Q1.Sodium carbonate reacts with dilute hydrochloric acid:

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

A student investigated the volume of carbon dioxide produced when different masses of sodium car-bonate were reacted with dilute hydrochloric acid.

This is the method used.

1. Place a known mass of sodium carbonate in a conical flask. 2. Measure 10 cm3 of dilute hydrochloric acid using a measuring cylinder. 3. Pour the acid into the conical flask. 4. Place a bung in the flask and collect the gas until the reaction is complete.

(a) The student set up the apparatus as shown in the figure below.

Identify the error in the way the student set up the apparatus.

Describe what would happen if the student used the apparatus shown. (2)

(b) The student corrected the error.

The student’s results are shown in the table below.

The result for 0.29 g of sodium carbonate is anomalous.

Suggest what may have happened to cause this anomalous result. (1)

(c) Why does the volume of carbon dioxide collected stop increasing at 95.0 cm3? (1)

(d) What further work could the student do to be more certain about the minimum mass of sodium

Mass of sodium carbonate

in g Volume of carbon dioxide gas

in cm3

0.07 16.0

0.12 27.5

0.23 52.0

0.29 12.5

0.34 77.0

0.54 95.0

0.59 95.0

0.65 95.0

M1.(a) (delivery) tube sticks into the acid 1

the acid would go into the water or the acid would leave the flask or go up the delivery tube

ignore no gas collected 1

(b) any one from:

• bung not put in firmly / properly • gas lost before bung put in • leak from tube

1

(c) all of the acid has reacted 1

(d) take more readings in range 0.34 g to 0.54 g 1

take more readings is insufficient

ignore repeat

1 (e) The carbon dioxide was collected at room temperature and pressure.

The volume of one mole of any gas at room temperature and pressure is 24.0 dm3.

How many moles of carbon dioxide is 95.0 cm3?

Give your answer in three significant figures.

................................................................ mol (2)

(f) Suggest one improvement that could be made to the apparatus used that would give more accu-rate results.

Give a reason for your answer. (2)

(g) One student said that the results of the experiment were wrong because the first few bubbles of gas collected were air.

A second student said this would make no difference to the results.

Explain why the second student was correct. (2)

1(e) 95 24000

1

0.00396

or

3.96 × 10−3

1

accept 0.00396 or 3.96 × 10−3 with no working shown for 2 marks

(f) use a pipette / burette to measure the acid 1

because it is more accurate volume than a measuring cylinder or greater precision than a measuring cylinder or use a gas syringe to collect the gas

so it will not dissolve in water

or use a flask with a divider

accept description of tube suspended inside flask

so no gas escapes when bung removed 1

(g) they should be collected because carbon dioxide is left in flask at end 1

and it has the same volume as the air collected / displaced

Q2.This question is about chemical analysis.

(a) A student has solutions of three compounds, X, Y and Z.

The student uses tests to identify the ions in the three compounds.

The student records the results of the tests in the table.

Identify the two ions present in each compound, X, Y and Z.

X ........................................................................................................................

Y ........................................................................................................................

Z ........................................................................................................................ (3)

(b) A chemist needs to find the concentration of a solution of barium hydroxide. Barium hydroxide solution is an alkali.

The chemist could find the concentration of the barium hydroxide solution using two different methods.

Method 1 • An excess of sodium sulfate solution is added to 25 cm3 of the barium hydroxide solution. A

precipitate of barium sulfate is formed.

• The precipitate of barium sulfate is filtered, dried and weighed.

• The concentration of the barium hydroxide solution is calculated from the mass of barium sulfate produced.

Method 2 • 25 cm3 of the barium hydroxide solution is titrated with hydrochloric acid of known concen-

tration.

• The concentration of the barium hydroxide solution is calculated from the result of the titra-tion.

Compare the advantages and disadvantages of the two methods.

(5)

(Total 8 marks)

Test

Compound Flame test Add sodium hy-

droxide solution

Add hydrochlo-ric acid and bari-um chloride so-

lution

Add nitric acid and silver nitrate

solution

X no colour green precipitate white precipitate no reaction

Y yellow flame no reaction no reaction yellow precipitate

Z no colour brown precipitate no reaction cream precipitate

M2.(a) X: Fe2+ / iron(II), SO4

2- / sulfate

allow iron(II) sulfate or FeSO4

1

Y: Na+ / sodium, I- / iodide

allow sodium iodide or NaI

1

Z: Fe3+ / iron(III), Br- / bromide

allow iron(III) bromide or FeBr3

correct identification of any two ions = one mark

correct identification of any four ions = two marks 1

(b) any five from:

allow converse arguments

method 1

• weighing is accurate • not all barium sulfate may be precipitated • precipitate may be lost • precipitate may not be dry • takes longer • requires energy

allow not all the barium hydroxide has reacted

method 2 • accurate • works for low concentrations

allow reliable / precise 5

[8]

Q3.A student investigated the reactions of copper carbonate and copper oxide with dilute hydrochloric acid.

In both reactions one of the products is copper chloride.

(a) Describe how a sample of copper chloride crystals could be made from copper carbonate and di-lute hydrochloric acid.

(4)

(b) A student wanted to make 11.0 g of copper chloride.

The equation for the reaction is:

CuCO3 + 2HCl → CuCl2 + H2O + CO2

Relative atomic masses, Ar: H = 1; C = 12; O = 16; Cl = 35.5; Cu = 63.5

Calculate the mass of copper carbonate the student should react with dilute hydrochloric acid to make 11.0 g of copper chloride.

Mass of copper carbonate = ........................................... g (4)

(c) The percentage yield of copper chloride was 79.1 %.

Calculate the mass of copper chloride the student actually produced.

Actual mass of copper chloride produced = .................... g (2)

(d) Look at the equations for the two reactions:

Reaction 1 CuCO3(s) + 2HCl(aq) → CuCl2(aq) + H2O(l) + CO2(g)

Reaction 2 CuO(s) + 2HCl(aq) → CuCl2(aq) + H2O(l)

Reactive formula masses: CuO = 79.5; HCl = 36.5; CuCl2 = 134.5; H2O = 18

The percentage atom economy for a reaction is calculated using:

Calculate the percentage atom economy for Reaction 2.

Percentage atom economy = ......................................... % (3)

(e) The atom economy for Reaction 1 is 68.45 %. Compare the atom economies of the two reactions for making copper chloride.

Give a reason for the difference. (1)

(Total 14 marks)

M3.(a) add excess copper carbonate (to dilute hydrochloric acid)

accept alternatives to excess, such as ‘until no more reacts’ 1

filter (to remove excess copper carbonate)

reject heat until dry 1

heat filtrate to evaporate some water or heat to point of crystallisation

accept leave to evaporate or leave in evaporating basin 1

leave to cool (so crystals form)

until crystals form 1

must be in correct order to gain 4 marks

(b) Mr CuCl2 = 134.5

correct answer scores 4 marks 1

moles copper chloride = (mass / Mr = 11 / 134.5) = 0.0817843866 1

Mr CuCO3= 123.5 1

Mass CuCO3 (=moles × M2= 0.08178 × 123.5) = 10.1(00) 1

accept 10.1 with no working shown for 4 marks

(c)

or

11.0 × 0.791 1

8.70 (g) 1

accept 8.70(g) with no working shown for 2 marks

(d) Total mass of reactants = 152.5 1

134.5

152.5

allow ecf from step 1 1

88.20 (%) 1

allow 88.20 with no working shown for 3 marks

(e) atom economy using carbonate lower because an additional product is made or carbon dioxide is

Q4. (a) The formula for the chemical compound magnesium sulphate is MgSO4.

Calculate the relative formula mass (Mr)of this compound. (Show your working.) (2)

(b) Magnesium sulphate can be made from magnesium and dilute sulphuric acid.

This is the equation for the reaction.

Mg + H2SO4 → MgSO4 + H2

Calculate the mass of magnesium sulphate that would be obtained from 4g of magnesium. (Show your working.)

Answer..................................... g (2)

Q6. (a) Iron powder is used in the manufacture of ammonia. Why is it used? (1)

(b) Ammonia is manufactured from nitrogen and hydrogen. The equation for the reaction between them is:

N2(g) + 3H2(g) 2NH3(g)

(i) Which two raw materials are used to make the hydrogen?

.......................................................... and ........................................................ (1)

(ii) Why does increasing the pressure increase the chance of molecules of nitrogen reacting with molecules of hydrogen?

(1)

(iii) Calculate the mass, in tonnes, of ammonia which could be produced from 560 tonnes of nitrogen.

The relative atomic masses are: H 1; N 14.

Show clearly how you get to your answer.

