Kinetics SQ Ans

7/23/2019 Kinetics SQ Ans

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Structure Question Answers

1(a) Average rate =

600)(6

0.8000.0125

mol s1 = 0.00219 mol s1 [1] (The negative sign means a

decrease in the amount of A.)

OR

Average rate of decrease in the amount of A =600)(6

0.01250.800

mol s

1 = 0.00219 mol s1 [1]

(b)

Correct labelling of axes [1]

Correct plot of graph [1]

(c)

A m o u n t o f A ( m

o l )

Time (min)



Initial rate =600)(0.4

0.800.50

mol s

1 = 0.0125 mol s1 [1]

(d)

A m o u n t o f A ( m o l )

Time (min)


Time (min)



Instantaneous rate =602.2)(4.0

0.160.030

mol s

1 = 0.00120 mol s1 [1]

2(a) The average rate of formation of hydrogen =

10

0.48 cm3 s

1 = 4.80 cm3 s1 [1]

(b) (i) The average rate of decrease of HCl =10

0.2 M s1 = 0.200 M s1 [1]

(ii) The average rate of decrease of HCl =10

1000

0.100.2

mol s1 = 2.00 × 103 mol s1 [1]

(c) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

From the equation, the mole ratio between HCl and MgCl2 is 2 : 1.

i.e. rate of decrease of HCl = 2 × rate of increase of MgCl2 [1]

The average rate of formation of magnesium chloride =2

100.2 3 mol s

1 = 1.0 × 103 mol s

1.

[1]

3(a) Number of moles of NaHCO3 =

30.160.120.10.23

5.3

mol = 0.0417 mol

Number of moles of H2SO4 = 3.0 ×1000

0.25 mol = 0.0750 mol [1]

From the equation, the mole ratio of NaHCO 3 to H2SO4 is 2 : 1.0.0750 mol of H2SO4 needs 0.150 mol of NaHCO3 to react completely.

∴ H2SO4 is in excess. [1]

(b) The average rate of decrease of sodium hydrogencarbonate =180

5.3 g s

1 = 0.0194 g s

1 [1]

(c) From the equation, the mole ratio of sodium hydrogencarbonate to sulphuric acid is 2 : 1.

Number of moles of H2SO4 reacted =2

0417.0 mol = 0.0209 mol

Change in concentration of H2SO4 =

1000

0.250209.0 M = 0.836 M [1]

Average rate of decrease of H2SO4 =180

836.0 M s

1 = 4.64 × 103 M s

1 [1]

(d) From the equation, the mole ratio of sodium hydrogencarbonate to carbon dioxide is 1 : 1.

Number of moles of CO2 produced = Number of moles of NaHCO3 reacted = 0.0417 mol [1]

Change in mass of CO2 = 0.0417 × (12.0 + 16.0 × 2) g = 1.83 g

Average rate of production of CO2 = 180

83.1 g s

1

= 0.0102 g s1

[1]



4(a) Number of moles of Cr 2O7

2 = 0.25 ×1000

0.25 mol = 6.25 × 10

3 mol

Number of moles of I = 0.50 ×1000

0.25 mol = 0.0125 mol [1]

The mole ratio of Cr 2O72 to I

is 1 : 6.

0.0125 mol of I can only react with 2.08 × 10

3 mol of Cr 2O72 completely.

∴ Acidified Cr 2O72 is in excess. [1]

(b) Since iodide is the limiting reagent, its concentration equals zero when the reaction is complete.

Initial concentration of I(aq) =

1000

0.250.251000

0.2550.0

M = 0.250 M [1]

Average rate of decrease of I(aq) concentration =40

250.0 M s1 = 6.25 × 103 M s1 [1]

(c) From the chemical equation, the mole ratio of Cr 2O72(aq) to I

(aq) is 1 : 6.

6 × rate of decrease of Cr 2O72 concentration = rate of decrease of I

[1]

Average rate of decrease of Cr 2O72(aq) concentration =

6

106.25 3

M s1 = 1.04 × 103 M s1

[1]

(d) From the chemical equation, the mole ratio of I(aq) to I2(aq) is 2 : 1.

rate of decrease of I concentration = 2 × rate of increase of I2 concentration [1]

Average rate of increase of I 2(aq) concentration =2

1025.6 3

M s1 = 3.13 × 10

3 M s1 [1]

5(a) Average rate of change in the amount of A =

0)min(6

0.8)mol(0.0125

[1]

= 0.13 mol min1 [1]

(b)



For correct labelling of axes [1]

For correct plot of graph [1]

(c)

Initial rate of reaction =0)min(1

0.8)mol38.0(

[1] = 0.42 mol min

1 [1]

(d)


Time (min)


Time (min)



Instantaneous rate of reaction at the third minute =2)min(4

0.17)mol04.0(

[1]

= 0.065 mol min1 [1]

6 (a) This is because the concentration of the reactant is the highest at the beginning of the reaction, thus

the reaction rate is also the highest. [1](b) Yes. The concentration of the reactant drops to zero from Y to Z , which shows the completion of

reaction. [2]

(c)

For correct drawing [1]

For correct labelling [1]

7 (a) The change in concentration of Br 2(aq) can be monitored by colorimetry. [1] Initially, the colour of


Time (min)

C o n c e n t r a t i o n o f a p r o d u c t

X

W

V

0 t Y Z Time



Br 2(aq) is brown. [1] When the reaction proceeds, the brown colour fades out. [1] The colour

intensity measured by the colorimeter is proportional to the concentration of Br 2(aq). [1]

(b)

Correct sketching [1]

Correct labelling of x-axis and y-axis [1]

8 (a) NaHCO3(s) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l) [1]

(b)

Correct diagram [1]

Correct labelling [1]

(c) (i) No. of moles of HCl = 2.0 ×1000

30 mol= 0.060 mol [1]

No. of moles of NaHCO3 =30.160.120.11.23

0.5

mol = 0.0595 mol [1]

From the equation, the mole ratio between HCl and NaHCO3 is 1 : 1.

∵ No. of moles of NaHCO3 < no. of moles of HCl [1]

∴ NaHCO3 is the limiting reactant. [1]

(ii) Average rate of the reaction =602

0.12.1

atm s1 = 1.67 × 103 atm s1 [1]

9 (a) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) [1]

(b)

to data-logger

interface and

computer baking soda

suction flask

dilute

hydrochloric

acid

Time

C o l o u r i n t e n s i t y



(correct diagram [1] correct labelling [1])

One of the products is in the gaseous form. [1]

(c) There can be a continuous monitoring of the progress of the reaction./It causes little disturbance to

the reaction. [1]

(d)

Correct graph trend [1]

Correct axes labels [1]

10 (a) 2NaN3(s) 3N2(g) + 2Na(s) [1]

(b) High. [1] Car accidents can happen instantaneously. The airbag must inflate almost simultaneously

upon a car crash. [1]

(c) Nitrogen is chemically inert. [1] It does no harm to passengers when an accident occurs.(d) To fasten the seat belts in cars. [1] (Accept other reasonable answers.)

