Amounts and Redox Problem Solving

This final lesson brings together everything you have learned about moles, equations, titrations, gas volumes, and redox chemistry. A-Level exam questions frequently require you to combine multiple concepts in a single multi-step calculation. The key to success is a systematic approach: identify what you know, write the balanced equation, and work step by step through the moles.

Strategy for Multi-Step Problems

flowchart TD
    A["Read the question — identify what you know and what you need"] --> B["Write the balanced equation"]
    B --> C["Identify the substance you can calculate moles for first"]
    C --> D{"What information is given?"}
    D -- Mass --> E["n = m / M"]
    D -- Volume + concentration --> F["n = c × V"]
    D -- Gas volume at RTP --> G["n = V / 24.0"]
    D -- "pV = nRT data" --> H["n = pV / RT"]
    E --> I["Use mole ratio to find moles of target substance"]
    F --> I
    G --> I
    H --> I
    I --> J["Convert moles to the required answer"]
    J --> K{"What does the question ask for?"}
    K -- Mass --> L["m = n × M"]
    K -- Concentration --> M["c = n / V"]
    K -- Volume of gas --> N["V = n × 24.0"]
    K -- Percentage --> O["(part / whole) × 100"]

Multi-Step Mole Calculations

Many exam questions present a chain of reactions or a scenario where you must move from one piece of information to the final answer through several intermediate steps.

Worked Example 1: Purity by Titration

A student dissolves 1.20 g of an impure sample of sodium carbonate (Na₂CO₃) in water and makes up the solution to 250 cm³. A 25.0 cm³ aliquot requires 18.5 cm³ of 0.100 mol dm⁻³ HCl for complete neutralisation. Calculate the purity of the sodium carbonate.

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Step 1: Moles of HCl = 0.100 × 0.0185 = 1.85 × 10⁻³ mol

Step 2: Moles of Na₂CO₃ in 25.0 cm³ = 1.85 × 10⁻³ / 2 = 9.25 × 10⁻⁴ mol (1:2 ratio)

Step 3: Moles of Na₂CO₃ in 250 cm³ = 9.25 × 10⁻⁴ × 10 = 9.25 × 10⁻³ mol

Step 4: Mass of Na₂CO₃ = 9.25 × 10⁻³ × 106.0 = 0.981 g

Step 5: Purity = (0.981 / 1.20) × 100 = 81.7%

Key skill: The scaling factor (×10) accounts for the fact that 25.0 cm³ is one-tenth of the 250 cm³ total volume.

Back Titrations

A back titration is used when the substance being analysed cannot be titrated directly — for example, because it is insoluble or reacts too slowly. The method is:

React the analyte with a known excess of one reagent.
Titrate the leftover (unreacted) reagent with a standard solution.
Calculate the amount of reagent that reacted with the analyte by subtraction.

Worked Example 2

2.00 g of an impure sample of CaCO₃ is added to 50.0 cm³ of 1.00 mol dm⁻³ HCl (excess). After the reaction is complete, the excess HCl requires 24.0 cm³ of 0.500 mol dm⁻³ NaOH to neutralise. Calculate the percentage purity of CaCO₃.

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ HCl + NaOH → NaCl + H₂O

Step 1: Total moles of HCl added = 1.00 × 0.0500 = 0.0500 mol

Step 2: Moles of NaOH = 0.500 × 0.0240 = 0.0120 mol

Step 3: Moles of excess HCl = 0.0120 mol (1:1 ratio with NaOH)

Step 4: Moles of HCl that reacted with CaCO₃ = 0.0500 − 0.0120 = 0.0380 mol

Step 5: Moles of CaCO₃ = 0.0380 / 2 = 0.0190 mol (1:2 ratio)

Step 6: Mass of CaCO₃ = 0.0190 × 100.0 = 1.90 g

Step 7: Purity = (1.90 / 2.00) × 100 = 95.0%

Worked Example 3 — Back Titration with an Antacid

A 0.500 g antacid tablet (containing Mg(OH)₂) was dissolved in 25.0 cm³ of 0.500 mol dm⁻³ HCl. The excess acid required 10.5 cm³ of 0.200 mol dm⁻³ NaOH to neutralise. Find the percentage of Mg(OH)₂ in the tablet.

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O HCl + NaOH → NaCl + H₂O

Step 1: Total HCl = 0.500 × 0.0250 = 0.0125 mol Step 2: Moles NaOH = 0.200 × 0.0105 = 0.00210 mol = excess HCl Step 3: HCl reacted with Mg(OH)₂ = 0.0125 − 0.00210 = 0.0104 mol Step 4: Moles Mg(OH)₂ = 0.0104 / 2 = 0.00520 mol Step 5: Mass Mg(OH)₂ = 0.00520 × 58.3 = 0.303 g Step 6: Percentage = (0.303 / 0.500) × 100 = 60.6%

Combining Gas and Solution Calculations

Some questions require you to move between gas volumes and solution concentrations within the same problem.