Mass of ammonia = ............................................ tonnes (3)

(Total 6 marks)

M4. (a) Mg S O4

24 + 32 + 16 (×4) or 64 / evidence of all Ar’s

gains 1 mark

but (Mr) = 120

gains 2 marks 2

(b) evidence that 24(g) magnesium would produce 120(g) mapesiurn sulphate

gains 1 mark

or correct scaling by 1/6

but 20(g) magnesium sulphate

gains 2 marks [credit error carried forward from (a) with full marks in (b)]

2

M6. (a) any one from

(as a) catalyst

or to mix with promoters

to speed up the reaction (process)

or process is quicker do not credit just it is quicker

to save energy to reduce costs

or process is cheaper do not credit just it is cheaper

larger surface area (than lumps of iron)

or larger surface area for the (catalysed) reaction (to take place)

1

(b) (i) water or steam and methane or natural gas or North Sea gas

both required either order 1

(ii) EITHER more (chance) of them colliding / coming into contact

do not credit just faster

OR volume of the product / ammonia less than / only half the volume of the reactants / the nitrogen and hydrogen

1

(iii) EITHER 680 (tonnes)

OR 28 (of nitrogen) → 34 (of ammonia)

accept any correct 14 : 17 ratio 1

560 (of nitrogen) → 34 × 20 (of ammonia)

Q5. (a) Ammonia is manufactured from nitrogen and hydrogen. The equation for the reaction between them is:

N2(g) + 3H2(g) 2NH3(g)

(i) What is the source of the nitrogen? (1)

(ii) Why does increasing the pressure increase the chance of molecules of hydrogen reacting with molecules of nitrogen?

(1)

(iii) The percentage yield of ammonia is the percentage, by mass, of the nitrogen and hydrogen which has been converted to ammonia. Calculate the mass, in tonnes, of ammonia which can be produced from 90 tonnes of hydrogen when the percentage yield is 50%. The rela-tive atomic masses are: H 1; N 14.

Show clearly how you get to your answer.

Mass = ................................................ tonnes (2)

(b) The percentage yield of ammonia depends on the temperature and pressure inside the reaction vessel. The set of graphs show this.

(i) MPa is the symbol for which unit? (1)

(ii) What is the percentage yield of ammonia produced at a temperature of 450 °C and a pres-sure of 20 MPa?

(1)

(iii) Suggest what changes the chemical engineers should make to both the temperature and the pressure to increase the percentage yield of ammonia.

Temperature .....................................................................................................

Pressure ............................................................................................................ (1)

(iv) How can the rate of ammonia production be increased without changing the temperature or

M5. (a) (i) atmosphere

or (fractional distillation of liquid) air 1

(ii) either more (chance) of them colliding/

not just ‘faster’

coming into contact or the volume of the product / the ammonia is less than / only half the volume of the reactants / the nitrogen and hydrogen

1

(iii) 3 × (1 ×2) of hydrogen → 2 × (14 +1 ×3) of ammonia

accept 6 parts of hydrogen →34 parts of ammonia or similar

i.e. candidate uses the atomic masses and works correctly from the equation

1

= 225 (tonnes/t)

unit not required 1

(b) (i) megapascal(s)

accept million pascal(s) 1

(ii) 28 (%)

accept any answer in the range 28.0 to 28.5 inclusive 1

(iii) reduce the temperature and increase the pressure

both required 1

(iv) either use a catalyst

accept use iron as a catalyst

accept use iron which has been more finely divided

accept use iron / catalyst with a bigger (surface) area

accept use a better catalyst 1

or remove the ammonia (as it is produced)

accept react the ammonia with or dissolve the ammonia in water (as it is produced)

Q7. ‘Iron tablets’ usually contain iron sulphate (FeSO4).

(a) This salt can be made by reacting iron with sulphuric acid.

Fe + H2SO4 → FeSO4 + H2

Calculate the mass of iron sulphate that could be obtained from 4 g of iron.

(Relative atomic masses: Fe = 56, H = 1, O = 16, S = 32)

Mass of iron sulphate = ........................... g (3)

(b) Under different conditions, another type of iron sulphate may form. Balance the symbol equation for this reaction.

Fe + H2SO4 → Fe2(SO4)3 + H2

(1) (Total 4 marks)

M7. (a) 10.86

accept answers between 10.64 to 10.9

if answer is incorrect allow 1 mark for rfm FeSO4 = 152 2 marks for 152 × 4/56

3

(b) 2 Fe + 3 H2SO4 → Fe2(SO4)3 + 3H2

accept other correct multiples for balancing 1

[4]

Q8. In this question you will need to use the following information:

The diagram shows a chemical reaction taking place in a conical flask.

The balanced equation for this reaction is:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

(a) Write a balanced ionic equation for this reaction. (2)

(b) Calculate the mass of magnesium required to produce 0.50 g of hydrogen. Show clearly how you work out your final answer and give the unit.

Mass = ............................... (2)

(c) (i) Draw a diagram to show how the electrons are arranged in a hydrogen molecule.

(1)

(ii) What is the name of the type of chemical bond between the hydrogen atoms in a hydrogen molecule?

(1)

(d) The chemical formula for hydrogen peroxide is H2O2.

Calculate, to the nearest whole number, the percentage, by mass, of hydrogen in hydrogen per-oxide. Show clearly how you work out your answer.

Percentage = ................................. % (2)

(Total 8 marks)

Relative atomic masses: H 1; O 16; Mg 24.

The volume of one mole of any gas is 24 dm>3 at room temperature and atmospheric pressure.

M8. (a) Mg + 2H+ → Mg2+ + H2

* reactants correct in every detail * products correct in every detail

if the spectator ions are sown then (1) mark should be credited but only if they are shown correctly on both sides e.g. Mg + 2H+ + 2CI- → Mg2+ + 2CI- + H2

2

(b) 24 (parts) of magnesium → 2 (parts) 1

of hydrogen or equally clear working (so) 6 grams/g (are needed) 1

unit required

(c) (i) two (and no more) atoms shown to be sharing their single electrons examples

do not credit if anything which contradicts the impression that these are hydrogen atoms

1

(ii) (single) covalent (bond) 1

(d) (×100) = 6 (just 6 is worth (1) mark) 1

× 100 = 6 or similar is (0)

do not credit 5.8823529 and the like 1

[8]

Q9. Iron is the most commonly used metal. Iron is extracted in a blast furnace from iron oxide using car-bon monoxide.

Fe2O3 + 3CO → Fe + 3CO2

(a) A sample of the ore haematite contains 70% iron oxide.

Calculate the amount of iron oxide in 2000 tonnes of haematite.

Amount of iron oxide = ......................................... tonnes (1)

(b) Calculate the amount of iron that can be extracted from 2000 tonnes of haematite. (Relative atomic masses: O = 16; Fe = 56)

Amount of iron = .................................................... tonnes (4)

(Total 5 marks)

M9. (a) 1400 1

(b) 980

correct answer gains full credit

160 tonnes Fe2O3 produces 112 tonnes Fe

if incorrect allow one mark for relative formula mass iron oxide = 160 allow e.c.f.

1400 tonnes Fe2O3 will produce 1400 / 160 × 112 tonnes Fe

use of 2000 tonnes Fe2O3 – deduct one mark only if working out is correct

4

[5]

Q10. The Haber process is used to make ammonia NH3. The table shows the percentage yield of ammonia at different temperatures and pressures.

(a) (i) Use the data in the table to draw two graphs on the grid below. Draw one graph for a tem-perature of 350°C and the second graph for a temperature of 500°C. Label each graph with its temperature.

(4)

(ii) Use your graphs to find the conditions needed to give a yield of 30% ammonia.

................................ °C and .............................. atmospheres (1)

(iii) On the grid sketch the graph you would expect for a temperature of 450°C. (1)

(b) (i) This equation represents the reaction in which ammonia is formed.

N2(g) + 3H2(g) 2NH3(g) + heat

What does the symbol in this equation tell you about the reaction? (1)

(ii) Use your graphs and your knowledge of the Haber process to explain why a temperature of 450°C and a pressure of 200 atmospheres are used in industry.

(5)

(c) (i) Ammonium nitrate is one type of artificial fertiliser.

Calculate the relative formula mass of ammonium nitrate NH4NO3. (Relative atomic masses: H = 1, N = 14, O = 16.)

