanol? Explain your answer.# (iii) Copy the graph in your answer book, showing curve ABC only. On the same graph, sketch and label two curves which

would be obtained if the process were carried out under the same conditions as those for the curve ABC, except that (1) no catalyst is used (label this curve as I), (2) a higher pressure is used (label this curve as II). Solution (i) (1) The initial rate at 400oC is greater than that at 200oC because the slope of AB is greater than that of AD.

(2) The system has attained equilibrium so that the rate of formation of CH3OH is equal to its rate of

decomposition.

(3) CO(g), H2(g) and CH3OH(g)

(4) The reaction is exothermic because the equilibrium position shifts to the left and this lowers the yield when

the temperature increases.

(ii) When more carbon monoxide is introduced into the system, the concentration of CO will be increased. This shifts

the equilibrium to the right according to the Le Chatelier’s principle. As a result, the yield of CH3OH will increase.

(iii)

17

9. HKCEE 1990 Q2b Two different samples of calcium carbonates (A and B), each weighing 0.8 g and containing inert impurities, were allowed

to react with excess dilute hydrochloric acid under the same laboratory conditions. The volumes of carbon dioxide gas evolved with time are shown in the graph below:

(i) Draw a diagram to show how the above experiment can be performed in the laboratory. (ii) Explain why the slope of the curve for sample A is steeper at X than at Y. (iii) From the two curves, deduce TWO differences between sample A and sample B. (iv) (1) What is the total volume of gas liberated from sample B? (2) Hence, calculate the percentage of calcium carbonate in sample B. [Relative atomic masses: C=12.0, O = 16.0, Ca = 40.0; Molar volume of gas under the laboratory conditions = 24 dm3] Solution (i)

(ii) The slope of the curve at X is steeper than that at Y, meaning that the rate of reaction at X is faster than at Y. This is

because as the reaction proceeds, the reactants (HCl and CaCO3) are used up so the rate at Y is lower than that at

X.

(iii) More CO2 is collected from B than from A, meaning that B has a higher purity than A. Initial slope of A is steeper

than that of B, indicating that the initial rate of A is greater than that of B. This is mainly due to the smaller particle

size of A.

(iv) (1) 120 cm3

(2) CaCO3 + 2HCl → CaCl2 + H2O + CO2

No. of moles of CO2 = 0.12 / 24 = 0.005 mol

Mass of CaCO3 in the sample = 0.005 × (40.0 + 12.0 + 16.0×3) = 0.5 g

Percentage of CaCO3 = (0.5 ÷ 0.8)×100% = 62.5%

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10. HKCEE 1994 Q8a

The rate of decomposition of hydrogen peroxide solution in the presence of manganese(IV) oxide was studies by means of the following experiment.

50.0 cm3 of a hydrogen peroxide solution was mixed with 0.5 g of powdered manganese(IV) oxide in a conical flask. The volumes of gas evolved at room temperature and pressure at different times are shown in the graph below.

(i) Write an equation for the decomposition of hydrogen peroxide.

(ii) Compare the rates of decomposition of the hydrogen peroxide sollution at points A, B and C, and explain why these rates are different.

(iii) Calculate the original molarity of the hydrogen peroxide solution.

(iv) If the experiment is repeated with an equal volume of the hydrogen peroxide solution and 1.0 g of powdered manganese(IV) oxide, would the shape of the curve obtained be the same? Explain your answer.

(Molar volume of gas at room temperature and pressure = 24.0 dm3) Solution (i) 2H2O2 → 2H2O + O2

(ii) The rate of decomposition is in the order A > B > C.

At A, the concentration of H2O2 is the highest and so it has the highest rate of decomposition.

Rate at B is lower than that at A because some of the H2O2 has been used up.

At C, the rate of decomposition is zero because all the H2O2 are used up.

(iii) Total volume of O2 evolved = 84 cm3 = 0.084 dm3

No. of moles of O2 evolved = 0.084 ÷24 = 3.5×10-3 mol

No. of moles of H2O2 = 2(3.5×10-3 mol) = 7×10-3 mol

Molarity of H2O2 = 7×10-3 mol ÷ 0.05 dm3 = 0.14 M

(iv) The shape of the curve obtained would be the same.

The rate of reaction is independent of the amount of catalyst added

Or The shape of the curve will not be the same. The catalytic effect of a solid depends on its surface area. An increase in mass of powdered MnO2 means an increase in

surface area of the catalyst. Rate of reaction is increased.

19

11. HKCEE 2005 Q10

The information below was found on the label of a brand of effervescent vitamin C tablets:

Each tablet contains 1000 mg of vitamin C.

Other ingredients: sodium hydrogencarbonate, citric acid, sugar and colourant (a) With the help of a chemical equation, explain why effervescence occurs when a tablet of the effervescent vitamin

C is added to water.