11(a) Rate =

020

027.030.0

g s1 = 0.0137 g s1 [2]

(b) Rate =3.24

0137.0 mol s

1 = 5.64 × 104 mol s

1 [2]

(c) Decrease [1]

(d) No change [1]

(e)

gas syringe

magnesium ribbon dilute h drochloric acid

V

o l u m e o f h y d r o g e n g a s

Time



Correct plotting [1]


(f) Since hydrogen gas is formed in the reaction, [1] the volume of hydrogen gas can be collected by a

gas syringe at different instants. [1]

OR

The pressure of the reaction system at different instants can be measured if the reaction occurs in a

closed container. [2](g)

OR

M a s s o f M g ( g )

Time (s)

V o l u m e o f h y d r o g e n

Time





12 (a) Pour the sodium thiosulphate solution into the beaker. Then place the beaker over a black cross

marked on the white tile. [1] Pour the hydrochloric acid quickly into the thiosulphate solution and

start the stopwatch at the same time. [1] Stir the mixture gently and observe the cross vertically

down through the solution. Record the time taken for the ‘blot out’ of the cross. [1]

Correct diagram [1]


(b) S2O32(aq) + 2H+(aq) H2O(l) + SO2(g) + S(s) [1]

P r e s s u r e i n

t h e c o n t a i n e r

Time

(c) Sulphur dioxide/SO2 [1]

(d) (i) rate crosstheof out' blot'for thetakentime

1 [1]

OR

The time for ‘blot out’ is inversely proportional to the average rate of reaction. [1]

beaker

sodium thiosulphate solution +

hydrochloric acid

cross



(ii) Rate =250

1 s

1= 0.00400 s1 [1]

(e) (i) Since sulphur dioxide is formed as a gaseous product in the reaction, [1] the volume of

sulphur dioxide collected can be measured by a gas syringe at different instants. [1]

(ii) The creamy yellow precipitate of sulphur makes the solution cloudy and affects the

transmittance of light through the solution, which is easily to be observed. [1] Since sulphur

dioxide is quite soluble in water, the volume of SO2(g) collected is not accurate. [1]

13 (a) Egg shells dissolve. [1]

Effervescence/Colourless bubbles evolve. [1]

(b) CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O(l) + CO2(g) [1]

(c) It is used to measure the change of pressure in the container during the reaction. [1]

(d) We can collect and store data over very short time intervals by using a data-logger. It can record

hundreds of readings for more accurate analysis at a later time. This is impossible or impractical

for a human operator. [2]

(e) (i) The initial increase in pressure of the flask is small since the reaction mixture has to be first

saturated with the CO2 formed. [1] After that, the curve is the steepest as the rate of reaction

is the highest. The curve then becomes less steep with time due to the decrease in rate of the

reaction. [1] Finally it becomes horizontal which indicates that the reaction stops. [1]

(ii) This pressure is the same as the pressure of the reaction system before the reaction starts. [1]

(f) Weigh the mass of the egg shells before the experiment. [1] Mix an excess known amount of HCl

with the egg shells. [1] When the reaction is complete, titrate the reaction mixture against standard

KOH(aq)/NaOH(aq). [1] The amount of unreacted HCl can be found and the mass of CaCO 3 reacted can be calculated from the titration result. [1] The formula below is used to calculate the

percentage by mass of CaCO3:

shellseggof Mass

CaCOof Mass 3 × 100% [1]

14 (a) Adding excess HCl immediately removes all NaOH in the reaction mixture [1] to quench the

reaction. [1]

(b) Titrate the mixture containing excess HCl against standard NaOH. [1] Number of moles of NaOH

present in the reaction mixture equals the difference between number of moles of excess HCladded and that of NaOH required to neutralize the remained acid. [1]

(c) (i) Pour the small portion of reaction mixture into ice-cold distilled water. [1]

(ii) The reaction does not stop as both reactants are still present in the mixture. [1]


(b) X is hydrogen gas. [1] It gives a ‘pop’ sound when a burning splint is brought near the mouth of a

test tube containing it. [1]

(c) Any TWO of the following: [2]

Friction between the piston and the syringe body slightly compresses the gas inside.

The gas syringe might not be air-tight.

Not all hydrogen gas released from the conical flask goes into the gas syringe.



Leakage of gas might occur through connections between any two apparatus.

Heat change of the reaction heats up the gas and expands it.

(Accept other reasonable answers)

(d) Measure the change of pressure in the container with a pressure sensor connected to a data-logger

in a closed system. [1] Hydrogen gas is the only gas among the reactants and products, the

pressure in the container increases as the reaction proceeds. [1]

OR

Measure the change in mass of the reaction mixture in an open system. [1] Hydrogen gas produced

can escape from the reaction mixture, so the mass of the reaction mixture decreases during the

reaction. [1]

(e) Yes. [1] There was no further increase in volume of gas after t = 5 min. [1]

(f)



(g)



16 (a) As the reaction proceeds, the solution changes from deep purple to pale pink (or colourless). [1]

3 )

V o l u m e o f X c o l l e c t e d ( c m

Time (min)

M a s s o f m a g n e s i u m s t r i p

Time (min)



(b) He is incorrect. The colour change from purple to pale pink (or colourless) is mainly due to the

formation of Mn2+ from MnO4. [1] The amount of water formed has a negligible volume in this

reaction. [1]

[Remarks:

Suppose 20 cm3 of solution is inside the test tube and the initial concentrations of MnO4 and

C2O42 are 0.20 M and 0.50 M respectively.

Number of moles of MnO4 present = 0.20 ×

1000

20 mol = 0.00400 mol

Number of moles of C2O42 present = 0.50 ×

1000

20 mol = 0.0100 mol

Mole ratio of MnO4 : C2O4

2 = 0.00400 : 0.0100 = 2 : 5

∴ both reactants just react completely.

Number of moles of H2O produced = 0.0160 mol

Molar mass of H2O = 1.0 × 2 + 16.0 g mol1 = 18.0 g mol1 and density of H2O = 1.0 g cm3

∴ the volume of H2O formed =0.1

0160.00.18 cm3 = 0.288 cm3, which is insignificant when

compared with the total volume of the solution.]

(c)


Correct sketching [1]17 (a) Use a colorimeter to follow the change in colour intensity [1] since MnO4

is purple in colour,

while other species in the reaction is colourless. [1]

(b) Hydrochloric acid. [1] HCl is a much stronger acid than ethanoic acid, so it can provide more H+

ions in aqueous medium, [1] which can better fulfil the high H+ ion concentration requirement of

MnO4 (ratio of MnO4

to H+ is 1 : 8). [1]

(c) The method is not suitable. [1] The main problem of this method is that the change in colour at the

end point is not obvious enough [1] since the colour of MnO 4 is much more intense than that of

phenolphthalein. [1]

18 (a) Collect the oxygen formed by gas syringe, [1] and record the volume collected at regular time

intervals. [1]

C o n c e n t r a t i o n o f M n O 4

Time



(b) Manganese(IV) oxide [1]

(c) At regular time intervals, [1] withdraw a small portion of the reaction mixture and filter out the

catalyst. [1] Then titrate the small portion of the reaction mixture against standard acidified

potassium permanganate solution. [1]

(d) 5H2O2(aq) + 2MnO4(aq) + 6H+(aq) 2Mn2+(aq) + 8H2O(l) + 5O2(g) [1]

19 (a) X should be the total mass of container and its contents. [1]

(b) Measure the change in mass of the reaction mixture by electronic balance. [1]

(c)

Correct labelling of axes [1]Correct points [1]

Correct curve [1]

