OCR GCSE Chemistry A: Monitoring chemical reactions (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 C5.1 Monitoring chemical reactions from OCR’s J248 specification for Chemistry A (Gateway Science) at GCSE (9–1), Version 4.0, dated August 2026: outcomes C5.1a–C5.1k, CM5.1i–CM5.1vi, and the PAG C6 and PAG C8 practical suggestions. Questions on Higher Tier only statements (C5.1a, c, d, e, f, g and k) are labelled; the rest suit both tiers. Topic C5 appears on the Foundation Tier in Paper 2 and on the Higher Tier in Paper 4.

Learn it first in the study guide or the revision notes; the course hub for OCR J248 chemistry and the printable J248 statement checklist place C5.1 in the course. Use Ar: Mg = 24.3, Al = 27.0, K = 39.1, Na = 23.0, Cr = 52.0; O = 16.0, N = 14.0, Cl = 35.5, C = 12.0, H = 1.0, I = 126.9. Molar gas volume = 24 dm³ at room temperature and pressure (RTP); Avogadro constant = 6.02 × 10²³ per mole.

Questions

1. (a) Define the atom economy of a reaction. [1] (b) Explain why a reaction that forms only one product has an atom economy of 100%. [1]

2. A student makes copper(II) sulfate crystals. The theoretical yield is 12.5 g. She collects 9.80 g of dry crystals.

(a) Calculate the percentage yield. [2] (b) Suggest two ways product could have been lost or not formed. [2] (c) Another student reports a yield of 13.1 g from the same method. Suggest why. [1]

3. Chromium can be extracted by heating chromium(III) oxide with aluminium:

Cr₂O₃ + 2Al → 2Cr + Al₂O₃

(a) Calculate the atom economy for making chromium. Give your answer to 3 significant figures. [3] (b) Suggest how the company could reduce the waste from this process. [1]

4. Describe how to find, by titration, the volume of dilute nitric acid that exactly neutralises 25.0 cm³ of sodium hydroxide solution. Include apparatus, indicator and how to get reliable results. [6]

5. (Higher Tier only) A solution is made by dissolving 8.30 g of potassium iodide, KI, in water and making it up to 500 cm³.

(a) Calculate the concentration in g/dm³. [1] (b) Calculate the concentration in mol/dm³. [2] (c) Calculate the mass of potassium iodide needed to make 250 cm³ of a 0.0400 mol/dm³ solution. [2]

6. (Higher Tier only) 25.0 cm³ of barium hydroxide solution is titrated with 0.150 mol/dm³ nitric acid.

Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O

Run	Rough	1	2	3
Titre (cm³)	23.40	22.85	22.95	22.90

(a) Identify the concordant titres and calculate the mean titre. [2] (b) Calculate the concentration of the barium hydroxide solution in mol/dm³. [4]

7. (Higher Tier only) Magnesium carbonate reacts with hydrochloric acid:

MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂

(a) Calculate the volume of carbon dioxide, in cm³ at RTP, formed from 2.53 g of magnesium carbonate. [3] (b) Calculate the minimum volume of 2.00 mol/dm³ hydrochloric acid needed to react with all of the 2.53 g. [2]

8. (Higher Tier only) Ammonia forms from nitrogen and hydrogen: N₂(g) + 3H₂(g) → 2NH₃(g). A mixture of 90 cm³ of nitrogen and 300 cm³ of hydrogen reacts completely, at constant temperature and pressure. Identify the gas in excess, and calculate the volume of ammonia formed and the volume of excess gas left over. [3]

9. (Higher Tier only) 0.729 g of magnesium ribbon is burned in air: 2Mg + O₂ → 2MgO.

(a) Calculate the theoretical mass of magnesium oxide. [3] (b) The student collects 1.04 g of magnesium oxide. Calculate the percentage yield to 3 significant figures. [1]

10. (Higher Tier only) A gas syringe collects 120 cm³ of carbon dioxide at RTP.

(a) Calculate the number of CO₂ molecules, in standard form. [2] (b) Calculate the total number of atoms in this sample. [1]

11. (Higher Tier only) A student makes a standard solution by dissolving 2.50 g of potassium hydrogencarbonate, KHCO₃, and making it up to 250 cm³ in a volumetric flask. She titrates 25.0 cm³ portions with hydrochloric acid, using methyl orange.

