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1.3.3Physical Chemistry

Reacting Volumes — AQA AS Chemistry Revision Notes & Topical Questions

5 min read 5 parts Lesson 16 of 102

📚 AQA AS Chemistry Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5Gas Volumes and Moles

The amount of a gas can be found from its volume without knowing its mass.

Avogadro’s law
  • Avogadro’s law Equal volumes of different gases, measured at the same temperature and pressure, contain the same number of molecules or moles.
Volume ratio
  • Volume ratio Because volume is proportional to moles, the ratio of gas volumes in a reaction equals the ratio of the coefficients in the balanced equation.
Molar gas volume
  • Molar gas volume At room temperature and pressure (about 298 K and 100 kPa) one mole of any gas occupies roughly 24.0 dm³,
    • so this value is given in many questions.
Same conditions
  • Same conditions The simple volume-ratio method only works when every gas is measured at one fixed temperature and pressure,
    • and only for substances that are gases.

Part 2 of 5Two Calculation Routes

1
Write the balanced equation with state symbols so you can see which substances are gases and which are solids or liquids.
2
For gas-only volume problems, scale the volumes directly using the coefficient ratio, for example 1 : 2 : 1 means 10 cm³ reacts with 20 cm³ to give 10 cm³.
3
For a gas produced from a solid or solution, first convert mass or concentration into moles of the limiting reagent.
4
Use the equation ratio to find moles of gas, then multiply by 24.0 dm³ mol⁻¹ to find the volume, or divide a volume by 24.0 to find moles.
5
Check units carefully: 1 dm³ equals 1000 cm³, so convert before substituting and quote the answer to suitable significant figures.

Part 3 of 5Typical Contexts

Typical Contexts
Combustion of gases
  • Combustion of gases Gaseous fuels burn in excess oxygen,
    • so volumes of oxygen used and carbon dioxide formed follow directly from the equation, once cooled and with water condensed.
Gas from carbonates
  • Gas from carbonates Adding acid to a carbonate releases carbon dioxide,
    • which can be collected in a gas syringe and compared with the volume predicted.
Limiting reagent
  • Limiting reagent When one gas is not in excess, the volume of product is set by the reactant that runs out first,
    • and leftover gas must be added to the final total.
Percentage yield
  • Percentage yield The measured volume of gas collected divided by the theoretical volume, multiplied by 100, gives the percentage yield for a gas-producing reaction.
QuantitySymbolUnit
Gas volumeVdm³ or cm³
Amount of gasnmol
Molar gas volumeVm24.0 dm³ mol⁻¹ at RTP

Part 4 of 5Worked Example

Question

  • Calcium carbonate of mass 2.50 g reacts completely with excess dilute hydrochloric acid: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g).
  • Calculate the volume of carbon dioxide at RTP.
  • (Mr of CaCO₃ = 100.1)

  1. Moles of CaCO₃ = 2.50 ÷ 100.1 = 0.02498 mol.
  2. The equation shows CaCO₃CO₂ = 1 : 1, so moles of CO₂ = 0.02498 mol.
  3. Volume = 0.02498 × 24.0 = 0.5995 dm³.
  4. Convert0.5995 dm³ × 1000 = 599.5 cm³.

Answer About 600 cm³ (0.600 dm³) of carbon dioxide.

Part 5 of 5Exam Checklist & Tips

1.3.3 – Key Points for the Exam

  • Avogadro’s law Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
  • Molar gas volume The volume occupied by one mole of a gas under stated conditions, about 24.0 dm³ at RTP.
  • RTP Room temperature and pressure, taken as about 298 K and 100 kPa in these calculations.
  • Limiting reagent The reactant that is completely used up first and so decides how much product forms.

Common Mistakes

  • Using 24.0 dm³ mol⁻¹ for a gas measured at a different temperature or pressure without correcting for the conditions.
  • Forgetting to convert cm³ to dm³ by dividing by 1000, giving an answer a thousand times too large.
  • Including liquid water in a gas-volume ratio; at room temperature it is not a gas.
  • Applying the volume ratio to solids or aqueous substances, which have no molar gas volume.

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