(d) Connect a gas syringe to the set-up [1] to measure the change in volume of gas with time. [1]

OR

Connect a pressure sensor with a data-logger to the set-up [1] to measure the change of pressure

inside the container with time. [1]

20 (a) Measure the change in colour intensity. [1]

(b) Measure the change in colour intensity. [1]

Measure the change in volume of the gas. [1]

Measure the change in pressure of the reaction system. [1]

Time (min)

T o t a l m a s s o f c o n t a i n e r a n d c o n t e n t s



Titrimetric analysis [1]

(c) Measure the change in pressure of the reaction system. [1]

Titrimetric analysis. [1]

21 (a) Adding excess HCl immediately removes all NaOH in the reaction mixture [1]. Since one of the

reactants is removed, the reaction can be quenched. [1]

(b) The reaction mixture can be poured into ice-cold distilled water to slow down the reaction

progress. [1]

(c) The reaction is not completely stopped by pouring the reaction mixture into an ice-cold distilled

water. [1]

(d) Titrate the mixture with excess HCl against standard NaOH solution. [1] The number of moles of

NaOH present in the reaction mixture before quenching is the difference between the number of

moles of excess HCl added and the number of moles of NaOH required to neutralize the excess

acid. [1]


(b) X is hydrogen gas. [1] It burns with a ‘pop’ sound when it is tested with a burning splint. [1]

(c) Friction between the plunger and the gas syringe slightly compresses the gas inside. [1]/There may

be leakage of gas through junctions between any two apparatus. [1]/Heat given out by the reaction

heats up the gas and expands it. [1]

(d) Yes. With a pressure gauge, the gas produced will not be allowed to expand but rather builds up a

pressure inside the flask. [1] The increase in pressure is directly proportional to the amount of gas

present. [1]

(e) Yes. Since there is no further increase in volume of gas X after the sixth minute, we can deducethat the reaction has completed. [1]

(f)


Time (min)

V o l u m e o f X c o l l e c t e d ( c m

3 )



For correct plot of graph [1]

(g)

For ability to sketch an upside down graph of (f) [1]

23 (a) As the reaction is going on, the solution turns gradually from purple to pale pink (or colourless).

[1] There is effervescence due to the formation of carbon dioxide gas. [1]

(b) No. The colour change of the solution from purple to pale pink (or colourless) is mainly due to

formation of Mn2+(aq) ions from MnO4(aq) ions instead. [1] The volume of water formed in this

reaction is too small to give an observable dilution effect. [1]

(c)

Further explanation

Suppose the test tube contains 20 cm3 of solution, with initial [MnO4(aq)]

and [C2O42(aq)] of 0.2 M and 0.5 M respectively.

Number of moles of C2O42 present = 0.5 mol dm3

1000

20 dm3 = 0.01 mol

Number of moles of MnO4 present = 0.2 mol dm

3 1000

20 dm3 = 0.004

mol

Consider the mole ratio of MnO4 to C2O4

2, the amount of C2O42 is just

enough for reacting with all MnO4

.Hence, no. of moles of H2O produced = 0.016 mol

Since molecular mass of H2O = 1.0 × 2 + 16.0 g mol1 = 18.0 g mol1, and

density of water = 1 g cm3; therefore the volume of H2O formed =

3

1

cmg1

mol016.0molg0.18

= 0.288 cm3 which is insignificant with respect to

the total volume of the reaction mixture.

M a s s o f m a g n e s i u m r i b b o n ( g )

Time (min)



For correct sketch of graph [1]


24 (a)

Correct diagram [1]


(b) Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) [1]

(c) All the magnesium metal was dissolved./No more gas bubbles were evolved. [1](d) (i) The initial rate of experiment I is higher than that of experiment II [1] since HCl is a stronger

acid than CH3COOH. [1]

(ii) The total volume of hydrogen collected in experiment I is equal to that in experiment II [1]

since the amount of the limiting reactant (magnesium) in both experiments is the same. [1]

(e)

C o n c e n t r a t i o n o f M n O 4 −

( a q ) ( m o l ) d m − 3 )

Time (min)

dilute acid (in excess)

stopwatch

gas syringe

gas collected

plungerrubber connecting tubing

magnesium

ribbon

cotton

thread




Correct sketching of curve I [1]

Correct sketching of curve II [1]

(f) The reaction time is shorter than that of experiments I and II. [1]

This is because sulphuric acid is a dibasic acid, thus the concentration of H+ is higher than that in

HCl and CH3COOH. [1] Therefore, the reaction rate would be higher when sulphuric acid was

used. [1]

25 (a) S2O32(aq) + 2H+(aq) S(s) + SO2(g) + H2O(l) [1]

(b)

Trialt

1 (s

1)

1 0.0278

2 0.0208

3 0.0175

4 0.0147

5 0.00826

6 0.00633

[2]

(c)

II

I

Time

V o l u m

e o f H 2 ( g )





(d) As [HCl] increases,t

1 also increases. [1] The relationship between rate of reaction and [HCl]

should be similar because reactionof rate1

t . [1]

(e) An increase in [HCl] will increase the frequency of effective collisions, [1] so the reaction rate will

be higher. [1]

26 (a) CaCO3(s) + 2H+

(aq) Ca2+

(aq) + H2O(l) + CO2(g) [1](b) The rate decreases with time. [1]

The concentration of the reactants decreases with time so the number of effective collisions

decreases. [1]

Draw a tangent to the curve at time t .

Rate of reaction at time t = slope of the tangent [1]

(c)

Correct sketching of curve (i) [1]

Correct sketching of curve (ii) [1]

(i) Volume of CO2 produced is smaller as the amount of HCl is halved. [1]

[HCl] (mol dm3)

1 / t ( s 1 )

(i)

(ii)

original

Time

V o l u m e o f C O 2



The curve is less steep since the concentration of HCl is lower. [1]

(ii) Equal volume of CO2 is produced since amount of HCl is unchanged. [1]

The curve is less steep since the concentration of HCl is lower. [1]

27 (a) It is incorrect. [1] When the pressure inside the container is higher than 1 atm, the temperature of

boiling water exceeds 100C. [1]

(b) The boiling point of water is raised (>100C) when the the pressure inside the container is higher

than atmopheric pressure. [1] An increase in temperature raises the average kinetic energy of the

reactant particles. [1] The higher chance of collisions increases the number of effective collisons

and hence the reaction rate. [1] Thus the cooking time is shortened.

(c) At a higher temperature, a larger portion of viruses and bacteria can be killed. [1]

28

(a) 2H2O2(aq) 2H2O(l) + O2(g) [1]

(b) Catalyst. [1]

(c) To increase the surface area, [1] so the reaction rate can be increased. [1]

(d) It can be reused [1] since a catalyst remains chemically unchanged at the end of the reaction. [1]

(i.e. it still has the catalytic activity.)

