Q1.Sodium carbonate reacts with dilute hydrochloric acid, Lecture notes of Electronics

Download Q1.Sodium carbonate reacts with dilute hydrochloric acid and more Lecture notes Electronics in PDF only on Docsity! 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 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) 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) 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) 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] 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) 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] 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 (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 (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. 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] 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) 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] 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) 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- 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.