OCR GCSE Chemistry A: Improving processes and products (J248) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sub-topic C6.1 Improving processes and products from OCR’s Chemistry A (Gateway Science) specification for GCSE (9–1), code J248, in its Version 4.0 issue of August 2026: learning outcomes C6.1a to C6.1s, mathematical learning outcomes CM6.1i to CM6.1iv, and the practicals PAG C1, PAG C2 and PAG C6. Questions on C6.1c to C6.1f are labelled Higher Tier only; all others suit both tiers. The content is examined in Paper 2 (Foundation) or Paper 4 (Higher).

For the rest of J248, use the course hub and the printable statement checklist.

Questions

1. Tin is extracted from tin(IV) oxide, SnO₂, by heating it with carbon: SnO₂ + 2C → Sn + 2CO. Relative atomic masses: Sn = 119, O = 16.

(a) Explain, using the reactivity series, why carbon can extract tin. [1] (b) Calculate the percentage by mass of tin in SnO₂. [2] (c) A crushed ore contains 3.0% SnO₂ by mass. Calculate the maximum mass of tin from 2000 kg of this ore. [1] (d) The plant obtains 41.2 kg of tin from 2000 kg of ore. Calculate the percentage yield. [1]

2. In PAG C1, a student heats a mixture of black copper(II) oxide and carbon powder in a crucible with a lid.

(a) Describe what the student sees when the reaction is complete. [1] (b) Identify the reducing agent. [1] (c) Suggest why the lid is kept on the crucible while it cools. [1]

3. Aluminium is extracted by electrolysis of aluminium oxide dissolved in molten cryolite.

(a) Explain why aluminium is not extracted by heating its oxide with carbon. [1] (b) Write an equation showing how aluminium ions become aluminium atoms. [1] (c) Suggest why aluminium smelters are often built near hydroelectric power stations. [2]

4. In PAG C2 a student electrolyses copper sulfate solution with inert electrodes, weighing the cathode every 5 minutes.

Time / min	0	5	10	15	20
Mass gained / g	0	0.07	0.15	0.21	0.28

(a) Use a line of best fit through the origin to calculate the rate at which copper is deposited. Give the unit. [2] (b) Predict the mass gained after 30 minutes, assuming the rate stays the same. [1] (c) Name the gas formed at the anode and describe a test for it. [2] (d) Sodium chloride solution is electrolysed next. Name the cathode product and explain why sodium does not form. [2]

5. (Higher Tier only) A waste tip of rock containing 0.3% copper lies next to farmland and a river. Evaluate bacterial extraction and phytoextraction for this site. [4]

6. (Higher Tier only) In the Haber process, nitrogen and hydrogen form ammonia: N₂ + 3H₂ ⇌ 2NH₃. The forward reaction is exothermic. Use these values for a model plant:

Conditions	350 °C, 200 atm	450 °C, 200 atm	550 °C, 200 atm	450 °C, 100 atm	450 °C, 300 atm
Equilibrium yield / %	48	28	14	16	36

(a) Use the data to describe how temperature affects the yield, then explain the pattern. [2] (b) Explain why 450 °C is used rather than 350 °C. [2] (c) Going from 200 to 300 atm adds 8 percentage points of yield. Suggest why the plant stays at 200 atm. [2] (d) In the Contact process, 2SO₂ + O₂ ⇌ 2SO₃, the plant runs close to atmospheric pressure. Suggest why. [1]

7. (Higher Tier only) A volume-of-gas against time curve is drawn for a reaction.