(e)

Correct diagram [1]


(f) The reaction should involve the release of a gaseous product. [1]There is a change in mass of the reaction mixture. [1]

29 (a) 4H+(aq) + 4I(aq) + O2(g) 2I2(aq) + 2H2O(l) [1]

(b) Add a few drops of starch solution into a test tube containing a certain amount of potassium iodide

solution and a fixed volume of sodium thiosulphate. [1] Add a certain amount of sulphuric acid to

another test tube. [1] Place both test tubes at room temperature for 15 minutes. [1] Then pour the

sulphuric acid into the test tube containing sodium thiosulphate, potassium iodide and starch

solution quickly. Start the stopwatch at the same time. [1] The time taken for the occurrence of a

dark blue colour is measured. Repeat the steps at different temperatures. [1]

(c) (i)

MnO2

wool

MnO2

H2O2(aq)

electronic balance





(ii) It is an exponential curve. [1]

This is because the reaction goes much faster at higher temperature. [1]

30 (a) (i) The rate is decreasing. [1]

Concentrations of the reactants decrease with time. Effective collisions become less frequent.

[1]

(ii) Reactant(s) was/were used up./Reaction is complete. [1]

(iii) 140 cm3 [1]

(iv) 60 s [1]

(b) (i) The rate will increase. [1] An increase in temperature raises the average kinetic energy of thereactant particles. They have more energy and move faster. [1] The number of effective

collisions increases so the reaction rate becomes higher. [1]

(ii) The rate will be lower [1] since large lumps of solid ammonium chloride have a smaller

surface area. [1]

31 (a) CaCO3(s) + 2CH3COOH(aq) (CH3COO)2Ca(aq) + CO2(g) + H2O(l) [1]

Temperature

R e a c t i o n r a t e

(b)

Correct curves [3]

3

2

1

Time

D e c r e a s e i n m a s s



(c) (i) Experiment 2 gives a higher initial rate than experiment 1. [1] The powder provides a larger

surface area than the fragments. [1] The higher chance of collision increases the number of

effective collisions and hence the reaction rate. [1]

(ii) Experiment 3 gives a higher initial rate than experiment 2. [1] The increase in temperature

raises the average kinetic energy of the reactant particles. [1] They have more energy and

move faster. [1] The higher chance of collision increases the number of effective collisions

and hence the reaction rate. [1]

32 (a) Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) [1]

(b) The reactant particles become more crowded when the concentration increases. [1] They have a

higher chance to collide and this increases the number of effective collisions. [1] As a result, the

reaction rate is higher. [1]

(c) (i) Sodium carbonate will react with some of the sulphuric acid. [1] The concentration of

hydrogen ions decreases. [1] Therefore, the rate of production of hydrogen gas decreases. [1]

(ii) Potassium sulphate solution dilutes the reaction mixture, [1] so the concentration of hydrogen

ions decreases. [1] Therefore, the rate of production of hydrogen gas decreases. [1]

(iii) Hydrogen chloride gas will dissolve into the reaction mixture to give H+ and Cl ions, [1] so

the concentration of hydrogen ions increases [1] and the rate of production of hydrogen gas is

higher. [1]

(d) The reaction rate will be lower [1] since the surface area of zinc granules is smaller than that of

zinc powder. [1]

33 (a) Amount of limestone/Concentration of hydrochloric acid/Experimental temperature (Any TWO)

[2](b)

B A

Time (min)

V o l u m e o f g a s ( c m

3 )



(Correct labelling of axes [1] Correct curve A [1] Correct curve B [1])

Experiment A [1]

(c) (i) The surface area of aluminium dust is much larger, [1] so it is much easier to react with air to

form the corresponding oxide, by which enormous heat is given out in a short time. [1]

(ii) Twigs have larger surface area than logs. [1]

34 (a) Any TWO of the following:

Measure volume of carbon dioxide/CO2/gas produced. [1]

[1]

Measure pH of the reaction mixture. [1]

[1]

Measure mass of the chemicals/apparatus. [1]

[1]

(b) Any TWO of the following:

Use powdered MgCO3. [1]

Surface area of MgCO3 is increased. [1]

Time

M a s s

p H

Time

Time

V o l u m e o f C O 2



Increase (reaction) temperature/heat. [1]

Average kinetic energy of reactant particles is increased, so effective collisions occur more

frequently. [1]

Increase acid concentration. [1]

The reactant particles become more crowded, so the number of effective collision increases. [1]

(c) (i) Remain unchanged. [1]

MgCO3 was already in excess. [1]

(ii) Remain unchanged. [1]

The same quantities of reactants are used. [1]

35 (a) 2S2O32(s) + I2(aq) S4O6

2(aq) + 2I(aq) [1]

(b) Number of moles of Na2S2O3(aq) =30.1621.3220.23

30

mol = 0.190 mol [1]

Number of moles of I2(aq) = 0.50 ×1000

250 mol = 0.125 mol [1]

The mole ratio of S2O32 to I2 is 2 : 1.

0.190 mol of S2O32 only needs 0.095 mol of I2 to react completely.

∴ I2 is in excess and Na2S2O3 is the limiting reagent. [1]

(c) Use I2 solution with a higher concentration./Use powdered Na 2S2O3./Increase the temperature.

(Any TWO) [2]

(d) Colour intensity. [1]36 (a) Sodium reacts with hydrochloric acid explosively [1] so the experiment is very difficult to control.

[1]

(b)

Correct assignment of five curves [5]

(c) (i) Powdered slaked lime has a much larger surface area, [1] so it can neutralize acid soils much

faster. [1]

Time

V o l u m e o f H 2 ( g )

1

2

5

4

3



(ii) The surface area of the ingredients increases, [1] so the cooking process can be faster. [1]

(iii) The temperature inside the refrigerator is low, [1] so the reactions that lead to deterioration of

food proceed slower. [1]

37 (a) The initial rate is lower since the concentration of acid is lower. [1] Half the amount of hydrogen is

formed since half the amount of acid is used. [1]

(b) The initial rate is higher as the temperature of acid is raised. [1] Equal amount of hydrogen is

produced since the amount of acid is unchanged. [1]

(c) The initial rate is higher as copper(II) ions act as a catalyst to the reaction. [1] Equal amount of

hydrogen is formed since the amount of acid is unchanged. [1]

[Remarks: The addition of copper(II) sulphate promotes the release of electrons from zinc. Zn(s)

Zn2+(aq) + 2e]

(d) The initial rate is lower since the surface area of zinc decreases. [1] Equal amount of hydrogen is

produced since the amount of acid is unchanged. [1]

38

New conditionChange in reaction

rate, if anyReason

Using 5 g of powdered

zincIncreased [1]

Surface area of the solid

reactant increased. [1]

Using 3 g of zinc

granules

[Remarks: Assume that

the size of granules

remains unchanged.]

Decreased [1]Surface area of the solid

reactant decrease. [1]

Using 100 cm3 of 2.0 M

ethanoic acidDecreased [1]

A weaker acid was used

instead, so the ionization

was less completed. [1]

Changing the temperature

to 40CIncreased [1]

Increase in temperature

will speed up the reaction.

[1]

Adding a few drops of

aqueous copper(II) nitrateIncreased [1]

CuSO4 acted as a catalyst

to speed up the reaction.

[1]

[Remarks: The addition of

copper(II) nitrate promotes

the release of electrons

from zinc. Zn(s)

Zn

2+

(aq) + 2e

]39 (a)




Correct curve [1]

Initial rate =60)04.2(

010.00

mol s1 = 6.94 × 105 mol s1 (The negative sign means a decrease

in the amount of X) [1]

(b)

Correct sketching [1]The reaction rate is higher when a catalyst is used, so the new curve is steeper initially and

becomes horizontal earlier. [1] The amount of X remained is unchanged since a catalyst has no

effect on the yield of the product. [1]

(c) Use powdered catalyst [1] to increase the surface area of the catalyst [1] and hence the reaction

rate.

