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GCSE/Chemistry/AQA· Higher tier

C3.9Using moles to balance equations and gas volumes (HT): mole ratios with gases and volumes at room conditions

Notes

Mole ratios with gases and gas volumes (HT)

For gases, the volume at the same temperature and pressure is proportional to the number of moles (Avogadro's law). At room temperature and pressure (RTP, 20 °C and 1 atm), one mole of any gas occupies 24 dm³.

Avogadro's law

Equal volumes of any gas at the same T and P contain the same number of molecules.

So at RTP: 1 mole = 24 dm³ (or 24,000 cm³).

This is true for any gas — H₂, O₂, CO₂, He — regardless of its molecular size.

Working with gas volumes

moles of gas = volume (dm³) ÷ 24 (at RTP) volume (dm³) = moles × 24 (at RTP)

Worked exampleWorked examples

Example 1

What volume does 0.25 mol of CO₂ occupy at RTP?

V = 0.25 × 24 = 6 dm³

Example 2

How many moles in 480 cm³ of O₂ at RTP?

V = 480/1000 = 0.48 dm³ moles = 0.48 / 24 = 0.02 mol

Example 3 — combined with stoichiometry

What volume of CO₂ is produced at RTP when 5.0 g of CaCO₃ is heated?

CaCO₃ → CaO + CO₂. M_r CaCO₃ = 100.

  • moles CaCO₃ = 5.0/100 = 0.05 mol
  • moles CO₂ = 0.05 mol (1:1)
  • V CO₂ = 0.05 × 24 = 1.2 dm³ (or 1200 cm³)

Example 4 — gas + gas

What volume of O₂ reacts with 100 cm³ of CH₄ at RTP?

CH₄ + 2O₂ → CO₂ + 2H₂O.

Mole ratio CH₄ : O₂ = 1 : 2. So volumes are in 1 : 2 ratio at the same T and P.

V O₂ = 2 × 100 = 200 cm³.

(For gas + gas problems you don't even need to convert to moles — volumes scale directly.)

Why volumes scale by moles for gases

Avogadro's law: at the same temperature and pressure, equal volumes of gas contain equal numbers of particles. Particle size and mass don't matter — gas particles are far apart, so each molecule "uses" the same volume.

Worked exampleWorked example — practical

In a school experiment, marble chips (CaCO₃) react with hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

A student uses 1.0 g of CaCO₃ in excess HCl. What volume of CO₂ is produced at RTP?

  • moles CaCO₃ = 1.0/100 = 0.01 mol
  • moles CO₂ = 0.01 mol
  • V = 0.01 × 24 = 0.24 dm³ = 240 cm³

Common mistakes

  • Using 24 cm³ instead of 24 dm³. 1 mole = 24,000 cm³ if you're working in cm³.
  • Forgetting the mole ratio. A 1 : 2 ratio means the gas volume of one is twice the other.
  • Working at non-RTP without adjusting. The 24 dm³/mol value only applies at RTP. (Outside GCSE, use ideal gas equation pV = nRT.)
  • Mixing units in a single calculation.

Links

Builds on C3.4 (moles), C3.5 (mole ratios). Useful in C5 (energy from gases), C6.1 (rates of gas-producing reactions) and C9.1–C9.5 (atmospheric chemistry).

AI-generated · claude-opus-4-7 · v3-deep-chemistry

Practice questions

Try each before peeking at the worked solution.

  1. Question 12 marks

    Volume of gas (H)

    (H1) Calculate the volume of 0.5 mol of nitrogen gas at RTP.

    [Higher tier — 2 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-chemistry

  2. Question 22 marks

    Moles from volume (H)

    (H2) Calculate the moles in 600 cm³ of oxygen at RTP.

    [Higher tier — 2 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-chemistry

  3. Question 33 marks

    From mass to gas vol (H)

    (H3) Calculate the volume of CO₂ at RTP produced from 4.0 g of CaCO₃.
    CaCO₃ → CaO + CO₂; M_r CaCO₃ = 100.

    [Higher tier — 3 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-chemistry

  4. Question 42 marks

    Gas-gas ratio (H)

    (H4) What volume of O₂ at RTP is needed to burn 50 cm³ of methane?
    CH₄ + 2O₂ → CO₂ + 2H₂O.

    [Higher tier — 2 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-chemistry

  5. Question 53 marks

    Hydrogen volume (H)

    (H5) Calculate the volume of H₂ released at RTP when 0.48 g of Mg reacts with excess HCl.
    Mg + 2HCl → MgCl₂ + H₂; A_r Mg = 24.

    [Higher tier — 3 marks]

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    AI-generated · claude-opus-4-7 · v3-deep-chemistry

  6. Question 62 marks

    Avogadro's law (H)

    (H6) State Avogadro's law for gases.

    [Higher tier — 2 marks]

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    AI-generated · claude-opus-4-7 · v3-deep-chemistry

  7. Question 73 marks

    Mass from gas vol (H)

    (H7) A student collects 240 cm³ of CO₂ at RTP from heating CaCO₃. What mass of CaCO₃ was used?

    [Higher tier — 3 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-chemistry

Flashcards

C3.9 — Gas volumes and moles

10-card SR deck on RTP gas volumes (HT).

10 cards · spaced repetition (SM-2)