Explain each application through reactivity, solubility or acid–base chemistry, then use balanced equations to calculate amounts.
Magnesium as a reducing agent in titanium extraction
Titanium(IV) chloride reacts with magnesium to form titanium and magnesium chloride. Two Mg atoms each lose two electrons; Ti changes from +4 to 0 and gains four electrons. Magnesium is the reducing agent because it supplies the electrons.
Industrial context: the oxide is converted to volatile TiCl₄, which can be purified by distillation before reduction under an inert atmosphere. Argon avoids unwanted reactions with air, and moisture is excluded because TiCl₄ reacts with water. Carbon reduction can introduce titanium carbide, making it unsuitable when pure titanium is required.
Focus on the specified Mg/TiCl₄ equation and electron transfer. Do not turn this industrial route into an absolute claim that titanium cannot ever be made by electrochemical methods.
Why different compounds suit different uses
Mg(OH)₂ neutralises excess stomach acid. It is sparingly soluble, but as dissolved hydroxide reacts with acid, further solid can dissolve. The chemical equation explains its antacid action; it is not a dosing instruction.
Ca(OH)₂ is used to neutralise acidic soil. Its hydroxide ions remove H⁺, raising the pH. The useful property is basicity, not the production of a gas.
BaSO₄ is used as an X-ray contrast material because it strongly attenuates X-rays and has very low solubility, limiting release and absorption of toxic Ba²⁺. Do not replace it with a soluble barium salt. “Barium is safe” is the wrong explanation: the specific compound and its insolubility matter.
Removing sulfur dioxide from flue gases
Sulfur dioxide is an acidic oxide. Basic calcium oxide, or calcium carbonate, removes it from waste gases. The products below contain sulfite, SO₃²⁻, not sulfate, SO₄²⁻. Oxidation by oxygen can then convert sulfite to sulfate.
Keep the carbonate and oxide equations distinct: using CaCO₃ produces CO₂, whereas the CaO equation does not. A question giving a final sulfate product may require oxygen in the overall equation.
Carbon dioxide and limewater
Limewater is a solution of calcium hydroxide. Carbon dioxide produces white calcium carbonate, turning it cloudy. With prolonged bubbling of excess CO₂, the cloudiness can disappear as soluble calcium hydrogencarbonate forms.
Observation: cloudiness or a white precipitate. Inference: CO₂ is present. Keep observations distinct from chemical explanations.
The mole ratio comes before mass
For any Group 2 calculation, first balance the equation, convert the measured quantity to moles, apply the coefficient ratio, then convert to the requested mass, volume or concentration. Label the substance at each stage.
Antacid example: 0.2915 g pure Mg(OH)₂ is 0.2915 ÷ 58.3 = 0.00500 mol. It neutralises 0.0100 mol HCl, equivalent to 50.0 cm³ of 0.200 mol dm⁻³ HCl. The acid:hydroxide ratio is 2:1, not 1:1.
If a tablet contains inert ingredients, use the mass of Mg(OH)₂, not automatically the total tablet mass. In a back titration, subtract leftover acid from initial acid before applying the ratio.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. Explain the role of magnesium in TiCl₄ + 2Mg → Ti + 2MgCl₂.Show answer
Mg changes from oxidation state 0 to +2 and donates electrons. It reduces Ti from +4 to 0 and is therefore the reducing agent.
Q2. Using Ar(Mg) = 24.3 and Ar(Ti) = 47.9, find the theoretical titanium mass from 4.86 g Mg and excess TiCl₄.Show answer
n(Mg) = 0.200 mol. n(Ti) = 0.100 mol from the 2:1 ratio. Mass(Ti) = 4.79 g.
Q3. Why is insolubility part of the explanation for using BaSO₄ in an X-ray examination?Show answer
Very low solubility limits the release of soluble toxic Ba²⁺ and its absorption. The barium-containing material attenuates X-rays, allowing contrast.
Q4. How many moles of CaCO₃ are required to remove 3.20 g SO₂? Use Mr(SO₂) = 64.0.Show answer
n(SO₂) = 3.20 ÷ 64.0 = 0.0500 mol. CaCO₃:SO₂ is 1:1, so 0.0500 mol CaCO₃ is needed.
Q5. Why does Mg(OH)₂ require two moles of HCl per mole, but CaO produces no CO₂ when removing SO₂?Show answer
Mg(OH)₂ contains two OH groups, each neutralising one H⁺: two HCl are needed. CaO + SO₂ → CaSO₃ contains no carbon, so it cannot produce CO₂. Carbon dioxide is produced when CaCO₃ is used instead.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- Chemrevise — AQA 2.2 Group 2 (October 2025) — Primary coverage checklist, pp1–4; qualifications explained in the notes.
- AQA 7405 specification — 3.2.2 and relevant Required Practical 4 work.
- AQA 7404/1 June 2019 mark scheme — Q07.1, p13: titanium extraction.
- AQA 7404/1 June 2019 examiner report — Q07, p5: extraction equation and identifying which component reacts.
- AQA 7404/1 2020 mark scheme — Q05.2, p16: oxidation-state explanation.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.