40 (a) Colorimetry. [1] Since the reaction mixture changes from colourless to brown as the reaction

proceeds, [1] there is a change in colour intensity. [1] A colorimeter can be used to follow the

progress of the reaction.

OR

Titrimetric analysis. [1] A small portion of the reaction mixture can be extracted and quenched by

Time (min)

A m o u n t o f X ( m o l )

‘b’

Time (min)

A m o u n t o f

X

m o l



ice-cold distilled water at regular time intervals. [1] Then the amount of iodine in it can be found

out by titration against standard S2O32. [1]

(b) Increase the concentration of S2O82(aq)/Increase the concentration of I

(aq)/Increase the

temperature of the reaction mixture (Any TWO [2])

(c) (i) It is because the reaction involves collisions of ions of the same charge. [1]

(ii) 2Fe3+(aq) + 2I(aq) 2Fe2+(aq) + I2(aq) [1]

(iii) Peroxodisulphate ions are reduced by Fe(II) ions. [1]

2Fe2+(aq) + S2O82(aq) 2Fe3+(aq) + 2SO4

2(aq) [1]

(iv) It can increase the reaction rate. [1]

It remains chemically unchanged at the end of a reaction. [1]

(d) (i) Zymase [1]

(ii) Vanadium(V) oxide/V2O5 [1]

(iii) Manganese(IV) oxide/MnO2 [1]

41 (a) The rate of production of oxygen gas can be monitored by measuring the change in mass of H2O2-

(aq) with time when it undergoes decomposition./by measuring the change in volume of oxygen

gas collected with time. [1]

(b) Any TWO of the following:

Increase the concentration of the H2O2(aq) solution. [1]

Increase the temperature of the H2O2(aq) solution. [1]

Add a catalyst (MnO2) to the reaction mixture. [1]

(c)


(d) The biological catalyst involved in the decomposition is called enzyme. [1] Initially, the activity of

enzyme increases with temperature. [1] Its activity is the highest at 37C. After 37C, its activity

decreases with further increase in temperature. [1] At a high temperature of 80C, the enzyme is

denatured and its activity becomes zero. [1]

42 (a) N2(g) + 3H2(g) 2NH3(g) [1]

Finely divided iron [1]

(b) The collision theory states that a chemical reaction only occurs when the reactant particles undergo

effective collisions. [1] Effective collisions can only be caused by reactant particles colliding at

high speed [1] and in the right orientation. [1]

original

‘c’

Time

V o l u m e o f o x y g e n g a s



(c) The rate of reaction is proportional to the number of effective collisions per unit time [1] between

reactant particles.

(d) Reactant particles carry different kinetic energies/are moving at different speeds. [1] Lowering the

minimum energy barrier would allow larger proportion of particles carrying enough kinetic energy

to cause effective collisions. [1] This increases the number of effective collisions per unit time and

hence the reaction rate. [1]

(e) He is partly correct. [1]

A catalyst can increase the rate of reaction so time can be saved and cost of production can be

reduced. [1] However, a catalyst has no effect on the yield of a reaction. [1]

43 (a) Catalyst is a substance that changes (usually increases) the reaction rate, [1] but itself remains

chemically unchanged at the end of the reaction. [1]

(b) To increase the surface area of the catalyst [1] and hence the reaction rate. [1]

(c) Vanadium(V) oxide/V2O5 [1]

Oxidation state = +5 [1]

(d) The student is partly correct. [1]

The presence of enzymes can cause browning of some fruits and vegetables. [1] However, the

enzymes only speed up the chemical reactions which lead to the formation of a brown substance in

fruits and vegetables. [1]

(e) Manufacture of beer/wine/cheese/soy sauce/yogurt. [2] (Any TWO)

44 (a) 1. [1] The curve is the steepest at the start. [1]

(b) 1. [1] The more concentrated the acid, the higher will be the rate of reaction. [1]

(c) 2. [1] The curve is flattened at the highest level, indicating the largest amount of hydrogen gasevolved. [1]

(d) (i) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

Number of moles of HCl used in Experiment 2 = MV mol

Since the volume of H2 produced is halved in Experiment 3, number of moles of HCl used is

also halved, i.e. equal to2

MV mol. [1]

∴ volume of HCl needed =

2

2 M

MV

cm3 = V cm3 [1]

(ii) Higher temperature of HCl/Mg with larger surface area/Higher concentration of HCl but

using less volume to keep the number of moles of HCl halved (Any TWO) [2]

45 (a) (i) Decomposition is a process in which a compound breaks down into two or more substances,

usually as a result of heating. [1]

(ii) It should be regenerated at the end of the reaction [1] and it is used in only trace amount. [1]

(iii)



Correct diagram [1]


(b) (i) Number of moles of hydrogen peroxide =1000

100 × 2.0 mol = 0.200 mol [1]

(ii) Mole ratio of H2O2 : O2 = 2 : 1

Number of moles of oxygen (O2) formed =2

200.0 mol = 0.100 mol [1]

(iii) Volume of oxygen produced = 0.100 × 24.0 dm3 = 2.40 dm3 [1]

46 (a) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) [1]

(b) From the equation, volume ratio of C2H4(g) : O2(g) = 1 : 3

C2H4(g) was the limiting reactant.

Volume ratio of C2H4 : CO2 = 1 : 2Volume of CO2 produced = 2 × 100 cm3 = 200 cm3 [1]

Volume of O2 remained = 400 3 × 100 cm3 = 100 cm3 [1]

47 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)

Let the volume of C2H2(g) be x cm3, then the volume of C2H4(g) is (10.0 x) cm3.

Volume of O2(g) reacted with C2H2(g) =2

5 x cm3 = 2.5 x cm3 [1]

Volume of O2(g) reacted with C2H4(g) = 3 × (10.0

x) cm

3

= 30.0

3 x cm

3

[1]Therefore, 30.0 2.5 x (30.0 3 x) = 2.0 [1]

x = 4.0 [1]

Volume of C2H2(g) = 4.0 cm3

Volume of C2H4(g) = 6.0 cm3

48C xH y(g) + ( x +

4

y)O2(g) xCO2(g) +

2

yH2O(l) [1]

Volume of carbon dioxide formed = 110 30.0 cm3 = 80.0 cm3

C xH y(g) was the limiting reactant.

