IB DP Chemistry Reactivity 2.1: How much? The amount of chemical change – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

Use this set to test Reactivity 2.1, “How much? The amount of chemical change”, from the IB’s DP Chemistry guide for first assessment 2025 (issued February 2023). Each understanding from Reactivity 2.1.1 to Reactivity 2.1.5 is tested. There is no AHL content, so every question suits SL and HL, for sessions from 2025 onwards.

Revise first with the study guide and revision notes. For the rest of the course, see the course page, the syllabus checklist and our diagnostics.

Data for every question: Ar values Ag 107.87, Al 26.98, C 12.01, Cl 35.45, Cu 63.55, Fe 55.85, H 1.01, Mn 54.94, N 14.01, O 16.00. Molar volume of an ideal gas at STP = 22.7 dm³ mol⁻¹. Give numerical answers to three significant figures except where a question says otherwise.

Questions

1. Solid sodium hydrogencarbonate reacts with dilute sulfuric acid, forming aqueous sodium sulfate, water and carbon dioxide. Write the balanced equation with state symbols. [2]

2. Manganese can be extracted by heating trimanganese tetroxide with aluminium powder:

3Mn₃O₄(s) + 8Al(s) → 4Al₂O₃(s) + 9Mn(l)

Calculate the mass of manganese produced from 50.0 g of Mn₃O₄, and the mass of aluminium needed to react with it completely. [4]

3. Hydrogen peroxide solution decomposes in the presence of a catalyst:

2H₂O₂(aq) → 2H₂O(l) + O₂(g)

Calculate the volume of oxygen, measured at STP, released when 50.0 cm³ of hydrogen peroxide solution, [H₂O₂] = 1.20 mol dm⁻³, decomposes completely. [3]

4. 25 cm³ of ethene is mixed with 100 cm³ of oxygen and ignited. The products return to the starting conditions, where water is a liquid.

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l)

Identify the reactant in excess, state the volume of it left over, and state the volume of carbon dioxide formed. [3]

5. A 10.00 cm³ sample of phosphoric acid is titrated with sodium hydroxide solution, [NaOH] = 0.125 mol dm⁻³. The end point is reached after 24.80 cm³ has been added.

H₃PO₄(aq) + 3NaOH(aq) → Na₃PO₄(aq) + 3H₂O(l)

(a) Calculate [H₃PO₄] in the sample. [3] (b) Calculate the concentration of sodium phosphate in the solution at the end point. [2]

6. 30.0 cm³ of silver nitrate solution, [AgNO₃] = 0.200 mol dm⁻³, is mixed with 25.0 cm³ of calcium chloride solution, [CaCl₂] = 0.100 mol dm⁻³.

2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)

(a) Identify the limiting reactant. Show your working. [2] (b) Calculate the theoretical yield of silver chloride in grams. [2] (c) Calculate the concentration of silver ions left in solution after the precipitate forms. Assume the volumes add. [2]

7. A student makes a salt from an acid and an insoluble base, then filters and crystallises it.

(a) Distinguish between the theoretical yield and the experimental yield of the salt. [1] (b) Suggest two reasons why the experimental yield could be lower than the theoretical yield. [2]

8. Urea, CO(NH₂)₂, is made industrially from ammonia and carbon dioxide:

2NH₃(g) + CO₂(g) → CO(NH₂)₂(s) + H₂O(l)

In a trial run, 34.0 kg of ammonia reacts with 66.0 kg of carbon dioxide. 51.0 kg of urea is collected.

(a) Show that ammonia is the limiting reactant. [3] (b) Calculate the theoretical yield of urea in kg. [2] (c) Calculate the percentage yield. [1] (d) Calculate the mass of carbon dioxide, in kg, left unreacted if all the ammonia reacts. [2] (e) Calculate the atom economy of this process for urea. [2]

9. Iron(III) oxide can be reduced to iron by three different reducing agents.

Route A: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) Route B: Fe₂O₃(s) + 2Al(s) → 2Fe(l) + Al₂O₃(s) Route C: Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(g)

(a) Calculate the atom economy for iron in each route. Give your answers to 1 decimal place. [3] (b) State how the atom economy of a route relates to the waste it produces. [1] (c) Route C has the highest atom economy. Suggest two other factors a steelmaker would weigh before choosing it. [2]

10. A strip of aluminium foil of mass 0.270 g is placed in 40.0 cm³ of copper(II) chloride solution, [CuCl₂] = 0.300 mol dm⁻³, and left until no further change occurs.