(1)

(ii) Use your answer to part (c)(i) to help you calculate the percentage by mass of nitrogen pre-

PRESSURE (ATMOSPHERES)

PERCENTAGE (%) YIELD OF AMMONIA AT 350°C

PERCENTAGE (%) YIELD OF AMMONIA AT 500°C

50 25 5

100 37 9

200 52 15

300 63 20

400 70 23

500 74 25

M10. (a) (i) both scales (must be sensible) (use at least half the paper ) plots for 350°C (to accuracy of +/- 1/2 square) plots for 500°C (to accuracy of +/- 1/2 square) lines of best fit (sensible smooth curves) (ignore below 50 atm.) (must not join the dots and each curve must be a single line)

for 1 mark each 4

(ii) read accurately from their graph (must be 350 °C and pressure read to +/– half square from their graph)

for one mark 1

(iii) smooth curve drawn between 350oC and 500 °C - must be of similar shape to the other curves - a dashed line would be accepted here but would not be accepted for part (i)

for one mark 1

(b) (i) reversible reaction (owtte) / equilibrium / equilibria / reaction goes in both directions etc.

for one mark 1

(ii) maximum of 2 marks from each section up to a maximum total of 5

effect of temperature (max. 2 marks) best yield at low temperature / poor yield at high temperature reaction too slow at low temperature / fast at high temperature

effect of pressure (max. 2 marks) high yield at high pressure (owtte) / low yield at low pressure ideas to do with cost / safety factor of using higher pressures

evaluation (max. 2 marks) formation of ammonia favoured at low temperature because reaction is exothermic formation of ammonia favoured at high pressure because more reactant molecules than product molecules actual temperature and / or pressure used are a compromise between good yield and reasonable rate ammonia removed / unreacted nitrogen and hydrogen recycled so rate more important than yield catalyst used (not a wrongly named catalyst)

for 1 mark each 5

(c) (i) NH4NO3 = 14 + (4 × 1) + 14 + (3 × 16) = 80 (ignore units)

for one mark 1

(ii) ecf (error carried forward from part (i)) look for (28/80) for first mark

gains 1 mark

but 35% (% sign not needed)

special case of (14/80 × 100 = 17.5%) gains one mark

Q11. Limestone is a useful mineral. Every day, large amounts of limestone are heated in limekilns to pro-duce lime. Lime is used in the manufacture of iron, cement and glass and for neutralising acidic soils.

CaCO3 CaO + CO2

(i) The decomposition of limestone is a reversible reaction. Explain what this means. (2)

(ii) Calculate the mass of lime, CaO, that would be produced from 250 tonnes of limestone, CaCO3.

Relative atomic masses: C 12; O 16; Ca 40.

Mass of lime = ........................................ tonnes (3)

(Total 5 marks)

Q15. This cake recipe is taken from a cookery book.

When sodium hydrogencarbonate is heated in an oven, it forms carbon dioxide gas.

2 NaHCO3 Na2CO3 + H2O + CO2

A teaspoonful of baking soda contains a mass of 11 g of sodium hydrogencarbonate. Calculate the mass of carbon dioxide that could be made from 11 g of sodium hydrogencarbonate. Show clearly how you work out your final answer.

Relative atomic masses: H = 1; C = 12; O = 16; Na = 23.

Mass of carbon dioxide = ............................................... g (Total 3 marks)

Soda Cake

• Mix the flour and butter and add the sugar, currants and flavouring. • Then add the beaten egg. • Add a little milk with a teaspoonful of baking soda (sodium hydrogencarbonate) and mix it in well. • Bake in a moderate oven for about 30 minutes.

M11. (i) a reaction in which the products can be changed back to reactants

accept a reaction that can go forwards or backwards 1

under certain conditions 1

(ii) Mr CaCO3 = 100 1

Mr CaO = 56 1

mass of CaO = 140 (tonnes) 1

mark consequentially [5]

Q12. 280 000 tonnes of magnesium are produced in the world each year. The pie chart below shows the ways in which magnesium is used.

(a) (i) Use the pie chart to calculate the percentage of magnesium used to make aluminium alloys.

....................................... % (1)

(ii) How many tonnes of magnesium are used to make aluminium alloys each year?

....................................... tonnes (1)

(b) Magnesium is produced by the electrolysis of molten magnesium chloride. The reactions which take place at the electrodes are represented by the equations below.

Mg2+ + 2e– → Mg

2Cl– – 2e– → Cl2

(i) Calculate the mass of chlorine produced when one kilogram of magnesium is made. (Relative atomic masses: Mg = 24, Cl = 35.5)

(3)

(ii) Give a use for chlorine. (1)

(Total 6 marks)

M12. (a) (i) 45%

for 1 mark 1

(ii) 126 000 (consequential on (i))

for 1 mark 1

(b) (i) Cl2 = 71 1 × 71/24 or correct mathematical attempt

for 1 mark

(If Cl2 wrong take figure given)

for 1 mark

= 2.96 kg

gains 3 marks

(or alternative methods) (if units not given - 3 marks. If units wrong - 2 marks)

3

(ii) any sensible eg. bleach/disinfectant/antiseptics/kill bacteria/ sterilise water/solvents/refrigerents/CFCs/PVC (not water treatment or warfare)

for 1 mark 1

[6]

Q13. Titanium is a transition metal used as pins and plates to support badly broken bones. Titanium is extracted from an ore that contains the mineral titanium oxide. This oxide is converted into titanium chloride. Titanium chloride is heated with sodium to form titanium metal. This reaction takes place in an atmosphere of a noble gas, such as argon.

4Na(s) + TiCl4(l) → Ti(s) + 4NaCl(s)

Calculate the mass of titanium that can be extracted from 570 kg of titanium chloride.

Relative atomic masses: Cl 35.5; Ti 48.

Mass of titanium = ............................ kg (Total 3 marks)

Q14. Petrol is a mixture of hydrocarbons such as octane, C8H18

When petrol is burned in a car engine, a large amount of carbon dioxide is produced.

This car uses 114 g of petrol to travel one mile.

Calculate the mass of carbon dioxide produced when this car travels one mile.

Assume that petrol is octane and that combustion is complete.

(Relative atomic masses: H = 1; C = 12; O = 16)

The combustion of octane can be represented by this equation.

C8H18 + 12 O2 → 8CO2 + 9H2O

Mass of carbon dioxide = ........................ g (Total 3 marks)

M13. 144

accept TiCl4 = 190 for 1 mark

accept another correct step in calculation eg 570/190 = 3 for 1 mark

[3]

M14. 352 g gains 3 marks

(moles C8H18 = 114 / 114 = 1 mole) moles CO2 = 8 (1)

Q16. Silicon is an important element used in the electronics industry.

(a) Silicon can be made by heating a mixture of sand (silicon dioxide) with magnesium powder.

The equation for this reaction is shown below.

SiO2 (s)+ 2Mg (s) → 2MgO (s) + Si (s)

Calculate the mass of silicon dioxide needed to make 1 g of silicon.

Relative atomic masses: O = 16; Si = 28

Mass = ........................................................g (3)

(b) The resulting mixture of magnesium oxide and silicon is added to a beaker containing hydrochloric acid. The silicon is then filtered from the solution.

(i) The magnesium oxide reacts with the hydrochloric acid and forms magnesium chloride

(MgCl2) solution and water.

magnesium oxide + hydrochloric acid → magnesium chloride solution + water

Write a balanced symbol equation for this reaction, including state symbols. (2)

(ii) The gases produced are a mixture of several silicon hydrides.

One of the gases produced in the reaction is the silicon hydride with the formula SiH4. The structure of this molecule is similar to methane, CH4.

Draw a diagram to show the bonding in a molecule of SiH4. Represent the electrons as dots and crosses and only show the outer shell (energy level) electrons.

(1)

(iii) A sample of a different silicon hydride was found to contain 1.4 g of silicon and 0.15 g of hydrogen.

Calculate the formula of this silicon hydride. You must show all your working to gain full marks.

Relative atomic masses: H = 1; Si = 28 (4)

M16. (a) Mr (SiO2) = 60

if Mr incorrect ecf for max 2 1

60 g SiO2 → 28 g Si

correct answer for 3 marks 1

2.14 g SiO2 → 1 g Si

allow 2, 2.1, 2.14 (or anything rounding to 2.14), 2.16 or 2.2

a unit is not required but an incorrect unit loses the third mark

OR Mr (SiO2) = 60 (1)

moles if silicon needed = = 0.0357

mass of SiO2 needed = 0.0357 × 60 (1)

= 2.14 g (1)

allow 2, 2.1, 2.14 (or anything rounding to 2.14), 2.16 or 2.2

OR Mr (SiO2) = 60 (1)

mass SiO2 = 1 × (1)

= 2.14 g (1)

allow 2, 2.1, 2.4 (or anything rounding to 2.14), 2.16 or 2.2 3

(b) (i) MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)

penalise incorrect symbols correctly balanced equation for 1 mark state symbols for 1 mark

allow correct multiples / fractions 2

or

ignore inner shell electrons of silicon allow correct drawings without symbols

Q17.This question is about iron and aluminium.

(a) Iron is extracted in a blast furnace. Figure 1 is a diagram of a blast furnace.

(i) Calcium carbonate decomposes at high temperatures.

Complete the word equation for the decomposition of calcium carbonate.

calcium carbonate .................................. + ......................................... (2)

(ii) Carbon burns to produce carbon dioxide.

The carbon dioxide produced reacts with more carbon to produce carbon monoxide.

Balance the equation.

C(s) + CO2(g) .......... CO(g) (1)

(iii) Carbon monoxide reduces iron(III) oxide:

Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)

Calculate the maximum mass of iron that can be produced from 300 tonnes of iron(III) ox-ide.