KHCO₃ + HCl → KCl + H₂O + CO₂

Run	Rough	1	2	3
Titre (cm³)	21.30	20.75	20.85	20.80

(a) Calculate the concentration of the KHCO₃ solution in mol/dm³, to 3 significant figures. [2] (b) Work out the mean of the concordant titres. [1] (c) Find the hydrochloric acid concentration, in mol/dm³. [3]

12. (Higher Tier only) Chloroethane, C₂H₅Cl, can be made by two routes.

Route A: C₂H₄ + HCl → C₂H₅Cl Route B: C₂H₆ + Cl₂ → C₂H₅Cl + HCl

	Route A	Route B
Yield of chloroethane	90%	40%
Conditions	catalyst, moderate rate	ultraviolet light, fast
Other products	none	saleable HCl; also C₂H₄Cl₂ and other chlorinated products
Cost of organic starting material	higher (ethene)	lower (ethane)

(a) Calculate the atom economy of each route. Give route B’s to 3 significant figures. [3] (b) Using all the data, evaluate which route a manufacturer should choose. [5]

Answers

1. (a) The percentage of the total mass of reactants that ends up in the desired product (from the balanced equation) [1] (b) All the reactant atoms end up in the desired product, so none is wasted as a by-product [1] [2] Examiner insight: “The amount of product made” describes yield; the definition needs reactant mass ending up in the desired product.

2. (a) (9.80 ÷ 12.5) × 100 [1] = 78.4% [1] (b) Any two: product left on apparatus or filter paper; some stayed dissolved after crystallising; incomplete reaction [1] [1] (c) The crystals were still wet (or impure), so the mass is too high and the yield would be over 100% [1] [5] Examiner insight: “Human error” is too vague to credit; name a specific step where product is lost.

3. (a) Total Mr of reactants = 152.0 + (2 × 27.0) = 206.0 [1]; desired product 2Cr = 2 × 52.0 = 104.0 [1]; atom economy = 104.0 ÷ 206.0 × 100 = 50.5% [1] (b) Find a use for, or sell, the aluminium oxide by-product [1] [4] Examiner insight: Using 52.0 for Cr instead of 2 × 52.0 gives 25.2%; always multiply by the coefficients in the balanced equation.

4. Indicative points, one mark each, up to six:

• measure 25.0 cm³ of the alkali with a pipette and filler, into a conical flask [1]
• add an indicator such as phenolphthalein or methyl orange [1]
• fill a burette with the acid and take readings at eye level from the meniscus bottom [1]
• add acid while swirling, on a white tile, drop by drop near the end point [1]
• stop when the colour change stays (pink to colourless with phenolphthalein) [1]
• repeat after a rough run until titres are concordant (within 0.10 cm³); average these [1] [6] Examiner insight: Naming universal indicator costs the indicator point because its gradual colour change gives no sharp end point; give one specific indicator and its colour change.

5. (a) 8.30 ÷ 0.500 = 16.6 g/dm³ [1] (b) Amount = 8.30 ÷ 166.0 = 0.0500 mol [1]; concentration = 0.0500 ÷ 0.500 = 0.100 mol/dm³ [1] (c) Amount = 0.0400 × 0.250 = 0.0100 mol [1]; mass = 0.0100 × 166.0 = 1.66 g [1] [5] Examiner insight: Dividing by 500 instead of 0.500 gives answers 1000 times too small; write the volume in dm³ first.

6. (a) Concordant titres: 22.85, 22.95 and 22.90 cm³ (rough excluded) [1]; mean = 22.90 cm³ [1] (b) Amount of HNO₃ = 0.150 × 0.02290 = 0.003435 mol [1]; ratio Ba(OH)₂ : HNO₃ = 1 : 2 [1], so amount of Ba(OH)₂ = 0.0017175 mol [1]; concentration = 0.0017175 ÷ 0.0250 = 0.0687 mol/dm³ [1] [6] Examiner insight: Missing the 1 : 2 ratio doubles the answer to 0.137 mol/dm³; state the ratio in words or numbers so the step is visible.

