OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2022: Question 21

14 marks · Medium difficulty · Structured Questions

Analyze Group 2 reactions and compounds involving magnesium, barium oxide, and calcium minerals through observations, half-equations, calculations of mass, ionic equations, and reaction equations.

Practise this question

Question

A structured chemistry exam question with three main parts. Part (a) asks about magnesium and calcium reacting with dilute hydrochloric acid, requiring two observations and two half-equations. Part (b) presents a barium oxide sample dissolved in water to form barium hydroxide with a pH of 13.12, asking to determine the mass of barium oxide used to 3 significant figures and write an ionic equation for its reaction with sulfuric acid. Part (c) discusses calcium minerals like limestone and huntite, requiring a calculation of the mass of limestone needed to make a specific mass of a complex fertilizer to 3 significant figures, and to construct an equation with state symbols for the reaction of huntite with dilute hydrochloric acid.
Question text

21 This question is about the reactions of Group 2 metals and their compounds.

(a) A student adds magnesium to dilute hydrochloric acid in one test tube.

The student adds calcium to dilute hydrochloric acid in a second test tube.

A redox reaction takes place in each test tube.

(i) Suggest two observations from the student’s experiment that would show that calcium

is more reactive than magnesium.

1 …

2 …

… [1]

(ii) Write half-equations for the reaction of magnesium with hydrochloric acid.

Oxidation half-equation: …

Reduction half-equation: …

[2]

(b) A sample of barium oxide is added to distilled water at 25 °C.

A colourless solution forms containing barium hydroxide, Ba(OH)2.

The solution is made up to 250.0 cm3 with distilled water.

The pH of this solution is 13.12.

(i) Determine the mass of barium oxide that was used.

Give your answer to 3 significant figures.

mass of barium oxide = … g [5]

(ii) 10 cm3 of dilute sulfuric acid is added to 10 cm3 of the colourless solution of Ba(OH) .

Write an ionic equation, including state symbols, for the reaction.

… [1]

(c) Limestone and huntite are two calcium minerals.

(i) A typical sample of limestone contains 95.0% by mass of calcium carbonate, CaCO3.

Fertiliser Z, Ca NH (NO ) •10H O (M = 1080.5 g mol–1) can be made from limestone.

54 3 11 2 r

Calculate the mass, in g, of limestone needed to make 1.50 kg of fertiliser Z.

Give your answer to 3 significant figures.

mass of limestone = … g [3]

(ii) Huntite is a carbonate mineral with the chemical formula Mg3Ca(CO3)4.

Huntite reacts with dilute hydrochloric acid to produce bubbles of a gas and a colourless

solution.

Construct the equation for the reaction. Include state symbols.

… [2]

Mark scheme

Show the mark scheme The official mark scheme detailing answers for question 21 parts a, b, and c. It provides expected observations, oxidation and reduction half-equations, step-by-step calculations for determining the mass of barium oxide and limestone using pH and molar masses, an ionic equation for barium hydroxide and sulfuric acid, and a balanced equation with state symbols for huntite reacting with hydrochloric acid.

AO

Question Answer Marks Guidance

element

21 (a) (i) Ca fizzes faster 1 AO2.3 CARE Both needed for 1 mark.

AND

Ca dissolves/disappears more quickly ORA

ALLOW AW

IGNORE finishes first

IGNORE more bubbles (need idea of rate)

IGNORE exothermic

21 (ii) 2

In half equations,

ALLOW the use of e for e–

Oxidation Mg → Mg2+ + 2e– AO2.6

×2 ALLOW Mg - 2e– → Mg2+

Reduction 2H+ + 2e– → H

OR H+ + e– → ½H IGNORE state symbols even is wrong BUT half

equations MUST only have species that change.

For charges on half equations,

ALLOW Mg+2 for Mg2+

OR H+1 for H+

If BOTH half equations are correct but shown with

oxidation and reduction the wrong way around, award

1 mark from the 2 marks for half equations

(b) (i) FIRST CHECK THE ANSWER ON ANSWER LINE 25 5 AO2.4 ALLOW ECF and 3SF throughout.