(2 marks)

(b) An experiment was carried out to study the action of water on a tablet of the effervescent vitamin C using the

set-up as shown below. The graph shows the results obtained in the experiment.

(i) Find, from the graph, the mass of gas liberated from the reaction of the tablet with water. (You may

assume that the gas liberated is NOT soluble in water.)

(ii) At the end of the experiment, the sodium hydrocarbonate in the tablet had been completely used up.

Calculate the mass of sodium hydrogencarbonate present in the tablet.

(iii) Suggest ONE advantage of using a data-logger in this experiment.

(iv) The experiment was repeated using warm water instead of cold water. Sketch, on the same graph, the

results that would be obtained in the repeated experiment.

(6 marks)

20

Solution

21

12. HKCEE 2009 Q10

In an experiment, a data-logger with pressure sensor was used to study the rate of decomposition of hydrogen peroxide

(H2O2) in the presence of manganese(IV) oxide. The relation between the pressure and time measured is shown in the

curve below.

(a) The decomposition of hydrogen peroxide gives water and oxygen. After the experiment, it was found that the

manganese(IV) oxide used did not undergo any chemical change.

(i) State the function of manganese(IV) oxide.

(ii) Explain why a pressure sensor could be used in this experiment.

(iii) Write a chemical equation for the decomposition of hydrogen peroxide. Hence discuss the changes, if any, in the

oxidation numbers of hydrogen and oxygen in the reaction.

(5 marks)

(b) (i) Explain why the respective rates of decomposition of hydrogen peroxide differ at pointsA, B and C on the curve.

(ii) On the graph above, sketch a curve that should be obtained if the initial concentration of the hydrogen peroxide

is half of its original value, while all other conditions remain unchanged.

(4 marks)

Solution

22

13. HKCEE 2002 Q8a

Sulphur dioxide is formed when coal is burnt in a power station.

(i) The coal used in the power station contains 1.5% of sulphur by mass. Calculate the volume of sulphur dioxide

released, measured at room temperature and pressure, when 1.0 kg of the coal is burnt.

(You may assume that all the sulphur in coal is converted to sulphur dioxide upon burning.)

(ii) State ONE environmental problem associated with the emission of sulphur dioxide into the atmosphere.

(iii) Suggest ONE measure to reduce the emission of sulphur dioxide from the power station.

(iv) Particulates are also present in the flue gas generated in the power station.

(1) State ONE environmental problem associated with the discharge of particulates into the atmosphere.

(2) Suggest ONE way to remove particulates from the flue gas.

(Relative atomic masses: O = 16.0, S = 32.0;

molar volume of gas at room temperature and pressure = 24 dm3)

(7 marks)

Solution

23

14. HKCEE 2004 Q8a

Coral consists mainly of calcium carbonate. An experiment was carried out to determine the percentage by mass of calcium carbonate in a sample of coral using the set-up shown below:

(i) Write down a chemical equation for the reaction of calcium carbonate with dilute hydrochloric acid. (ii) The mass of the sample used was 0.36g. At the end of the experiment, 78cm3 of carbon dioxide was collected at

room temperature and pressure. Calculate

(1) the number of moles of carbon dioxide collected; and (2) the percentage by mass of calcium carbonate in the sample.

(iii) Assuming that there was no leakage of gas in the set-up, suggest ONE source of error in the experiment. (Molar volume of gas at room temperature and pressure = 24dm3; relative atomic masses: C = 12.0, O = 16.0, Ca = 40.0)

(7 marks)

Solution

24

15. HKCEE 2006 Q12

You are provided with the following materials:

magnesium ribbon and 2M hydrochloric acid

Design an experiment to determine the molar volume of hydrogen at room temperature and pressure.

(You may use apparatus commonly available in a school laboratory.)

(9 marks)

Solutions

25

16. HKCEE 2007 Q12

Organic compound Z contains carbon, hydrogen and oxygen only. Analysis of Z gives the following results:

(I) 1.0 g of Z contains 0.401 g of carbon, 0.068 g of hydrogen and 0.531 g of oxygen.

(II) 1.0 g of Z, upon complete vaporisation, occupies 400 cm3 at room temperature and pressure.

(III) There are no observable changes when potassium carbonate solution is added to Z.

(IV) Brown colour of bromine remains unchanged when several drops of bromine in organic solvent are added to Z.

(Molar volume of gas at room temperature and pressure = 24 dm3)

(a) Calculate the empirical formula of Z. (2 marks)

(b) Deduce the molecular formula of Z. (2 marks)

(c) (i) Suggest a possible structure of Z. Explain your answer.

(ii) Give the systematic name for the compound represented by the structure you suggested in (i).

(2 marks)

Solution

26