Relative atomic masses (Ar): O = 16; Fe = 56

Maximum mass = .............................. tonnes (3)

M17.(a) (i) calcium oxide

in either order 1

carbon dioxide

accept correct formulae 1

(ii) C(s) + CO2(g) → 2CO(g)

allow multiples 1

(iii) 210 (tonnes)

award 3 marks for the correct answer with or without working

allow ecf for arithmetical errors

if answer incorrect allow up to 2 marks for any of the steps below:

160 → 112

300 → 112 / 160 × 300

or

moles Fe2O3 = 1.875 (× 106) or 300 / 160

moles of Fe = 3.75 (× 106) or 2 × moles Fe2O3

mass Fe = moles Fe × 56

105 (tonnes) scores 2 (missing 1:2 ratio)

420 (tonnes) scores 2 − taken Mr of iron as 112 3

17 (b) Aluminium is extracted by electrolysis, as shown in Figure 2.

Figure 2

(i) Why can aluminium not be extracted by heating aluminium oxide with carbon? (1)

(ii) Explain why aluminium forms at the negative electrode during electrolysis. (3)

(iii) Explain how carbon dioxide forms at the positive electrodes during electrolysis. (3)

(Total 13 marks)

(b) (i) aluminium is more reactive than carbon or carbon is less reactive than aluminium

must have a comparison of reactivity of carbon and aluminium

accept comparison of position in reactivity series. 1

(ii) (because) aluminium ions are positive

ignore aluminium is positive 1

and are attracted / move / go to the negative electrode / cathode 1

where they gain electrons / are reduced / Al3+ + 3e− → Al

accept equation or statements involving the wrong number of electrons. 1

(iii) (because) the anodes or (positive) electrodes are made of carbon / graphite 1

oxygen is produced (at anode) 1

which reacts with the electrodes / anodes

do not accept any reference to the anodes reacting with oxygen from the air

equation C + O2 CO2 gains 1 mark (M3) 1

[13]

Q18. This question is about calcium hydroxide.

Ancient artworks and monuments can be protected from acid rain if the surface is sprayed with calcium hydroxide nanoparticles.

By Svilen Enev (Own work) [GFDL or CC-BY-SA-3.0], via Wikimedia Commons

(a) Calcium hydroxide has the formula Ca(OH)2

Why are there two hydroxide ions for each calcium ion in the formula? (1)

(b) The calcium hydroxide is used in the form of nanoparticles.

What are nanoparticles? (1)

(c) A student added water to calcium oxide to make calcium hydroxide.

The equation for the reaction is shown below.

CaO + H2O → Ca(OH)2

Calculate the maximum mass of calcium hydroxide which could be made from 2.00 g of calcium oxide.

Relative atomic masses (Ar): H = 1; O = 16; Ca = 40.

Maximum mass of calcium hydroxide = ............................... g (3)

M18. (a) because calcium is +2 and hydroxide is –1

accept to balance the charges

or to make the compound neutral (in terms of charges)

allow calcium needs to lose 2 electrons and hydroxide needs to gain one electron

1

(b) particles of size 1-100 nm

allow clear comparison to ‘normal’ size particles

or particles with a few hundred atoms / ions

or particles with a high surface area (to volume ratio)

or as different properties to ‘normal’ size particles of the same substance 1

(c) Mr CaO = 56 and

Mr Ca(OH)2= 74 1

2/56 (x74) or 0.036 (x74) or

allow ecf from step 1

74/56 (x2) or 1.3(214…) (x2) 1

2.6(428…) in range 2.6 to 2.96

correct answer with or without working gains 3 marks

allow ecf carried through from step 1

ignore final rounding to 3 1

Q19. Aluminium is extracted from aluminium oxide.

(a) The formula of aluminium oxide is Al2O3

The relative formula mass (Mr) of aluminium oxide is 102.

Calculate the percentage of aluminium in aluminium oxide.

Relative atomic masses (Ar): O = 16; Al = 27.

Percentage of aluminium = ................................ % (2)

(b) Aluminium is extracted from aluminium oxide using electrolysis.

The diagram shows a cell used for the extraction of aluminium.

(i) The electrolyte contains cryolite.

Explain why. (2)

(ii) Oxygen is formed at the positive electrode. Complete and balance the equation for this reac-tion.

... O2- → O2 + ............. (2)

(iii) The positive electrode in the cell is used up during the process.

Explain why. (2)

(Total 8 marks)

M19. (a) 52.9(411765) / 53

correct answer with or without working = 2 marks

if answer incorrect allow 2 x 27= 54 or 27/102 x 100 or 26.5 for 1 mark 2

(b) (i) because it lowers the melting point (of the aluminium oxide)

allow lowers the temperature needed

do not accept lowers boiling point 1

so less energy is needed (to melt it)

accept so that the cell / equipment does not melt 1

(ii) 2 O2– on left hand side

accept correct multiples or fractions 1

4e– on right hand side

accept –4e– on left hand side 1

(iii) because the electrode reacts with oxygen or

because the electrode burns 1

to form carbon dioxide or

electrode made from carbon / graphite 1

Q20.Some students were investigating the rate at which carbon dioxide gas is produced when metal car-bonates react with an acid.

One student reacted 1.00 g of calcium carbonate with 50 cm3, an excess, of dilute hydrochloric acid.

The apparatus used is shown in Diagram 1.

Diagram 1

Dilute hydrochloric acid

(a) Complete the two labels for the apparatus on the diagram. (2)

(b) The student measured the volume of gas collected every 30 seconds.

The table shows the student’s results.

(i) Diagram 2 shows what the student saw at 60 seconds.

Diagram 2

What is the volume of gas collected?

Volume of gas = .................... cm3

(1)

Time in

seconds Volume of carbon dioxide

collected in cm3

30 104

60

90 198

120 221

150 232

180 238

210 240

240 240

M20.(a) left hand: (conical) flask

do not accept round bottomed flask or container which is not a flask

1

right hand: beaker / trough

accept plastic box 1

(b) (i) 157 1

(ii) all calcium carbonate used up or reaction stopped

do not accept all acid used up 1

20 (c) Another student placed a conical flask containing 1.00 g of a Group 1 carbonate (M2CO3) on a balance.

He then added 50 cm3, an excess, of dilute hydrochloric acid to the flask and measured the mass of carbon dioxide given off.

The equation for the reaction is:

M2CO3 + 2HCl 2MCl + H2O + CO2

The final mass of carbon dioxide given off was 0.32 g.

(i) Calculate the amount, in moles, of carbon dioxide in 0.32 g carbon dioxide.

Relative atomic masses (Ar): C = 12; O = 16

Moles of carbon dioxide = .................... moles (2)

(ii) How many moles of the metal carbonate are needed to make this number of moles of car-bon dioxide?

Moles of metal carbonate = .................... moles (1)

(iii) The mass of metal carbonate used was 1.00 g.

Use this information, and your answer to part (c) (ii), to calculate the relative formula mass

(Mr) of the metal carbonate.

If you could not answer part (c) (ii), use 0.00943 as the number of moles of metal car-bonate. This is not the answer to part (c) (ii).

Relative formula mass (Mr) of metal carbonate = .................... (1)

(iv) Use your answer to part (c) (iii) to calculate the relative atomic mass (Ar) of the metal in the metal carbonate (M2CO3) and so identify the Group 1 metal in the metal carbonate.

If you could not answer part (c) (iii), use 230 as the relative formula mass of the metal car-bonate. This is not the answer to part (c) (iii).

To gain full marks, you must show your working.

Relative atomic mass of metal is .........................................................

Identity of metal ..................................................................................... (3)

(d) Two other students repeated the experiment in part (c).

(i) When the first student did the experiment some acid sprayed out of the flask as the metal carbonate reacted.

Explain the effect this mistake would have on the calculated relative atomic mass of the metal.

(3)

(ii) The second student used 100 cm3 of dilute hydrochloric acid instead of 50 cm3.

Explain the effect, if any, this mistake would have on the calculated relative atomic mass of

(c) (i) 0.007(272727…)

correct answer with or without working gains 2 marks

if answer incorrect, allow (0.32 / 44) for 1 mark 2

(ii) 0.007(272727…)

allow ecf from (c)(i) 1

(iii) (Mr = mass / moles = 1 / 0.00727…) = 137.5 or 138

allow ecf from (c)(ii)

if use 0.00943 moles then = 106

if use 0.007 allow 143 (142.857) 1

(iv) (138) – 60 (= 78)

23 / 85 1

(78 / 2) = 39 1

potassium

sodium / rubidium

identity of metal ecf on Ar, but must be Group 1

If no working max 1 mark 1

(d) (i) (relative atomic mass) would decrease 1

because the mass lost greater 1

so moles carbon dioxide larger or moles metal carbonate greater 1

(ii) no change 1

because the acid (already) in excess 1

so the amount carbon dioxide lost is the same 1

[17]

Q21.Saturated hydrocarbons, for example methane and octane, are often used as fuels.