Volume ratio of C xH y : CO2 = 1 : 4

Mole ratio of C xH y to CO2 should be 1 : 4.

manganese(IV)

oxide

gas syringe

H2O2(aq)

oxygen gas



Therefore, x = 4 [1]

Volume of O2 reacted = 150 30.0 cm3 = 120 cm3

Mole ratio of O2 to CO2 = volume ratio of O2 to CO2

4

44

y

= 0.80

120

Solving the equation, y = 8 [1]

∴ the molecular formula of A is C4H8. [1]

49(a) Number of moles of H2 present =

24.0

104.80 3

mol = 200 mol [1]

(b) 2H2(g) + O2(g) 2H2O(l) [1]

(c) From the equation, mole ratio of H2 : O2 = 2 : 1

By Avogadro’s Law, volume ratio of H2 : O2 = 2 : 1

Volume of oxygen consumed =2

104.80 3 dm3 = 2.40 × 103 dm3 [1]

Mole ratio of H2 : H2O = 1 : 1

Number of moles of water formed = 200 mol

Mass of water formed = 200 × (1.0 × 2 + 16.0) g = 3600 g [1]

Volume of water formed =0.1

3600 cm3 = 3600 cm3 [1]

(d) Volume of air required =

21%

102.40 3

dm3 = 11 429 dm3 [1]

50 (a) CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) [1]

(b)

Time (s) 5 10 15 20 25 30 35

No. of moles of gas produced

(10 4

mol)4.17 6.25 7.5 7.92 8.33 8.33 8.33

(c)

[2]





51 (a) 2H2(g) + O2(g) 2H2O(l) [1]

(b) From the equation, mole ratio of H2 : O2 = 2 : 1 [1]

Volume ratio of H2 : O2 = 2 : 1 [1]The volume of oxygen needed = 100 cm3

∴ the excess volume of oxygen = 200 cm3 100 cm3 = 100 cm3 [1]

52 (a) NH3(g) + HBr(g) NH4Br(s) [1]

(b) From the equation, mole ratio of NH3 : HBr = 1 : 1 [1]

Volume ratio of NH3 : HBr = 1 : 1 [1]

The volume of HBr needed = 45 cm3

∴ the excess volume of HBr = 60 cm3 45 cm3 = 15 cm3 [1]

53 (a) The initial rate =09.0

120

cm3 min1 = 13.3 cm3 min1 (The negative sign means the rate of

decrease in amount of reactant.) [1]

(b) The reaction took 7 minutes to complete. The final volume of oxygen was 1.41 cm3. [1]

The average rate =07

0.1241.1

cm3 min

1 = 1.51 cm3 min1 (The negative sign means the rate of

decrease in amount of reactant.) [1]

(c) (i) From the equation, the mole ratio of H2 : O2 is 2 : 1.

Volume of O2 reacted = 12.0 1.41 cm3 = 10.6 cm3

∴ volume of H2 reacted = 2 × 10.6 cm3 = 21.2 cm3

Volume of H2 remained = 24.0 21.2 cm3 = 2.80 cm3 [1]

N u m b e r o f m o l e s o f g a s ( c m

3 )

Time s



Mole ratio of H2O : O2 is 2 : 1.

Volume of H2O formed = 2 × 10.6 cm3 = 21.2 cm3 [1]

∴ total volume of gases = 21.2 + 1.41 + 2.80 cm3 = 25.4 cm3 [1]

(ii) Percentage change of volume =

%100

0.120.24

0.120.244.25

= 29.4% [1]

54 (a) Collect the oxygen formed using a gas syringe [1] and record the volume collected at regular time

intervals. [1] A graph of volume of oxygen collected against time is plotted. [1]

OR

At regular time intervals, pipette a small portion of the reaction mixture to a conical flask and

quench the small portion with ice-cold distilled water. [1] Then titrate it with standard potassium

permanganate solution. [1] A graph of concentration of hydrogen peroxide against time is plotted.

[1]

(b) Number of moles of H2O2 = 1.0 × 1000

100

mol = 0.100 mol

Mole ratio of H2O2 : O2 = 2 : 1

∴ theoretical yield of oxygen =2

100.0 mol = 0.0500 mol [1]

Actual yield of oxygen = 0.0500 × 86.7% mol = 0.0434 mol [1]

Volume of oxygen produced = 0.0434 × 24.0 dm3 = 1.04 dm3 [1]

(c) Manganese(IV) oxide/MnO2(s) [1]

55 (a) SO2(g) + H2O(l) H2SO3(aq) [1]

H2SO3(aq) 2H+(aq) + SO32

(aq) [1](b) CaCO3(s) + 2H+(aq) Ca2+(aq) + CO2(g) + H2O(l) [1]

Marbles contain calcium carbonate, which can react with acid and be washed away by rain water.

[1]

(c) Number of moles of SO2(g) =0.24

1600 mol = 66.7 mol [1]

Number of moles of CaCO3(s) = Number of moles of SO2(g) = 66.7 mol [1]

Mass of CaCO3(s) = 66.7 × (40.1 + 12.0 + 16.0 × 3) g = 6680 g [1]

56 (a) (i) CH3CH2CH3(g) + 4O2(g)

CO2(g) + 2CO(g) + 4H2O(l) [1]

(ii) Number of moles of CH3CH2CH3(g) =80.130.12

95.4

mol = 0.113 mol [1]

Mole ratio of CH3CH2CH3(g) to CO(g) is 1 : 2.

∴ number of moles of CO(g) = 2 × 0.113 mol = 0.226 mol [1]

Volume of CO(g) produced = 0.226 × 24.0 dm3 = 5.42 dm3 [1]

(b) 2CO(g) + O2(g) 2CO2(g) [1]

(c) (i) Mole ratio of CO(g) : O2(g) = 2 : 1

Number of moles of O2(g) = 2

226.0 mol = 0.113 mol [1]

Volume of O2(g) = 0.113 × 24.0 dm3 = 2.71 dm3 [1]



(ii) Mass of O2(g) = 16.0 × 2 × 0.113 g = 3.62 g [1]

57 (a) 2NaN3(s) 2Na(s) + 3N2(g) [1]

(b) Number of moles of nitrogen gas =00024

1020.5 4

mol = 2.17 mol [1]

(c) Mole ratio of NaN3 : N2 = 2 : 3

Number of moles of NaN3 =3

217.2 mol = 1.45 mol [1]

Mass of NaN3 needed = 1.45 × (23.0 + 14.0 × 3) g = 94.3 g [1]

(d) Average rate =3

4

1040

1000

1020.5

dm3 s1 = 1300 dm3 s

1 [1]

58 (a) 2H2(g) + O2(g) 2H2O(l) [1]

2CO(g) + O2(g) 2CO2(g) [1]

(b) Concentrated KOH is used to absorb carbon dioxide.

Volume of carbon dioxide produced = 51.0 41.0 cm3 = 10.0 cm3 [1]

(c) Mole ratio of CO : CO2 = 1 : 1

Volume of carbon monoxide in Town gas = 10.0 cm3 [1]

(d) ΔV = 51.0 (40.0 + 40.0) cm3 = 29.0 cm3 (negative sign means a decrease in volume) [1]

(e) (i) 10.0 cm3 of carbon monoxide reacted with 5.00 cm3 of oxygen to form 10.0 cm3 of carbon

dioxide.

Decrease in volume caused = 10.0 + 5.00 10.0 cm3 = 5.00 cm3 [1]

(ii) Decrease in volume caused by H2 = total decrease in volume decrease in volume caused by CO

= 29.0 5.00 cm3

= 24.0 cm3 [1]

(f) (i) Method 1:

Let the volume of H2 be x cm3.

Volume of unreacted oxygen + volume of unreacted town gas + volume of carbon dioxide

formed = total volume of final gaseous mixture

(40.0 2 x 5.00) + (40.0 x 10.0) + 10.0 = 51.0 [1]

x = 16.0 [1]

∴ volume of H2 = 16.0 cm3

Method 2:

Decrease in volume caused by H2 = 24.0 cm3

Since water existed as liquid, this volume was contributed by H 2 and O2 only.