(a) A tangent at 10 s passes through (0 s, 12 cm³) and (25 s, 62 cm³). Calculate the rate at 10 s. [2] (b) The initial rate is 3.6 cm³/s. Rate is directly proportional to the pressure of the reacting gas. Predict the initial rate when the pressure is doubled. [1] (c) Describe how the curve changes at a higher temperature with the same amounts of reactants. [1]

8. Ammonium phosphate, (NH₄)₃PO₄, is used in fertilisers. Relative atomic masses: N = 14, H = 1, P = 31, O = 16.

(a) Give the three elements supplied by NPK fertilisers, and one property every fertiliser compound must have. [2] (b) Work out the nitrogen content of ammonium phosphate as a percentage by mass. [2] (c) Describe how a factory site makes ammonium sulfate from air, natural gas and sulfur. [2] (d) Give two differences between making ammonium sulfate in a school laboratory and in industry. [2]

9. A council compares two garden benches using these illustrative energy values (MJ).

Stage	Timber bench	Aluminium bench
Raw materials and manufacture	600	2400
Transport	40	30
Use (wood treatment over its life)	150	0
Disposal	20	−900 (recycled)
Expected lifetime	10 years	25 years

(a) Calculate the total energy for each bench. [2] (b) Use the lifetimes to decide which bench uses less energy per year. [2] (c) Give one reason not to decide on energy figures alone. [1]

10. Old car tyres can be shredded into rubber crumb for children’s play-area surfaces.

(a) Explain why this is a viable use for old tyres. [2] (b) Suggest two factors a council would consider before starting a tyre-recycling scheme. [2]

11. This question is about alloys and corrosion.

(a) A 250 g brass fitting is 70% copper. Name the other metal and calculate its mass. [2] (b) A bicycle chain left outside rusts. State the two conditions needed and explain how oiling prevents it. [2] (c) Magnesium blocks are wired to a buried steel pipe. Explain how they protect it. [2] (d) Name the alloy used in aircraft bodies that is mainly aluminium with about 4% copper. [1]

12. A delivery drone needs arms that are strong, light and will not shatter in a hard landing.

Material	Density / g/cm³	Tensile strength / MPa	Brittle?
Nylon (polymer)	1.15	75	no
Glass-fibre composite	1.9	300	no
Aluminium alloy	2.7	270	no
Glass	2.5	40	yes
Clay ceramic	2.3	10	yes

(a) Calculate strength ÷ density for the three materials that are not brittle. [2] (b) Recommend a material and justify your choice using the data. [2]

Answers

1. (a) Carbon is more reactive than tin, so it can take the oxygen from tin(IV) oxide [1]. (b) Mr = 119 + 32 = 151 [1]; 119 ÷ 151 × 100 = 78.8% [1] (c) SnO₂ = 0.030 × 2000 = 60.0 kg; tin = 60.0 × 119 ÷ 151 = 47.3 kg [1] (d) 41.2 ÷ 47.3 × 100 = 87.1% [1] Examiner insight: Carry unrounded values forward; rounding 78.8% to 79% early shifts the yield.

2. (a) Pink-brown (copper-coloured) solid among the black powder [1]. (b) Carbon [1] (c) Keeps air out so the hot copper is not oxidised back to copper oxide [1]. Examiner insight: “Copper forms” is not an observation; describe the colour seen.

3. (a) Aluminium is above carbon in the reactivity series; carbon cannot take oxygen from its oxide [1]. (b) Al³⁺ + 3e⁻ → Al [1] (c) Electrolysis needs huge amounts of electricity [1]; hydroelectric power supplies it cheaply [1]. Examiner insight: Electrons are gained at the cathode and charges must balance; a reversed equation is unlikely to gain credit.

4. (a) Gradient = 0.28 ÷ 20 [1] = 0.014 g/min [1] (b) 0.014 × 30 = 0.42 g [1] (c) Oxygen [1]; relights a glowing splint [1]. (d) Hydrogen [1]; sodium is more reactive than hydrogen, so hydrogen ions are discharged instead [1]. Examiner insight: Give g/min with the value in (a); a bare number leaves the rate incomplete.

5. Bacterial extraction works on ore this low grade, with little energy and no new mine [1]. But toxic leachate could escape into the river or farmland [1]. Phytoextraction produces no liquid waste and cleans the tip, but is slow, and burning the plants releases carbon dioxide [1]. Judgement: phytoextraction is safer beside a river, if the company can wait [1]. Examiner insight: “Evaluate” needs a judgement about the stated site; pros and cons with no conclusion limit the credit.