(a) Methane can be represented as:

(i) The formula of methane is ................................................................. . (1)

(ii) Draw a ring around the correct answer to complete the sentence.

(1)

(iii) Draw a ring around the correct answer to complete the sentence.

(1)

(b) (i) The complete combustion of petrol produces carbon dioxide, water vapour and sulfur diox-ide.

Name three elements petrol must contain.

1 ............................................................................................................

2 ............................................................................................................

3 ............................................................................................................ (3)

(ii) The exhaust gases from cars can contain oxides of nitrogen.

Complete the sentence.

Nitrogen in the oxides of nitrogen comes from ................................... . (1)

(iii) The sulfur dioxide and oxides of nitrogen from cars cause an environmental problem.

Name the problem and describe one effect of the problem.

Name of problem ..................................................................................

Effect of problem ..................................................................................

............................................................................................................... (2)

In a saturated hydrocarbon molecule all of the bonds are

double.

ionic.

single.

The homologous series that contains methane and octane is called the

alcohols.

alkanes.

alkenes.

M21.(a) (i) CH4

allow H4C

do not allow lower-case h

do not allow superscript 1

(ii) single 1

(iii) alkanes 1

(b) (i) carbon / C

any order 1

hydrogen / H

allow phonetic spelling 1

sulfur / sulphur / S 1

(ii) air / atmosphere 1

(iii) acid rain 1

damages trees / plants or kills aquatic organisms or damages buildings / statues or causes respiratory problems

allow harmful to living things 1

21 (c) When a fuel burns without enough oxygen, there is incomplete combustion.

One gaseous product of incomplete combustion is carbon monoxide.

Name one solid product of incomplete combustion.

................................................................................ (1)

(d) A student investigated how well different hydrocarbon fuels would heat up 100 g of water.

Her hypothesis was:

The apparatus the student used is shown in the diagram.

She burned each hydrocarbon fuel for 2 minutes.

Her results are shown in the table.

The student investigated only hydrocarbons.

Look carefully at her results.

How well do the student’s results support her hypothesis?

Give reasons for your answer. (4)

The more carbon atoms there are in a molecule of any fuel, the better the fuel is.

Name of hydrocar-

bon fuel

Number of car-bon atoms in a

molecule of hydrocarbon

fuel

Temperature change of water in °C

after 2

minutes

Temperature change per g

of fuel

burned

Observations

Pentane 5 30 60 no smoke

Hexane 6 40 57 very small amount of smoke

Octane 8 55 55 small amount of smoke

Decane 10 57 52 large amount of smoke

Dodecane 12 60 43 very large amount of smoke

The more carbon atoms there are in a molecule of any fuel, the better the fuel is.

(c) carbon / C

accept soot / particulates / charcoal 1

(d) any four from:

• (supports hypothesis) because when the fuel contained more carbon the temperature of the water went up more / faster (in 2 minutes)

• (does not support hypothesis as) temperature change per gram decreases as the number of carbons increases

• (does not support hypothesis) because the more carbon in the fuel the more smoke or the dirtier / sootier it is

• only tested hydrocarbons / alkanes / fuels with between 5 and 12 carbon atoms • valid, justified, conclusion

accept converse statements 4

21 (e) A 0.050 mol sample of a hydrocarbon was burned in excess oxygen.

The products were 3.60 g of water and 6.60 g of carbon dioxide.

(i) Calculate the number of moles of carbon dioxide produced.

Relative atomic masses: C = 12; O = 16.

...............................................................................................................

...............................................................................................................

Moles of carbon dioxide = ...................................... (2)

(ii) When the hydrocarbon was burned 0.20 mol of water were produced.

How many moles of hydrogen atoms are there in 0.20 mol of water?

...............................................................................................................

Moles of hydrogen atoms = ................................... (1)

(iii) The amount of hydrocarbon burned was 0.050 mol.

Use this information and your answers to parts (e) (i) and (e) (ii) to calculate the molecular formula of the hydrocarbon.

If you could not answer parts (e) (i) or (e) (ii) use the values of 0.20 moles carbon dioxide and 0.50 moles hydrogen. These are not the answers to parts (e) (i) and (e) (ii).

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

Formula = ............................................................... (2)

(Total 19 marks)

(e) (i) 0.15

correct answer with or without working gains 2 marks

if answer incorrect, Mr carbon dioxide = 44 gains 1 mark

allow 0.236 / 0.24 / 0.2357142 (ecf from Mr of 28) for 1 mark 2

(ii) 0.4(0) 1

(iii) C3H8

correct formula with or without working scores 2 marks

0.15 / 0.05 = 3

allow ecf from (e)(i)

and

0.4 / 0.05 = 8 (1)

allow ecf from (e)(ii)

allow 1 mark for correct empirical formula from their values

If use ‘fall-back-values:

0.50 / 0.05 = 10

and

0.20 / 0.05 = 4

1 mark

C4H10

1 mark

if just find ratio of C to H using fall-back values, get C2H5 allow 1 mark 2

[19]

Q22. Air bags are used to protect the passengers in a car during an accident. When the crash sensor detects an impact it causes a mixture of chemicals to be heated to a high temperature. Reactions take place which produce nitrogen gas. The nitrogen fills the air bag.

(a) The mixture of chemicals contains sodium azide (NaN3) which decomposes on heating to form sodium and nitrogen.

2NaN3 → 2Na + 3N2

A typical air bag contains 130 g of sodium azide.

(i) Calculate the mass of nitrogen that would be produced when 130 g of sodium azide decom-poses.

Relative atomic masses (Ar): N = 14; Na = 23

Mass of nitrogen = ..................................... g (3)

(ii) 1 g of nitrogen has a volume of 0.86 litres at room temperature and pressure.

What volume of nitrogen would be produced from 130 g of sodium azide?

(If you did not answer part (a)(i), assume that the mass of nitrogen produced from 130 g of sodium azide is 80 g. This is not the correct answer to part (a)(i).)

...............................................................................................................

Volume = ..................................... litres (1)

(b) The sodium produced when the sodium azide decomposes is dangerous. The mixture of chemicals contains potassium nitrate and silicon dioxide which help to make the sodium safe.

(i) Sodium reacts with potassium nitrate to make sodium oxide, potassium oxide and nitrogen. Complete the balancing of the equation for this reaction.

10Na + ..........KNO3 → Na2O + K2O + N2

(1)

(ii) The silicon dioxide reacts with the sodium oxide and potassium oxide to form silicates.

Suggest why sodium oxide and potassium oxide are dangerous in contact with the skin.

M22. (a) (i) 84 / 84.5 / 83.98

correct answer with or without working gains 3 marks

(moles of NaN3 =) 130/65 (1)

moles of nitrogen = 3 (1)

mass of nitrogen = 3 x 28 = 84 (1)

or

2 x (23 + (3 x 14)) (1)

3 x (2 x14) (1)

or

2NaN3 = 130 (1)

3N2 = 84 (1)

if answer is incorrect then look for evidence of correct working.

allow ecf from previous stage

1 mark lost for each mistake in the working if they do not have the cor-rect answer.

3

(ii) 72 / 72.24 / 72.2

allow ecf from part (i) × 0.86

or

ignore working

69 or 68.8 1

(b) (i) 2 and 5 1

(ii) any one from:

• corrosive / burns

• alkaline / basic

do not accept acidic

• attacks / destroys / damages living tissue / cells

allow irritant

ignore reference to reactivity

ignore reference to silicates

ignore harmful / toxic 1

[6]

Q23. A student carried out a titration to find the concentration of a solution of hydrochloric acid. The fol-lowing paragraph was taken from the student’s notebook.

I filled a burette with hydrochloric acid. 25.0 cm3 of 0.40 mol/dm3 potassium hydroxide was added to a flask. 5 drops of indicator were added. I added the acid to the flask until the indicator

changed colour. The volume of acid used was 35.0 cm3.

(a) What piece of apparatus would be used to measure 25.0 cm3 of the potassium hydroxide solu-tion?

............................................................................................................................. (1)

(b) Name a suitable indicator that could be used.

............................................................................................................................. (1)

(c) Calculate the number of moles of potassium hydroxide used.

Moles of potassium hydroxide = ............................................ mol (2)

(d) Calculate the concentration of the hydrochloric acid. The equation for the reaction is:

KOH + HCl → KCl + H2O

Concentration of hydrochloric acid = ........................................ mol/dm3

(2)

M23. (a) pipette / burette 1

(b) named indicator eg methyl orange / phenolphthalein

not universal accept litmus but not litmus paper

1

(c)

2 for correct answer 1

= 0.01 1

(d) 1KOH ≡ 1 HCl

0.01 moles HCl in 35 cm3

1

= 0.29

2 for correct answer 0.3 = (1) (with correct working = (2))

1

Q24. A student carried out a titration to find the concentration of a solution of sulphuric acid. 25.0 cm3 of the sulphuric acid solution was neutralised exactly by 34.0 cm3 of a potassium hydroxide solution of

concentration 2.0 mol/dm3. The equation for the reaction is:

2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)

(a) Describe the experimental procedure for the titration carried out by the student. (4)

(b) Calculate the number of moles of potassium hydroxide used.