Mole ratio of H2 to O2 = 2 : 1

Volume ratio of H2 to O2 = 2 : 1 [1]

∴ volume of H2 = 16.0 cm3 [1]

(ii) Volume of N2 = 40.0 16.0 10.0 cm3 = 14.0 cm3 [1]



59 (a)

Correct diagram [1]


(b)


Correct curve [1]

(c) The concentration of H+

ions is the highest and the surface area of calcium is the largest, [1] so thereaction rate is the highest [1] and the curve is the steepest.

(d) The limiting reactant, Ca, was used up [1] and the reaction rate became zero. [1]

(e) 1.1 min. [1]

(f)

gas syringe

hydrogen gas

watercalcium ribbon

X

V o l u m e o f H 2 ( g ) ( c m

3 )

Time (min)



Correct curve (i) [1]

Correct curve (ii) [1]

(g) Ca(s) + 2H2O(l) Ca(OH)2(aq) + H2(g)

Number of moles of hydrogen gas =00024

60 mol = 0.00250 mol [1]

Since the mole ratio of Ca(s) : H2(g) = 1 : 1,

Number of moles of Ca(s) = 0.00250 mol

Mass of Ca(s) = 0.00250 × 40.1 g = 0.100 g [1](h) Ca(s) + 2HCl(l) CaCl2(aq) + H2(g)

Since the amount of the limiting reactant (Ca) and the mole ratio of Ca to H 2 remain unchanged,

volume of hydrogen gas formed is still 60 cm3. [1]

Hydrochloric acid is more reactive than water towards metal as it is completely ionized to give H +

ions but water only ionizes slightly to give H+ ions. [1]

60(a) Number of moles of magnesium hydroxide produced =

20.120.163.24

6.116

mol = 2.00 mol

[1]From the equation, the mole ratio of magnesium nitride to magnesium hydroxide is 1 : 3.

Number of moles of magnesium nitride needed =3

00.2 mol = 0.667 mol [1]

Mass of magnesium nitride = 0.667 × (24.3 × 3 + 14.0 × 2) g = 67.3 g [1]

(b) The mole ratio of magnesium nitride to ammonia is 1 : 2, so the number of moles of ammonia =

0.667 × 2 mol = 1.33 mol [1]

Volume of ammonia = 1.33 × 24.0 dm3 = 31.9 dm3 [1]

(c) NH3(g) + HCl(aq) NH4Cl(aq) [1]

The mole ratio of ammonia to hydrochloric acid is 1 : 1.

∴ number of moles of HCl needed = 1.33 mol [1]

(ii)

(i) X

Time (min)

V o l u m e o f H 2 ( g

) ( c m

3 )



Volume of 2.0 M HCl needed =0.2

33.1 dm3 = 0.665 dm3 [1]

61 (a) CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(s) [1]

(b) Number of moles of ethyne =00024

650 mol = 0.0271 mol [1]

From the chemical equation, the mole ratio of calcium carbide to ethyne is 1 : 1.

Number of moles of calcium carbide needed = 0.0271 mol [1]

Mass of calcium carbide needed = 0.0271 × (40.1 + 12.0 × 2) g = 1.74 g [1]

(c) C2H2(g) + 2H2(g) C2H6(g) [1]

The mole ratio of ethyne to hydrogen is 1 : 2, so the volume ratio of ethyne to hydrogen is 1 : 2.

Volume of hydrogen gas required = 650 × 2 cm3 = 1300 cm3 [1]

62(a) Number of moles of hydrogen gas =

00024

144 mol = 6.00 × 103 mol

The mole ratio of sodium hydroxide to hydrogen is 2 : 1.

Number of moles of sodium hydroxide formed = 2 × 6.00 × 103 mol = 0.0120 mol [1]

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) [1]

The mole ratio of HCl to NaOH is 1 : 1, so 0.0120 mole of HCl is needed.

Volume of HCl needed =5.0

0120.0 dm3 = 0.0240 dm3 = 24.0 cm3 [1]

(b) From the given equation, the mole ratio of the alloy to sodium hydroxide is 1 : 1, so 0.0120 mole

of sodium exists in the alloy. [1]

Mass of sodium = 0.0120 × 23.0 g = 0.276 g [1]

(c) Percentage by mass of lead =00.3

276.000.3 × 100% = 90.8% [1]

63 (a) The mass of sulphur burnt each day = 25 000 × 0.6 % = 150 tonnes [1]

(b) S(s) + O2(g) SO2(g) [1]

(c) (i) Mass of Sulphur = 150 × 1 × 106 g = 1.5 × 108 g [1]

No. of moles of sulphur = mol1.32

105.1 8

= 4.7 × 106 mol

From the equation, mole ratio of S : SO2 = 1 : 1 No. of moles of SO2 = 4.7 × 106 mol

Mass of SO2 produced = 4.7 × 106 mol × 64.1 g mol1 = 301.3 tonnes [1]

(ii) Volume of SO2 produced = 4.7 × 106 mol × 24.0 dm3 mol1 [1]

= 1.13 × 108 dm3 [1]

64 (a) 2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(l) [1]

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) [1]

(b) Number of moles of octane present =181.0812.0

75%100

mol = 0.658 mol [

2

1]

Number of moles of ethanol present =16.061.0212.0

25%100

mol = 0.543 mol [

2

1]



Mole ratio of C8H18 : O2 = 2 : 25

Number of moles of oxygen required to burn octane = 0.658 ×2

25 mol = 8.23 mol [

2

1]

Mole ratio of C2H5OH : O2 = 1 : 3

Number of moles of oxygen required to burn ethanol = 0.543 × 3 mol = 1.63 mol [ 2

1

]

Total number of moles of oxygen required to completely burn 100 g of the fuel = 8.23 + 1.63 mol

= 9.86 mol [2

1]

Volume of oxygen needed to completely burn 100 g of the fuel = 9.86 × 24.0 dm3 = 237 dm3 [2

1]

65 (a) Mole ratio of N2 to H2 is 1 : 3.

Number of moles of H2(g) reacted =00024

0.72 mol = 3.00 × 10

3 mol [1]

Number of moles of N2(g) used =3

1000.3 3

mol = 1.00 × 103 mol [1]

(b) Mole ratio of NH3(g) to N2(g) is 2 : 1.

Number of moles of NH3(g) = 2 × 1.00 × 103 mol = 2.00 × 103 mol [1]

Volume of NH3(g) produced = 2.00 × 103 × 24.0 dm3 = 0.0480 dm3 [1]

(c) No. [1]

The reaction is reversible. [1]

66 (a) Effervescence occurred./The metal dissolved./Gas bubbles were evolved. (Any ONE) [1](b) Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g) [1]

(c) He is incorrect. [1]

Mole ratio of Mg to H2 is 1 : 1.

Number of moles of Mg =24.3

0.70 mol = 0.0288 mol

Number of moles of H2 = number of moles of Mg = 0.0288 mol [1]

Volume of H2 = 0.0288 × 24.0 dm3 = 0.691 dm3 [1]

67 (a) C5H12(g) + 5H2O(g)

5CO(g) + 11H2(g) [1]2CO(g) + 2H2(g) CO2(g) + CH4(g) [1]

(b) Number of moles of CO =0.160.12

8.2

mol = 0.100 mol

Number of moles of H2 =20.1

5.1

mol = 0.750 mol [1]

2CO(g) + 2H2(g) CO2(g) + CH4(g)

Mole ratio of CO : H2 : CH4 = 2 : 2 : 1

CO is the limiting reactant.