7. (a) Mr(MgCO₃) = 84.3, so amount = 2.53 ÷ 84.3 = 0.0300 mol [1]; 1 : 1 ratio gives 0.0300 mol CO₂ [1]; volume = 0.0300 × 24 000 = 720 cm³ [1] (b) Amount of HCl = 2 × 0.0300 = 0.0600 mol [1]; volume = 0.0600 ÷ 2.00 = 0.0300 dm³ = 30.0 cm³ [1] [5] Examiner insight: Using 24 when cm³ is asked gives 0.720; match the molar volume to the unit.

8. 90 cm³ of N₂ needs 3 × 90 = 270 cm³ of H₂, so hydrogen is in excess [1]; ammonia formed = 2 × 90 = 180 cm³ [1]; excess hydrogen = 300 − 270 = 30 cm³ [1] [3] Examiner insight: Working from the 300 cm³ of hydrogen gives 200 cm³ of ammonia, impossible with so little nitrogen.

9. (a) Mg: 0.729 ÷ 24.3 = 0.0300 mol [1]; ratio 2 : 2, so 0.0300 mol MgO [1]; mass = 0.0300 × 40.3 = 1.21 g [1] (b) (1.04 ÷ 1.209) × 100 = 86.0% [1] [4] Examiner insight: Mr(MgO) is 40.3, not 2 × 40.3; the 2s cancel in the ratio.

10. (a) Amount = 120 ÷ 24 000 = 0.00500 mol [1]; molecules = 0.00500 × 6.02 × 10²³ = 3.01 × 10²¹ [1] (b) Each CO₂ has 3 atoms: 3 × 3.01 × 10²¹ = 9.03 × 10²¹ atoms [1] [3] Examiner insight: Writing 3.01²¹ instead of 3.01 × 10²¹ is not standard form; show the power of ten clearly.

11. (a) Mr(KHCO₃) = 100.1, amount = 2.50 ÷ 100.1 = 0.02498 mol [1]; concentration = 0.02498 ÷ 0.250 = 0.0999 mol/dm³ [1] (b) (20.75 + 20.85 + 20.80) ÷ 3 = 20.80 cm³ [1] (c) Amount of KHCO₃ in 25.0 cm³ = 0.0999 × 0.0250 = 0.002498 mol [1]; 1 : 1 ratio, so 0.002498 mol HCl [1]; concentration = 0.002498 ÷ 0.02080 = 0.120 mol/dm³ [1] [6] Examiner insight: Carry at least four significant figures through multistep working and round only the final answer.

12. (a) Route A has one product, so atom economy = 100% [1]. Route B: Mr(C₂H₅Cl) = 64.5 and total Mr of reactants = 30.0 + 71.0 = 101.0 [1]; 64.5 ÷ 101.0 × 100 = 63.9% [1] (b) Indicative points, one mark each, up to five:

• route A has the higher atom economy, so less waste [1]
• route A has the higher yield, 90% compared with 40% [1]
• route B forms a mixture of chlorinated products that must be separated, adding cost [1]
• for route B: saleable HCl, cheaper ethane and a fast reaction [1]
• a justified conclusion, such as route A because its yield and atom economy outweigh the cheaper starting material [1] [8] Examiner insight: An evaluation needs points on both sides and a conclusion tied to the data; listing only route A’s advantages limits the credit.

Where marks are usually lost

• Leaving volumes in cm³ when the formula needs dm³.
• Skipping a 1 : 2 or 2 : 1 mole ratio, as in questions 6 and 7.
• Including a rough titre in the mean.
• Mixing up 24 and 24 000 for the molar gas volume.
• Writing a bare number with no working on a multistep calculation.
• Choosing a route on atom economy alone when other data are given.

Next steps

• Monitoring chemical reactions revision notes
• Monitoring chemical reactions study guide
• Course hub for OCR J248 chemistry
• Printable J248 statement checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Aligned to OCR’s J248 specification for Chemistry A (Gateway Science) at GCSE (9–1): Version 4.0, dated August 2026, for first assessment in 2018. Sub-topic: C5.1 Monitoring chemical reactions.