.............................................................................................................................

Number of moles = ....................................... (2)

(c) Calculate the concentration of the sulphuric acid in mol/dm3.

Concentration = .................................. mol/dm3

(3) (Total 9 marks)

Q25. An oven cleaner solution contained sodium hydroxide. A 25.0 cm3 sample of the oven cleaner solution was placed in a flask. The sample was titrated with hydrochloric acid containing

73 g/dm3 of hydrogen chloride, HCI.

(a) Describe how this titration is carried out. (3)

(b) Calculate the concentration of the hydrochloric acid in mol/dm3.

Relative atomic masses: H 1; Cl 35.5

Answer = ................................... mol/dm3

(2)

(c) 10.0 cm3 of hydrochloric acid were required to neutralise the 25.0 cm3 of oven cleaner solution.

(i) Calculate the number of moles of hydrochloric acid reacting.

Answer = .......................................... mol (2)

(ii) Calculate the concentration of sodium hydroxide in the oven cleaner solution in mol/dm3.

Answer = ................................... mol/dm3

(2) (Total 9 marks)

M24. (a) any four from:

• sulphuric acid measure by pipette

or diagram

• potassium hydroxide in burette

or diagram

• if solutions reversed, award

• note initial reading

• use of indicator

• note final reading or amount used 4

(b)

1

= 0.068 1

(c) ½ or 0.5 moles H2SO4 react with 1 mole KOH 1

moles H2SO4 in 25.0 cm3 = 0.068 × 0.5 1

moles H2SO4 in 1 dm3 = = 1.36 mol/dm3

1

[9]

M25. (a) hydrochloric acid in burette 1

indicator 1

note volume at end / neutralisation point

titre must be HC1

1

(b) 1 mole HCl = 36.5g /36.5 1

= 2 moles / dm3

2 for correct answer

1

(c) (i)

allow e.c.f. ie their (b) ×

2 for correct answer

1

= 0.02 moles 1

(ii) 0.02 × = 0.8 mol / dm3

1

[9]

Q26. This label has been taken from a bottle of household ammonia solution.

Household ammonia is a dilute solution of ammonia in water. It is commonly used to remove grease from ovens and windows.

(a) The amount of ammonia in household ammonia can be found by titration.

25.0 cm3 of household ammonia is placed in a conical flask. Describe how the volume of dilute nitric acid required to neutralise this amount of household ammonia can be found accurately by titration. Name any other apparatus and materials used.

To gain full marks you should write down your ideas in good English. Put them into a sensible order and use correct scientific words.

(4)

(b) In an experiment, it was found that 25.0 cm3 of household ammonia was neutralised by 20.0 cm3 of dilute nitric acid with a concentration of 0.25 moles per cubic decimetre.

The balanced symbol equation which represents this reaction is

NH3(aq) + HNO3(aq) → NH4NO3(aq)

Calculate the concentration of the ammonia in this household ammonia in moles per cubic deci-metre.

Concentration = ............................. moles per cubic decimetre (2)

(c) The salt, ammonium nitrate, is formed in this reaction.

Describe, and give the result of, a chemical test which shows that ammonium nitrate contains ammonium ions.

(2) (Total 8 marks)

M26. (a)

must be a description of a titration no titration = 0 marks

Quality of written communication

for correct sequencing of 2 of first 3 bullet points i.e. 1 + 2 or 2 + 3 or 1 + 3

1

any three from:

• nitric acid in burette

do not accept biuret can be inferred from 3rd point

• add nitric acid until indicator changes (colour)

can be named acid-base indicator colour change does not have to be correct

• note (burette) volume used or final reading

• accuracy: e.g. repeat

accept white tile or dropwise near end or white background or swirling the flask or read meniscus at eye level

3

(b) e.g. formula method:

25 × MNH3 = 0.25 × 20 1

MNH3 = 0.2

correct answer alone = 2

OR

moles NH3 = moles HNO3

= × 0.25 = 0.005 moles (1)

concentration NH3

= = 0.2 (1) 1

(c) sodium hydroxide or potassium hydroxide or lithium hydroxide or calcium hydroxide

ignore mention of alkali 1

ammonia produced

accept gas produced turns (damp) (red) litmus blue (not blue litmus) or alkaline gas produced

any suitable named indicator e.g. UI with consequential marking white fumes / smoke with (concentrated) HCl

Q27. (a) This label has been taken from a bottle of vinegar.

Vinegar is used for seasoning foods. It is a solution of ethanoic acid in water.

In an experiment, it was found that the ethanoic acid present in a 15.000 cm3 sample of vinegar was neutralised by 45.000 cm3 of sodium hydroxide solution, of concentration 0.20 moles per cubic decimetre (moles per litre).

The equation which represents this reaction is

CH3COOH + NaOH → CH3COONa + H2O

Calculate the concentration of the ethanoic acid in this vinegar:

(i) in moles per cubic decimetre (moles per litre);

Concentration =................................... moles per cubic decimetre (2)

(ii) in grams per cubic decimetre (grams per litre).

Relative atomic masses: H = 1; C = 12; O = 16.

Concentration = .................................. grams per cubic decimetre (2)

M27. (a) (i) e.g. moles NaOH = moles of acid

or formula:

0.2 × = 0.009

15M1 = 0.2 × 45 1

rounding to 0.01 loses mark

= 0.009 × = 0.6(M)

M1 = 0.6(M)

ecf for arithmetical error

correct answer 2 marks 1

(ii) 36

ecf – (a)(i) × 60

correct answer 2 marks

0.6 × 60 gets 1 mark relative formula mass of ethanoic acid = 60 for 1 mark 0.6 × incorrect molar mass gains second mark only

2

27 (b) The flow diagram shows some reactions of ethanoic acid.

Give the name of:

(i) gas A,

............................................................................................................................ (1)

(ii) alkali B,

............................................................................................................................ (1)

(iii) ester C,

............................................................................................................................ (1)

(iv) catalyst D,

............................................................................................................................ (1)

(v) carboxylic acid salt E.

............................................................................................................................ (1)

(Total 9 marks)

(b) (i) A = hydrogen / H2

1

B = sodium hydroxide / NaOH or

sodium oxide / Na2O 1

(iii) C = ethyl ethanoate (acetate) /

CH3COOC2H5 / CH3CO2C2H5

1

(iv) D = (concentrated) sulphuric acid /

H2SO4

do not accept dilute sulphuric acid 1

E = sodium ethanoate (acetate) / CH3COONa / CH3CO2Na 1

[9]

Q28. Four labels have come off four bottles.

Describe and give the results of the chemical tests that you would do to identify which bottle contained which substance.

(Total 5 marks)

M28. any series of chemical tests that work should be given credit

each mark is for test + result + inference

identifying all 4 substances unambiguously with no errors gains 5 marks

e.g.

• Flame test: yellow / orange

Na+ sodium sulphate

ignore incorrect flame test colours for other compounds

1

• Add NaOH to remaining 3 samples:

no (white) ppt / ammonia

no need to test for ammonia 1

NH4+ ammonium sulphate (white) ppt magnesium ions

or aluminium ions 1

• add excess NaOH to the 2 samples which gave a (white) ppt:

ppt dissolves aluminium sulphate

ppt insoluble magnesium sulphate 2

or

• Add NaOH:

no ppt: ammonia NH4+ (1)

ammonium sulphate the other one is sodium sulphate (1)

(damp red) litmus* goes blue

NH3 ammonium sulphate the other one is sodium sulphate

• Add excess NaOH to the 2 samples which gave the white ppt (1)

ppt dissolves aluminium sulphate (1)

ppt insoluble magnesium sulphate (1)

(*) or UI/pH indicator goes blue/purple [5]

Q29. In 1916, during the First World War, a German U-boat sank a Swedish ship which was carrying a cargo of champagne. The wreck was discovered in 1997 and the champagne was brought to the sur-face and analysed.

(a) 25.0 cm3 of the champagne were placed in a conical flask.

Describe how the volume of sodium hydroxide solution needed to react completely with the weak acids in 25.0 cm3 of this champagne can be found by titration, using phenolphthalein indicator.

Name any other apparatus used. (4)

(b) The acid in 25.0 cm3 of the champagne reacted completely with 13.5 cm3 of sodium hydroxide of concentration 0.10 moles per cubic decimetre.

Calculate the concentration in moles per cubic decimetre of acid in the champagne.