Worked Example 4

What volume of 0.200 mol dm⁻³ sulfuric acid is needed to react with the ammonia produced when 1.20 g of ammonium chloride is heated with excess sodium hydroxide?

NH₄Cl + NaOH → NaCl + H₂O + NH₃ 2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Step 1: Moles of NH₄Cl = 1.20 / 53.5 = 0.02243 mol

Step 2: Moles of NH₃ = 0.02243 mol (1:1 ratio)

Step 3: Moles of H₂SO₄ = 0.02243 / 2 = 0.01121 mol (2:1 ratio)

Step 4: Volume of H₂SO₄ = 0.01121 / 0.200 = 0.05607 dm³ = 56.1 cm³

Worked Example 5 — Gas Volume from a Reaction

What volume of CO₂ at RTP is produced when 25.0 cm³ of 2.00 mol dm⁻³ HCl reacts with excess CaCO₃?

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Step 1: Moles HCl = 2.00 × 0.0250 = 0.0500 mol Step 2: Moles CO₂ = 0.0500 / 2 = 0.0250 mol (2:1 ratio) Step 3: Volume CO₂ = 0.0250 × 24.0 = 0.600 dm³ = 600 cm³

Redox Stoichiometry Problems

Worked Example 6

In acidic solution, potassium dichromate(VI) oxidises iron(II) ions. 25.0 cm³ of Fe²⁺ solution required 20.0 cm³ of 0.0167 mol dm⁻³ K₂Cr₂O₇. Calculate the concentration of Fe²⁺.

Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺

Step 1: Moles Cr₂O₇²⁻ = 0.0167 × 0.0200 = 3.34 × 10⁻⁴ mol

Step 2: Mole ratio Cr₂O₇²⁻ : Fe²⁺ = 1 : 6 Moles Fe²⁺ = 6 × 3.34 × 10⁻⁴ = 2.004 × 10⁻³ mol

Step 3: [Fe²⁺] = 2.004 × 10⁻³ / 0.0250 = 0.0802 mol dm⁻³

Worked Example 7 — Combined Redox and Mass

A 2.50 g sample of steel was dissolved in acid. The Fe²⁺ ions formed were titrated with 0.0200 mol dm⁻³ KMnO₄; the titre was 31.2 cm³. Calculate the percentage of iron in the steel.

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

Step 1: Moles MnO₄⁻ = 0.0200 × 0.0312 = 6.24 × 10⁻⁴ mol

Step 2: Moles Fe²⁺ = 5 × 6.24 × 10⁻⁴ = 3.12 × 10⁻³ mol

Step 3: Mass Fe = 3.12 × 10⁻³ × 56.0 = 0.175 g

Step 4: Percentage = (0.175 / 2.50) × 100 = 7.00%

Tricky Multi-Concept Problems

Worked Example 8 — Hydrated Salt Analysis

3.57 g of hydrated iron(II) sulfate (FeSO₄·xH₂O) was dissolved in dilute sulfuric acid and made up to 250 cm³. 25.0 cm³ aliquots were titrated with 0.0200 mol dm⁻³ KMnO₄; mean titre = 20.0 cm³. Find x.

Step 1: Moles MnO₄⁻ = 0.0200 × 0.0200 = 4.00 × 10⁻⁴ mol

Step 2: Moles Fe²⁺ in 25.0 cm³ = 5 × 4.00 × 10⁻⁴ = 2.00 × 10⁻³ mol

Step 3: Moles Fe²⁺ in 250 cm³ = 2.00 × 10⁻³ × 10 = 0.0200 mol

Step 4: 1 mol FeSO₄·xH₂O contains 1 mol Fe²⁺, so moles of hydrated salt = 0.0200 mol

Step 5: Mᵣ of FeSO₄·xH₂O = 3.57 / 0.0200 = 178.5

Step 6: Mᵣ of FeSO₄ = 56.0 + 32.0 + 64.0 = 152.0 Mass from xH₂O = 178.5 − 152.0 = 26.5 x = 26.5 / 18.0 = 1.47... Hmm, this should be a whole number. Let us check: if x = 1, Mᵣ = 170; if x = 7, Mᵣ = 278. With the given data, 178.5 does not yield a clean integer. In a real exam, the data would be chosen to give x = 7 (the common hydrate). The value of x for the well-known hydrate is x = 7 (FeSO₄·7H₂O, Mᵣ = 278.0), so if the mass were 5.56 g the calculation works perfectly:

Revised: Mᵣ = 5.56 / 0.0200 = 278.0. Then 278.0 − 152.0 = 126.0, x = 126.0 / 18.0 = 7.

This type of question tests your ability to combine a redox titration with the concept of hydrated salt formulae.