2Al(s) + 3CuCl₂(aq) → 2AlCl₃(aq) + 3Cu(s)

(a) Show that copper(II) chloride is the limiting reactant. [3] (b) Calculate the theoretical yield of copper in grams. [1] (c) The copper is filtered off and weighed straight away. Its mass is 0.791 g. Calculate the percentage yield. [1] (d) Suggest why the value in (c) is above 100 %. [1] (e) Calculate the mass of aluminium left unreacted. [2]

Answers

1. Correct formulas with 2NaHCO₃, 2H₂O and 2CO₂ [1]; all state symbols correct [1]. 2NaHCO₃(s) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l) + 2CO₂(g) [2] Examiner insight: Check each state symbol against the wording: the solid is (s), and water at room temperature is (l), not (aq).

2. M(Mn₃O₄) = 3(54.94) + 4(16.00) = 228.82 g mol⁻¹, so n = 50.0 ÷ 228.82 = 0.2185 mol [1]. n(Mn) = 0.2185 × 9/3 = 0.6555 mol [1]; m(Mn) = 0.6555 × 54.94 = 36.0 g [1]. n(Al) = 0.2185 × 8/3 = 0.5827 mol; m(Al) = 0.5827 × 26.98 = 15.7 g [1]. [4] Examiner insight: Show each ratio (9/3, 8/3) as its own step, so a slipped molar mass does not hide a correct method.

3. n(H₂O₂) = 1.20 × 0.0500 = 0.0600 mol [1]; n(O₂) = 0.0600 ÷ 2 = 0.0300 mol [1]; V = 0.0300 × 22.7 = 0.681 dm³ (681 cm³) [1]. [3] Examiner insight: Convert 50.0 cm³ to 0.0500 dm³ first; otherwise the answer is a thousand times too large.

4. O₂ needed = 25 × 3 = 75 cm³ [1]; oxygen is in excess, with 25 cm³ left over [1]; CO₂ formed = 25 × 2 = 50 cm³ [1]. [3] Examiner insight: Use the volume ratio directly; converting to moles and back only adds rounding.

5. (a) n(NaOH) = 0.125 × 0.02480 = 0.003100 mol [1]; n(H₃PO₄) = 0.003100 ÷ 3 = 0.001033 mol [1]; [H₃PO₄] = 0.001033 ÷ 0.01000 = 0.103 mol dm⁻³ [1]. [3] (b) n(Na₃PO₄) = 0.001033 mol; total volume = 10.00 + 24.80 = 34.80 cm³ [1]; concentration = 0.001033 ÷ 0.03480 = 0.0297 mol dm⁻³ [1]. [2] Examiner insight: In (b), divide by the combined volume at the end point, not by either starting volume.

6. (a) n(AgNO₃) = 0.200 × 0.0300 = 0.00600 mol; n(CaCl₂) = 0.100 × 0.0250 = 0.00250 mol [1]. AgNO₃ needed for all the CaCl₂ = 2 × 0.00250 = 0.00500 mol, which is less than 0.00600 mol, so CaCl₂ is limiting [1]. [2] (b) n(AgCl) = 2 × 0.00250 = 0.00500 mol; M(AgCl) = 143.32 g mol⁻¹ [1]; mass = 0.00500 × 143.32 = 0.717 g [1]. [2] (c) Ag⁺ left = 0.00600 − 0.00500 = 0.00100 mol [1]; volume = 55.0 cm³, so [Ag⁺] = 0.00100 ÷ 0.0550 = 0.0182 mol dm⁻³ [1]. [2] Examiner insight: In (a), a bare name is unsupported; compare amounts that allow for the 2 : 1 ratio.