Assume that 1 mole of sodium hydroxide reacts completely with 1 mole of acid.

Concentration = ......................... moles per cubic decimetre (2)

(c) Is analysis by titration enough to decide whether this champagne is safe to drink?

Explain your answer. (1)

(d) The graph shows how the pH of the solution changes during this titration.

Phenolphthalein is the indicator used in this titration. It changes colour between pH 8.2 and pH 10.0.

Methyl orange is another indicator. It changes colour between pH 3.2 and pH 4.4.

Suggest why methyl orange is not a suitable indicator for this titration. (2)

M29. (a) must be description of a titration no titration = no marks

NaOH in burette

do not accept biuret etc 1

add NaOH until (indicator) changes colour

if specific colour change mentioned, must be correct – colourless to pink / red or ‘goes pink / red’

do not accept ‘clear’ for colourless 1

note (burette) volume used or final reading

accept ‘work out the volume’ 1

one other point: eg repeat

accept: (white) tile or add dropwise / slowly or white background or swirling / mix or read meniscus at eye level or wash apparatus

1

(b) 0.054

for 2 marks

(0.1 × 13.5)/25 for 1 mark

(c) don’t know – insufficient evidence to decide

owtte

any sensible answer

or

depends on whether acid level is considered safe or unsafe

yes, safe – acid level low / weak acids / low compared with stomach acid

owtte

any sensible answer 2

no, unsafe – acid level (too) high / other substances or bacteria may be present / insufficient evidence to decide

owtte

any sensible answer 1

(d) (methyl orange) would have changed colour (well) before the end-point / pH7 / neutral

owtte 1

weak acid present

weak acid-strong base (titration)

allow methyl orange used for strong acid-weak base titration 1

[9]

Q32. Ammonium sulfate is an artificial fertiliser.

(a) (i) When this fertiliser is warmed with sodium hydroxide solution, ammonia gas is given off. Describe and give the result of a test for ammonia gas.

Test .......................................................................................................

Result ................................................................................................... (2)

(ii) Describe and give the result of a chemical test to show that this fertiliser contains sulfate

ions (SO42–).

Test .......................................................................................................

Result ................................................................................................... (2)

(b) Ammonium sulfate is made by reacting sulfuric acid (a strong acid) with ammonia solution (a weak alkali).

(i) Explain the meaning of strong in terms of ionisation. (1)

(ii) A student made some ammonium sulfate in a school laboratory.

The student carried out a titration, using a suitable indicator, to find the volumes of sulfuric acid and ammonia solution that should be reacted together.

Name a suitable indicator for strong acid-weak alkali titrations. (1)

M32. (a) (i) incorrect test or no test = 0 mark

testing the solution or using blue litmus = 0 mark

(test ammonia / gas with red) litmus

accept any acid-base indicator with correct result 1

(goes) blue

OR

(conc.) HCl (1)

white fumes / smoke / solid (1)

allow white gas / vapour

OR

(test ammonia / gas with) Universal Indicator (1)

blue / purple (1) 1

(ii) incorrect test or no test = 0 marks

add barium chloride / BaCl2 (solution)

do not accept H2SO4 added

or add barium nitrate / Ba(NO3)2 (solution)

allow Ba2+ solution / aqueous added 1

white precipitate / solid (formed)

allow white barium sulfate / BaSO4

ignore barium sulfate / BaSO4 alone 1

(b) (i) fully / completely ionised / dissociated or hydrogen ions fully dissociated

accept has more ions than weaker acid / alkali of same concentration

ignore strongly ionised

do not accept ions are fully ionised

ignore concentrated or reference to concentrations of ions 1

(ii) methyl orange

accept correct spelling only

accept any strong acid-weak base indicator

do not allow phenolphthalein / litmus / universal indicator 1

32 (iii) The student found that 25.0 cm3 of ammonia solution reacted completely with 32.0 cm3 of sulfuric acid of concentration 0.050 moles per cubic decimetre.

The equation that represents this reaction is:

Calculate the concentration of this ammonia solution in moles per cubic decimetre.

Concentration = .................................. moles per cubic decimetre (3)

(iv) Use your answer to (b)(iii) to calculate the concentration of ammonia in grams per cubic decimetre.

(If you did not answer part (b)(iii), assume that the concentration of the ammonia solution is 0.15 moles per cubic decimetre. This is not the correct answer to part (b)(iii).)

Relative formula mass of ammonia (NH3) = 17.

Concentration = .................................. grams per cubic decimetre (2)

(Total 11 marks)

2H2SO4(aq) + 2NH3(aq) → (NH4)2SO4 (aq)

32 (iii) 32 × 0.05/1000 or 0.0016 (mole H2SO4 )

accept (0.05 x 32) = (V x 25) or 0.05 x 32 / 25 1

(reacts with) 2 × 0.0016 or 0.0032 (mole NH3 in 25 cm3)

accept dividing rhs by 2 or multiplying lhs by 2 1

(0.0032 × 1000/25 =) 0.128

allow ecf from previous stage

correct answer 0.128 or 0.13 with or without working gains all 3 marks 1

(iv) 2.176 or 2.18

correct answer with or without working

or ecf from candidate’s answer to (b)(iii)

or 2.55 if 0.15 moles used

if answer incorrect or no answer

0.128 × 17 or 0.13 x 17

or their (b)(iii) × 17

or 0.15 × 17 gains 1 mark 2

Q33. Vinegar can be added to food. Vinegar is an aqueous solution of ethanoic acid.

Ethanoic acid is a weak acid.

(a) Which ion is present in aqueous solutions of all acids?

........................................................................................................................ (1)

(b) What is the difference between the pH of a weak acid compared to the pH of a strong acid of the same concentration?

Give a reason for your answer. (2)

(c) The diagram shows the apparatus used to find the concentration of ethanoic acid in vinegar.

(i) Why should phenolphthalein indicator be used for this titration instead of methyl orange? (1)

(ii) 25.00 cm3 of vinegar was neutralised by 30.50 cm3 of a solution of sodium hydroxide with a concentration of 0.50 moles per cubic decimetre.

The equation for this reaction is:

Calculate the concentration of ethanoic acid in this vinegar.

Concentration of ethanoic acid in this vinegar = .................... moles per cubic decimetre

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

M33. (a) Hydrogen / H+

ignore state symbols

ignore proton / H 1

(b) it = weak acid

pH of weak acid is higher than the pH of a strong acid

allow converse for strong acids

allow correct numerical comparison 1

any one from:

allow converse for strong acids

• only partially dissociated (to form ions)

allow ionises less

• not as many hydrogen ions (in the solution)

allow fewer H+ released 1

(c) (i) (titration of) weak acid and strong base 1

(ii) 0.61

correct answer with or without working gains 2 marks

if the answer is incorrect:

moles of sodium hydroxide = (30.5 × 0.5)/1000 = 0.01525 moles

or

(0.5 × 30.5/25) gains 1 mark 2

33 (d) The concentration of ethanoic acid in a different bottle of vinegar was 0.80 moles per cubic decimetre.

Calculate the mass in grams of ethanoic acid (CH3COOH) in 250 cm3 of this vinegar.

The relative formula mass (Mr) of ethanoic acid = 60.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

Mass of ethanoic acid = ...................................... g (2)

34 (c ii) Sodium hydroxide neutralises hydrochloric acid as shown in the equation:

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

The student found that 27.20 cm3 of 0.100 moles per dm3 sodium hydroxide neutralised 5.00 cm3 of hydrochloric acid.

Calculate the concentration of the hydrochloric acid in moles per dm3.

Give your answer to three significant figures.

Concentration of hydrochloric acid = .............................. moles per dm3

(3) (Total 14 marks)

33 (d) 12

correct answer with or without working gains 2 marks or even with incor-rect working.

if the answer is incorrect:

0.8 × 60 = 48g

or

evidence of dividing 48g (or ecf) by 4

or

0.8 × 0.25 = 0.2 mol

or

evidence of multiplying 0.2mol (or ecf) by 60

would gain 1 mark 2

[8]

(ii) moles of NaOH

0.10 × 0.0272 = 0.00272 moles

correct answer with or without working gains 3 marks 1

Concentration of HCl

0.00272 / 0.005 = 0.544

allow ecf from mp1 to mp2 1

correct number of significant figures 1

[14]

Q34.A student investigated the rate of reaction of magnesium and hydrochloric acid.

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

The student studied the effect of changing the concentration of the hydrochloric acid.

She measured the time for the magnesium to stop reacting.

(a) The student changed the concentration of the hydrochloric acid.

Give two variables that the student should control.

1 ....................................................................................................................

2 .................................................................................................................... (2)

(b) (i) The rate of reaction increased as the concentration of hydrochloric acid increased.

Explain why. (2)

(ii) Explain why increasing the temperature would increase the rate of reaction. (3)

(c) (i) The student had a solution of sodium hydroxide with a concentration of 0.100 moles per dm3.