If answer = 2.53(g) award 5 marks ×5 ALLOW calculation process in any order.

--------------------------------------------------------------- IGNORE rounding errors past 3SF

-----------------------------------------------------------

[H+] = 10–13.12 OR 7.58 … × 10–14 (mol dm–3)

Calculator: 7.58577575 × 10–14

1 × 10–14

[OH–] = OR 0.1318 … (mol dm–3)

7.58 … × 10–14

Calculator: 0.1318256739

ALLOW alternative approach using pOH for first 2

marks.

p[OH-] = 14 – 13.12 = 0.88

[OH-] = 10 -0.88 = 0.1318…..

– 3 0.1318….. Calculator: 0.03295641846

n(OH ) in 250 cm = OR 0.0329……. (mol) –

4 0.033(0) comes from [OH ] = 0.132

0.0329……. Calculator: 0.01647820923

n(Ba(OH)2) or n(BaO) = OR 0.0164….. (mol)

Mass of BaO = 0.0164….. × 153.3 = 2.53 (g) 3SF Calculator: 2.526109475

Common errors

4 marks

5.05g Not dividing by 2

2.82g Use of Mr for Ba(OH)2

5.06g rounds to 0.132 in M2 then not

dividing by 2

3 marks

5.65g not dividing by 2 and using Mr for

Ba(OH)2

AO3.2 ALLOW multiples

(ii) Ba2+(aq) +2H+(aq) + SO42 –(aq) +2OH-(aq) 26 1 ALLOW

→ BaSO (s) +2H O(l) H+(aq)+ OH-(aq) → H O(l)

42 2

OR

Ba2+(aq) + SO 2 –(aq) → BaSO (s)

(c) (i) FIRST CHECK THE ANSWER ON ANSWER LINE 3 AO2.6 ALLOW ECF throughout

If answer = 731(g) award 3 marks ×3 ALLOW calculation process in any order.

--------------------------------------------------------------- IGNORE rounding errors past 3SF

n(Z)

1500

n (Ca5NH4(NO3)11•10H2O) = 1080.5 OR 1.388246…

Mass of limestone

n(CaCO3) = 1.388246… × 5 OR 6.94123…

AND DO NOT ALLOW 100 for Mr of CaCO3

mass CaCO3 = 6.94123… × 100.1 OR 694.8 g Common errors

2 marks

146g no x 5 for moles of CaCO3

694.8 × 100 660g use of 95.0/100

mass limestone = = 731 g (3SF)

95.0 29.3g divide by 5 rather than x5

(ii) Mg3Ca(CO3)4 (s) + 8HCl(aq) → 27 2 AO2.6 ALLOW multiples

3MgCl2(aq) + CaCl2(aq)+ 4H2O(l) + 4CO2(g) ×2

Correct formulae M2 dependent on M1

Balanced AND state symbols IGNORE incorrect state symbol for Mg3Ca(CO3)4

TOTAL 13

How to answer it

Reactions of Group 2 Metals and Their Compounds

What this question tests

This multi-step question assesses your understanding of Group 2 trends in reactivity, construction of half-equations and ionic equations, advanced pH and stoichiometry calculations, and multi-stage percentage purity and mass determinations.

Question 21 (a)

Reactivity Trends & Half-Equations

✅ Correct Answers

  • (i) Calcium fizzes faster AND calcium dissolves/disappears more quickly. (Both needed for 1 mark).
  • (ii) Oxidation: Mg → Mg²⁺ + 2e⁻
  • (ii) Reduction: 2H⁺ + 2e⁻ → H₂ (or H⁺ + e⁻ → ½H₂ )

💡 Key Knowledge

  • Reactivity of Group 2 metals increases down the group as atomic radius increases and shielding increases, making it easier to lose outer shell electrons.
  • In half-equations, oxidation is the loss of electrons (electrons on the right), and reduction is the gain of electrons (electrons on the left).