7. (a) Theoretical yield is the maximum amount calculated from the limiting reactant; experimental yield is the mass of salt actually obtained [1]. [1] (b) One reason, such as salt left dissolved after crystallisation [1]; a second, such as salt lost on the filter paper or in transfer, or incomplete reaction [1]. [2] Examiner insight: “Human error” or “experimental error” is too vague; name a specific step where product is lost.

8. (a) n(NH₃) = 34.0 × 10³ ÷ 17.04 = 1995 mol; n(CO₂) = 66.0 × 10³ ÷ 44.01 = 1500 mol [1]. CO₂ needed for all the NH₃ = 1995 ÷ 2 = 997.7 mol [1]. 997.7 mol < 1500 mol available, so CO₂ is in excess and NH₃ is limiting [1]. [3] (b) n(urea) = 997.7 mol; M(urea) = 60.07 g mol⁻¹ [1]; theoretical yield = 997.7 × 60.07 = 59 930 g = 59.9 kg [1]. [2] (c) 51.0 ÷ 59.93 × 100 = 85.1 % [1]. [1] (d) CO₂ left = 1500 − 997.7 = 502.0 mol [1]; mass = 502.0 × 44.01 = 22 090 g = 22.1 kg [1]. [2] (e) Total reactant mass = 2(17.04) + 44.01 = 78.09 g mol⁻¹ [1]; atom economy = 60.07 ÷ 78.09 × 100 = 76.9 % [1]. [2] Examiner insight: Convert kg to g before n = m/M, then back to kg for the answer.

9. (a) M(Fe₂O₃) = 159.70 g mol⁻¹ and 2Fe = 111.70 g mol⁻¹. Route A: 111.70 ÷ (159.70 + 3 × 28.01) × 100 = 111.70 ÷ 243.73 × 100 = 45.8 % [1]. Route B: 111.70 ÷ (159.70 + 2 × 26.98) × 100 = 111.70 ÷ 213.66 × 100 = 52.3 % [1]. Route C: 111.70 ÷ (159.70 + 3 × 2.02) × 100 = 111.70 ÷ 165.76 × 100 = 67.4 % [1]. [3] (b) The lower the atom economy, the larger the share of reactant mass that ends up as by-products, so the more waste [1]. [1] (c) One factor, such as the cost and energy of making the hydrogen [1]; a second, such as the rate, the hazard of a flammable gas, or the percentage yield achieved [1]. [2] Examiner insight: Include both the 2 in front of Fe and the reducing agent’s coefficient; omitting either changes every value.

10. (a) n(Al) = 0.270 ÷ 26.98 = 0.01001 mol; n(CuCl₂) = 0.300 × 0.0400 = 0.0120 mol [1]. Al needed for all the CuCl₂ = 0.0120 × 2/3 = 0.00800 mol [1]. 0.00800 mol < 0.01001 mol, so aluminium is in excess and CuCl₂ is limiting [1]. [3] (b) n(Cu) = 0.0120 mol; mass = 0.0120 × 63.55 = 0.763 g [1]. [1] (c) 0.791 ÷ 0.7626 × 100 = 104 % [1]. [1] (d) The copper was not dried, so its mass includes water or solution [1]. [1] (e) Al left = 0.01001 − 0.00800 = 0.00201 mol [1]; mass = 0.00201 × 26.98 = 0.0542 g [1]. [2] Examiner insight: A yield above 100 % needs a physical cause: name the extra mass, such as water or leftover aluminium.

Where marks are usually lost

• Altering a formula to balance an equation, or leaving out state symbols.
• Using the mole ratio the wrong way up, especially for ratios such as 9/3, 8/3 or 2/3.
• Putting cm³ into n = cV, or using one solution’s volume when the question wants the combined volume.
• Naming a limiting reactant from masses or raw moles instead of moles divided by coefficients.
• Calculating a yield from the reactant in excess.
• Mixing kilograms and grams within one calculation.
• Missing a coefficient when totalling reactant molar masses for atom economy.

Next steps

• Reactivity 2.1 revision notes
• Reactivity 2.1 study guide
• Every DP Chemistry topic
• All free 10-minute diagnostics
• Syllabus tick-list to print
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Official syllabus

Diploma Programme Chemistry guide, first assessment 2025 (International Baccalaureate Organization, February 2023). The questions follow its Reactivity 2.1 understandings.