Number of moles of CH4 formed =2

100.0 mol = 0.0500 mol [1]



Mass of CH4 formed = 0.0500 × (12.0 + 1.0 × 4) g = 0.800 g [1]

Volume of CH4 produced = 0.0500 × 24.0 dm3 = 1.20 dm3 [1]

(c) (i) Fe2O3(s) + 3H2(g) 2Fe(s) + 3H2O(l) [1]

(ii) Number of moles of Fe2O3 =316.0255.8

100

mol = 0.627 mol [1]

Mole ratio of Fe2O3 : H2 = 1 : 3

Number of moles of H2 required = 0.627 × 3 mol = 1.88 mol [1]

Volume of H2 needed = 1.88 × 24.0 dm3 = 45.1 dm3 [1]

(iii) Volume of town gas needed =%49

1.45 dm3 = 92.0 dm3 [1]

68 (a) 2NH3(g) + 3CuO(s) N2(g) + 3Cu(s) + 3H2O(l) [1]

(b) (i) Number of moles of NH3 used =00024

100 mol = 4.17 × 103 mol [1]

(ii) Mole ratio of NH3 : N2 = 2 : 1

∴ number of moles of the gaseous product formed =2

1017.4 3

mol = 2.09 × 103 mol [1]

Volume of the gaseous product formed = 2.09 × 103 × 24.0 dm3 = 0.0502 dm3 [1]

(c) Mole ratio of NH3 : CuO = 2 : 3

Number of moles of CuO reacted =2

31017.4 3

mol = 6.26 × 103 mol [1]

Mass of CuO reacted = 6.26 × 103 × (63.5 + 16.0) g = 0.498 g [1]

(d) Number of copper atoms reacted = 6.26 × 103

× 6.02 × 1023

= 3.77 × 1021

[1]69 (a) 3Cu(s) + 8HNO3(aq) 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l) [1]

(b) Number of moles of copper =5.63

100 mol = 1.57 mol

Mole ratio of Cu to NO is 3 : 2.

Number of moles of NO =3

257.1 mol = 1.05 mol [1]

Volume of NO = 1.05 × 24.0 dm3 = 25.2 dm3 [1]

(c) Number of copper atoms reacted = 1.57 × 6.02 × 10

23

= 9.45 × 10

23

[1](d) (i) Nitrogen monoxide reacted with oxygen in air to form nitrogen dioxide.

2NO(g) + O2(g) 2NO2(g) [1]

(ii) Mole ratio of NO : NO2 = 1 : 1

Volume ratio of NO : NO2 = 1 : 1 [1]

Volume of NO needed = 500 cm3 [1]

(iii) Mole ratio of O2 : NO2 = 1 : 2

Volume ratio of O2 : NO2 = 1 : 2 [1]

Volume of O2 needed = 2

500 cm

3

= 250 cm3

[1]



Volume of air needed =%20

250 cm3 = 1250 cm3 [1]

70(a) Number of moles of CO2 =

23

23

1002.6

1075.1

mol = 0.291 mol [1]

Volume of CO2 = 0.291 × 24.0 dm3 = 6.98 dm3 [1]

(b) (i) From the equation, mole ratio of ethane : oxygen = 2 : 7

15.0 dm3 of ethane needs 72

0.15 dm3 = 52.5 dm3 of oxygen to react completely.

∴ oxygen is in excess and ethane is the limiting reagent. [1]

(ii) On half way, volume of ethane used =2

0.15 dm3 = 7.50 dm3

Number of moles of ethane =0.24

50.7 mol = 0.313 mol [1]

By mole ratio, number of moles of water formed = 0.313 × 3 mol = 0.939 mol [1]

(iii) There is no ethane remained as it is the limiting reagent.

Volume of oxygen remained = 53.0 52.5 dm3 = 0.5 dm3 [1]

From the equation, volume ratio of carbon dioxide : ethane = 2 : 1

Volume of carbon dioxide formed = 15.0 × 2 dm3 = 30.0 dm3 [1]

Total volume of gases = 30.0 + 0.5 dm3 = 30.5 dm3 [1]

Total number of moles of gases =0.24

5.30 mol = 1.27 mol [1]

71 (a) One nitrogen molecule contains two atoms.

Number of nitrogen molecules =2

1024.3 24

= 1.62 × 1024

Number of moles of nitrogen molecules =23

24

1002.6

1062.1

mol = 2.69 mol [1]

Volume of nitrogen gas = 2.69 × 24.0 dm3 = 64.6 dm3 [1]

(b) Number of moles of FeO =0.168.55

40

mol = 0.557 mol [1]

Mole ratio of FeO to CO2 is 2 : 1.

Number of moles of CO2 produced =2

557.0 mol = 0.279 mol [1]

Volume of CO2 produced = 0.279 × 24.0 dm3 = 6.70 dm3 [1]

(c) Number of moles of atoms in 67.0 dm3 of carbon dioxide

= 30.24

0.67 mol

= 8.38 mol [1]

Number of moles of atoms in 36.0 dm3 of methane

= 50.24

0.36 mol



= 7.50 mol [1]

∴ 67.0 dm3 of carbon dioxide possesses more atoms. [1]

72 (a) CaCO3(s) + 2CH3COOH(aq) (CH3COO)2Ca(aq) + CO2(g) + H2O(l) [1]

(b) The decrease in mass was caused by the escape of CO2(g).

Number of moles of CO2(g) = 20.160.12

59.9

mol = 0.218 mol [1]

Volume of CO2(g) evolved = 0.218 × 24.0 dm3 = 5.23 dm3 [1]

(c) Mole ratio of CaCO3(s) : CO2(g) = 1 : 1

Number of moles of CaCO3(s) reacted = 0.218 mol

Mass of CaCO3(s) reacted = 0.218 × (40.1 + 12.0 + 16.0 × 3) g = 21.8 g [1]

Percentage of purity = %10025

8.21 = 87.2% [1]

(d) Aluminium forms a thin oxide layer (Al2O3) [1] when exposed to oxygen in air. This layer is quite

resistant to vinegar. [1]

73 (a) 2H2O2(aq) 2H2O(l) + O2(g) [1]

(b) Number of moles of H2O2 =20.1620.1

102

mol = 3.00 mol

Mole ratio of H2O2 : O2 = 2 : 1

Number of moles of gas collected =2

00.3 mol = 1.50 mol [1]

Volume of gas collected = 1.50 × 24.0 dm3 = 36.0 dm3 [1]

(c) Mole ratio of H2O2 : H2O = 1 : 1 Number of moles of H2O = 3.00 mol

Mass of H2O = 3.00 × (1.0 × 2 + 16.0) g = 54.0 g [1]

Volume of H2O =0.1

0.54 cm3 = 54.0 cm3 [1]

OR

Mass of O2 produced = 1.50 × 16.0 × 2 g = 48.0 g

By conservation of mass,

Mass of H2O = 102

48.0 g = 54.0 g [1]

Volume of H2O =0.1

0.54 cm3 = 54.0 cm3 [1]

(d) Total number of moles of O2 and H2O = 1.50 + 3.00 mol = 4.50 mol [1]

Total number of molecules = 4.50 × 6.02 × 1023 = 2.71 × 1024 [1]

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Kinetics SQ Ans