Common Mistakes in Multi-Step Problems

Mistake	How to spot it	Fix
Forgetting to scale up from aliquot to total volume	Answer is too small by factor of 10	Multiply: total volume / aliquot volume
Using the wrong mole ratio	Answer off by factor of 2, 3, 5, etc.	Write and check the balanced equation
Confusing which substance is from the burette vs. flask	Everything inverted	Burette = known concentration added
Not converting units (cm³ ↔ dm³, kPa ↔ Pa)	Off by factor of 1000	Convert before substituting
Rounding intermediate values	Final answer drifts	Keep 4+ s.f. until the end
In back titrations: subtracting the wrong way round	Negative moles (impossible)	Total − excess = amount that reacted

Exam Strategy Summary

Step	What to do
1	Read the question twice. Identify all given data.
2	Write the balanced equation(s).
3	Find the substance you can calculate moles for directly.
4	Use the mole ratio to move to the target substance.
5	Convert moles to whatever the question requires (mass, concentration, volume, percentage).
6	Check your answer: does it have reasonable units and magnitude?
7	Give your answer to the appropriate number of significant figures (usually 3 s.f.).

Practice Problems (Work Through These)

Problem 1: 0.500 g of an alloy of zinc and copper was dissolved in excess HCl. Only the zinc reacts: Zn + 2HCl → ZnCl₂ + H₂. The hydrogen gas collected occupied 120 cm³ at RTP. Find the percentage of zinc in the alloy.

Solution outline: n(H₂) = 0.120/24.0 = 0.00500 mol → n(Zn) = 0.00500 mol → mass Zn = 0.00500 × 65.4 = 0.327 g → % Zn = (0.327/0.500) × 100 = 65.4%

Problem 2: 25.0 cm³ of hydrogen peroxide solution was diluted to 250 cm³. 25.0 cm³ of the diluted solution was acidified and titrated with 0.0200 mol dm⁻³ KMnO₄; titre = 25.0 cm³. Find the concentration of the original H₂O₂.

The relevant equation: 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 8H₂O + 5O₂

Solution outline: n(MnO₄⁻) = 0.0200 × 0.0250 = 5.00 × 10⁻⁴ mol → n(H₂O₂) in 25.0 cm³ = (5/2) × 5.00 × 10⁻⁴ = 1.25 × 10⁻³ mol → in 250 cm³ = 0.0125 mol → this came from 25.0 cm³ of original → [H₂O₂] = 0.0125/0.0250 = 0.500 mol dm⁻³

These combined problems are where marks are won and lost at A-Level. The students who score highest are those who approach every problem with the same systematic method: equation → moles → ratio → answer.

A-Level Deep Dive: Amounts and Redox Problem Solving

Spec mapping

Edexcel 9CH0 specification synoptic across Topics 3, 5 and 14, this lesson integrates the mole concept, balanced equations, oxidation states, half-equations, redox titrations and concentration calculations into multi-step problem solving (refer to the official specification document for exact wording). Examined across all three papers, but particularly Paper 3 (synoptic practical paper, 2 hours, with extended-response 12-mark questions). The data sheet provides relative atomic masses, molar volume at RTP, gas constant and standard electrode potentials. Multi-step synoptic questions are the highest-tariff items on the exam (10–14 marks) and discriminate strongly between A and A* candidates.

Worked example with full mark scheme

Question (12 marks): A 1.250 g sample of impure pyrolusite ore (mainly MnO $_2$ ) is treated with excess concentrated HCl. The MnO $_2$ oxidises Cl $^-$ to Cl $_2$ , which is collected. The remaining Mn $^{2+}$ solution is made up to 250 cm $^3$ . A 25.00 cm $^3$ aliquot is added to excess KI in acidic solution; the liberated I $_2$ is titrated with 0.1000 mol dm $^{-3}$ Na $_2$ S $_2$ O $_3$ , requiring 17.20 cm $^3$ to reach the starch endpoint.

(a) Write the balanced equation for MnO $_2$ + HCl. (2)

(b) Calculate the moles of S $_2$ O $_3^{2-}$ used in the titration. (1)

(d) Identify the species responsible for liberating I $_2$ from KI, and write the half-equation. (2)

(e) Calculate the moles of MnO $_2$ in the original sample. (3)

(f) Calculate the percentage purity of the pyrolusite ( $M_r(\text{MnO}_2) = 86.9$ ). (3)

Solution with mark scheme:

(a) From half-equations: $\text{MnO}_2 + 4\text{H}^+ + 2e^- \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O}$ and $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ .

Combine: $\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}$ .

M1 A1.

(b) $n(\text{S}_2\text{O}_3^{2-}) = 0.01720 \times 0.1000 = 1.720 \times 10^{-3}\,\text{mol}$ . B1.

$n(\text{I}_2) = 1.720 \times 10^{-3}/2 = 8.600 \times 10^{-4}\,\text{mol}$ . B1.