6. (a) Yield falls as temperature rises [1]; the forward reaction is exothermic, so higher temperature shifts equilibrium to the left [1]. (b) At 350 °C the yield is higher but the rate is too slow [1]; 450 °C is a compromise: acceptable yield at a faster rate [1]. (c) Higher pressure needs more energy for compression [1] and stronger, costlier equipment, so the extra yield does not pay; unreacted gas is recycled anyway [1]. (d) Yield is already high at low pressure, so extra pressure costs more than it gains [1]. Examiner insight: Compromise answers need both yield and rate; one side alone rarely earns full credit.

7. (a) (62 − 12) ÷ 25 [1] = 2.0 cm³/s [1] (b) 2 × 3.6 = 7.2 cm³/s [1] (c) Steeper start, levels off sooner, same final volume [1]. Examiner insight: Use the tangent’s own coordinates; two points on the curve give an average rate instead.

8. (a) Nitrogen, phosphorus and potassium [1]; soluble in water [1]. (b) Mr = 3 × 18 + 31 + 64 = 149 [1]; 42 ÷ 149 × 100 = 28.2% [1] (c) Nitrogen (air) and hydrogen (natural gas) make ammonia in the Haber process [1]; sulfur gives sulfuric acid in the Contact process, which reacts with the ammonia [1]. (d) Any two: laboratory batch, industry continuous [1]; laboratory uses bought dilute solutions, industry makes concentrated reactants on site; industry reuses reaction heat [1]. Examiner insight: The formula has three nitrogen atoms; using 14 instead of 42 gives a wrong answer.

9. (a) Timber: 600 + 40 + 150 + 20 = 810 MJ [1]; aluminium: 2400 + 30 + 0 − 900 = 1530 MJ [1] (b) Timber: 810 ÷ 10 = 81 MJ per year; aluminium: 1530 ÷ 25 = 61.2 MJ per year [1]. The aluminium bench uses less per year [1]. (c) Energy misses impacts such as forest loss or mining damage, which need value judgements [1]. Examiner insight: Subtract the negative disposal value; adding 900 reverses the conclusion.

10. (a) Large numbers of tyres give a reliable supply [1]; the surface does not need pure rubber, and tyres stay out of landfill [1]. (b) Any two: cost of collection and transport [1]; energy needed to shred; market demand for the crumb; landfill charges saved [1]. Examiner insight: “Good for the environment” is too vague; name the saving, such as landfill space.

11. (a) Zinc [1]; 0.30 × 250 = 75 g [1] (b) Oxygen and water [1]; oil is a barrier keeping them off the iron [1]. (c) Magnesium is more reactive than iron [1], so it oxidises in place of the steel (sacrificial protection) [1]. (d) Duralumin [1] Examiner insight: In (c), calling magnesium “a coating” misses the point; credit is for reactivity and for magnesium corroding instead.

12. (a) Nylon 75 ÷ 1.15 = 65.2; composite 300 ÷ 1.9 = 157.9 [1]; aluminium alloy 270 ÷ 2.7 = 100 [1] (b) Glass-fibre composite [1]: highest strength for its density (157.9 against 100 and 65.2) and not brittle [1]. Examiner insight: Quote the numbers; “it is strongest” does not use the data.

Where marks are usually lost

• Leaving electrons on the wrong side of an electrode half equation.
• (Higher Tier) Saying a catalyst raises the yield of ammonia.
• (Higher Tier) Explaining a compromise with yield only.
• Using one nitrogen atom when the formula contains three.
• Adding a negative LCA value instead of subtracting it.
• Giving one rusting condition instead of two.

Next steps

• Revision notes for this sub-topic
• All free 10-minute diagnostics
• Study guide with worked examples
• Book a free trial class
• OCR GCSE Chemistry hub
• Printable J248 checklist

Official syllabus

Source document, published by OCR: the GCSE (9–1) Chemistry A (Gateway Science) specification, code J248, Version 4.0, August 2026, with first assessment in 2018. Used here: sub-topic C6.1 Improving processes and products, with statements C6.1a–C6.1s, maths outcomes CM6.1i–CM6.1iv and the PAG C1, PAG C2 and PAG C6 practical suggestions.