🧠 Exam Technique

For part (a)(i), ensure you give two distinct observable changes comparing rates (e.g., speed of effervescence and rate at which the solid disappears). Generic statements like "more reactive" or "exothermic" will not score.

❌ Common Errors

Writing reduction as H₂ → 2H⁺ + 2e⁻ or mixing up oxidation and reduction halves. Note that state symbols are ignored in half-equations as long as the chemistry is correct.

Question 21 (b)

Barium Oxide Calculations & Ionic Equations

📐 Step-by-Step Calculation: Part (b)(i)

  1. Step 1: Find [H⁺] from pH.
    [H⁺] = 10⁻¹³.¹² = 7.5857 × 10⁻¹⁴ mol dm⁻³
  2. Step 2: Calculate [OH⁻] using Kw (1.00 × 10⁻¹⁴ at 25°C).
    [OH⁻] = (1.00 × 10⁻¹⁴) / (7.5857 × 10⁻¹⁴) = 0.1318 mol dm⁻³
  3. Step 3: Calculate moles of OH⁻ in 250.0 cm³.
    n(OH⁻) = 0.1318 × (250.0 / 1000) = 0.03296 mol
  4. Step 4: Relate moles of ions to moles of BaO.
    BaO + H₂O → Ba(OH)₂, so 1 mol of BaO produces 2 mol of OH⁻.
    n(BaO) = 0.03296 / 2 = 0.01648 mol
  5. Step 5: Calculate mass of BaO to 3 significant figures.
    Mr(BaO) = 137.3 + 16.0 = 153.3 g mol⁻¹
    Mass = 0.01648 × 153.3 = 2.53 g
Award 5 marks for the correct final answer of 2.53 g displayed on the answer line. ECF applies throughout.

✅ Correct Answers: Part (b)(ii)

Ionic equation: Ba²⁺(aq) + 2H⁺(aq) + SO₄²⁻(aq) + 2OH⁻(aq) → BaSO₄(s) + 2H₂O(l)

Acceptable alternatives include net equations like Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) or H⁺(aq) + OH⁻(aq) → H₂O(l) .

❌ Common Calculation Traps in (b)(i)

  • Forgetting to divide by 2 when converting moles of hydroxide ions into moles of barium oxide (losing 1 mark).
  • Using the M_r of Ba(OH)₂ instead of BaO.
  • Failing to round the final answer to 3 significant figures.
Question 21 (c)

Limestone, Purity, and Stoichiometry

📐 Step-by-Step Calculation: Part (c)(i)

  1. Step 1: Calculate moles of fertiliser Z required.
    n(Z) = Mass / Mr = 1500 g / 1080.5 g mol⁻¹ = 1.3882 mol
  2. Step 2: Use stoichiometry to find moles of CaCO₃ needed. formula shows 5 Ca per molecule of Z.
    n(CaCO₃) = 1.3882 × 5 = 6.9412 mol
  3. Step 3: Calculate pure mass of CaCO₃.
    Mass (pure) = 6.9412 mol × 100.1 g mol⁻¹ = 694.8 g
  4. Step 4: Scale up for 95.0% purity limestone.ခ
    Mass of limestone = (694.8 / 95.0) × 100 = 731 g (3 SF)
Award 3 marks for 731 g on the answer line.

✅ Correct Answers: Part (c)(ii)

Equation: Mg₃Ca(CO₃)₄(s) + 8HCl(aq) → 3MgCl₂(aq) + CaCl₂(aq) + 4H₂O(l) + 4CO₂ (g)

M1: Correct formulae for all reactants and products. M2: Balanced equation with correct state symbols.

🧠 Exam Technique for (c)(i)

Always inspect complex molecular formulae carefully. Here, the formula Ca₅NH₄(NO₃)₁₁·10H₂O contains 5 calcium atoms per formula unit, meaning 5 moles of CaCO₃ are consumed to make 1 mole of fertiliser Z.

Topics

Module 3: Periodic table and energy · Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · 3.1 The periodic table · 2.1 Atoms and reactions · 5.1 Rates, equilibrium and pH

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.