She wanted to check the concentration of a solution of hydrochloric acid.

She used a pipette to transfer 5.00 cm3 of the hydrochloric acid into a conical flask.

She filled a burette with the 0.100 moles per dm3 sodium hydroxide solution.

Describe how she should use titration to obtain accurate results.

Concentration of hydrochloric acid in moles per dm3

0.5 1.0 1.5 2.0

M34.(a) any two from:

• temperature (of the HCl) • mass or length of the magnesium • surface area of the magnesium • volume of HCl

2

(b) (i) (a greater concentration has) more particles per unit volume

allow particles are closer together 1

therefore more collisions per unit time or more frequent collisions. 1

(ii) particles move faster

allow particles have more (kinetic) energy 1

therefore more collisions per unit time or more frequent collisions 1

collisions more energetic (therefore more collisions have energy greater than the activation energy) or more productive collisions

1

(c) (i) add (a few drops) of indicator to the acid in the conical flask

allow any named indicator 1

add NaOH (from the burette) until the indicator changes colour or add the NaOH dropwise

candidate does not have to state a colour change but penalise an incor-rect colour change.

1

repeat the titration 1

calculate the average volume of NaOH or repeat until concordant results are ob-

Q37.Etching is a way of making printed circuit boards for computers.

© Dario Lo Presti/Shutterstock

Printed circuit boards are made when copper sheets are etched using iron(III) chloride solution. Where the copper has been etched, only plastic remains.

(a) Copper is a good conductor of electricity.

Explain why. (2)

(b) Iron(III) chloride can be produced by the reaction shown in the equation:

2 Fe + 3 Cl2 → 2 FeCl3

(i) Calculate the maximum mass of iron(III) chloride (FeCl3) that can be produced from 11.20 g of iron.

Relative atomic masses (Ar): Cl = 35.5; Fe = 56.

Maximum mass of iron(III) chloride = .............................. g (3)

(ii) The actual mass of iron(III) chloride (FeCl3) produced was 24.3 g.

Calculate the percentage yield.

(If you did not answer part (b)(i) assume that the maximum theoretical mass of iron(III)

chloride (FeCl3) is 28.0 g. This is not the correct answer to part (b)(i).)

Percentage yield = ..............................% (1)

(Total 6 marks)

M37.(a) copper has delocalised electrons

accept copper has free electronsignore sea of electrons or mobile elec-trons

1

(electrons) which can move through the metal / structure

allow (electrons) which can carry a charge through the metal / structure 1

(b) (i) (M r FeCl 3 =) 162.5

correct answer with or without working gains 3 marks

can be credited from correct substitution in step 2 1

or

2 (moles of) FeCl 3 = 325

or

112 → 325

allow ecf from step 1

accept 1

= 32.5

accept 32.48 1

(ii) 74.8

accept 74.77 - 75

accept ecf from (b)(i)

if there is no answer to part(i)

or

if candidate chooses not to use their answer then accept 86.79 - 87

Q38.Sodium hydroxide neutralises sulfuric acid.

The equation for the reaction is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

(a) Sulfuric acid is a strong acid.

What is meant by a strong acid? (2)

(b) Write the ionic equation for this neutralisation reaction. Include state symbols. (2)

(c) A student used a pipette to add 25.0 cm3 of sodium hydroxide of unknown concentration to a con-ical flask.

The student carried out a titration to find out the volume of 0.100 mol / dm3 sulfuric acid needed to neutralise the sodium hydroxide.

Describe how the student would complete the titration.

You should name a suitable indicator and give the colour change that would be seen. (4)

(d) The student carried out five titrations. Her results are shown in the table below.

Concordant results are within 0.10 cm3 of each other.

Use the student’s concordant results to work out the mean volume of 0.100 mol / dm3 sulfuric acid added.

Mean volume = .......................................................... cm3

(2)

(e) The equation for the reaction is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

Calculate the concentration of the sodium hydroxide.

Give your answer to three significant figures.

Concentration = ................................................ mol / dm3

(4)

(f) The student did another experiment using 20 cm3 of sodium hydroxide solution with a concentra-

tion of 0.18 mol / dm3.

Relative formula mass (Mr) of NaOH = 40

Calculate the mass of sodium hydroxide in 20 cm3 of this solution.

Mass = ................................................................ g (2)

Titration 1 Titration 2 Titration 3 Titration 4 Titration 5

Volume of 0.100 mol / dm3 sulfuric acid in cm3

27.40 28.15 27.05 27.15 27.15

M38.(a) (sulfuric acid is) completely / fully ionised 1

In aqueous solution or when dissolved in water 1

(b) H+(aq) + OH−(aq) → H2O(l)

allow multiples

1 mark for equation

1 mark for state symbols 2

(c) adds indicator, eg phenolpthalein / methyl orange / litmus added to the sodium hydroxide (in the conical flask)

do not accept universal indicator 1

(adds the acid from a) burette 1

with swirling or dropwise towards the end point or until the indicator just changes colour 1

until the indicator changes from pink to colourless (for phenolphthalein) or yellow to red (for methyl orange) or blue to red (for litmus)

1

(d) titrations 3, 4 and 5 or

1

27.12 cm3

accept 27.12 with no working shown for 2 marks 1

allow 27.1166 with no working shown for 2 marks

(e) Moles H2SO4 = conc × vol = 0.00271

allow ecf from 8.4 1

Ratio H2SO4:NaOH is 1:2 or

Moles NaOH = Moles H2SO4 × 2 = 0.00542 1

Concentration NaOH = mol / vol = 0.00542 / 0.025 = 0.2168 1

0.217 (mol / dm3)

accept 0.217 with no working for 4 marks 1

accept 0.2168 with no working for 3 marks

(f) × 0.18 = no of moles

or

0.15 × 40 g 1

0.144 (g) 1

accept 0.144g with no working for 2 marks [16]

Q40. Aspirin tablets have important medical uses.

A student carried out an experiment to make aspirin. The method is given below.

(a) The equation for this reaction is shown below.

C7H6O3 + C4H6O3 → C9H8O4 + CH3COOH salicylic acid aspirin

Calculate the maximum mass of aspirin that could be made from 2.00 g of salicylic acid.

The relative formula mass (Mr) of salicylic acid, C7H6O3, is 138

The relative formula mass (Mr) of aspirin, C9H8O4, is 180

Maximum mass of aspirin = .............................. g (2)

(b) The student made 1.10 g of aspirin from 2.00 g of salicylic acid.

Calculate the percentage yield of aspirin for this experiment.

(If you did not answer part (a), assume that the maximum mass of aspirin that can be made from 2.00 g of salicylic acid is 2.50 g. This is not the correct answer to part (a).)

Percentage yield of aspirin = .............................. % (2)

(c) Suggest one possible reason why this method does not give the maximum amount of aspirin. (1)

(d) Concentrated sulfuric acid is a catalyst in this reaction.

Suggest how the use of a catalyst might reduce costs in the industrial production of aspirin. (1)

1. Weigh 2.00 g of salicylic acid. 2. Add 4 cm3 of ethanoic anhydride (an excess). 3. Add 5 drops of concentrated sulfuric acid. 4. Warm the mixture for 15 minutes. 5. Add ice cold water to remove the excess ethanoic anhydride. 6. Cool the mixture until a precipitate of aspirin is formed. 7. Collect the precipitate and wash it with cold water. 8. The precipitate of aspirin is dried and weighed.

M40. (a) 2.61 / range 2.5 to 2.7

correct answer with or without or with wrong working gains 2 marks

(accept answers between 2.5 and 2.7)

if answer incorrect moles of salicylic acid = 2/138 = 0.0145 moles ie 2/138 or 0.0145 gains 1 mark or (180/138) × 2 gains 1 mark or 1 g → 180/138 = (1.304 g) gains 1 mark (not 1.304g alone)

2

(b) 42.1 range 40.7 to 42.3

accept correct answer with or without or with wrong working for 2 marks

ecf ie (1.1 / their answer from (a)) × 100 correctly calculated gains 2 marks

if answer incorrect percentage yield = 1.1 / 2.61 × 100 gains 1 mark

if they do not have an answer to part (a) or they choose not to use their answer then:

• yield = (1.1 / 2.5) × 100 (1)

• = 44

accept 44 for 2 marks with no working 2

(c) any one from:

• errors in weighing

• some (of the aspirin) lost

do not allow ‘lost as a gas’

• not all of the reactant may have been converted to product

eg reaction didn’t go to completion

allow loss of some reactants

• the reaction is reversible

accept other products / chemicals

• side reactions

ignore waste products

• reactants impure

• not heated for long enough

• not hot enough for reaction to take place 1

(d) any one from:

• use lower temperature

• use less fuel / energy

ignore references to use of catalyst

• produce product faster or speed up reaction

• more product produced in a given time (owtte)

• increased productivity

• lowers activation energy 1

[6]