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Reacting Gas Volumes

Learn and apply Reacting Gas Volumes in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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H1 Reacting Gas Volumes: Orientation

Reacting gas volumes are equation-ratio problems. When every gas is measured at the same temperature and pressure, equal amounts occupy equal volumes, so the gaseous coefficients become volume ratios.

H1 8873 scope
  • H1 8873 requires reacting gas-volume calculations, including combustion examples.
  • The H2 ideal-gas equation pV = nRT is not required and is not used in this lesson.

Definitions (Must Know)

  • A stoichiometric coefficient is the number before a formula in a balanced equation; it gives the reacting amount ratio.
  • A gas-volume ratio is the ratio of gaseous volumes measured at the same temperature and pressure.
  • The limiting reactant is consumed first and fixes the maximum amount or volume of product.
  • An excess reactant remains after the limiting reactant has been consumed.
  • A molar gas volume is the volume occupied by one mole under stated conditions. Use a numerical value only when the question supplies or defines the conditions.

Detailed Explanations

A. Why coefficient ratios become volume ratios

At fixed temperature and pressure, gas volume is proportional to amount. For:

N₂(g) + 3H₂(g) → 2NH₃(g)

the amount ratio 1:3:2 is also the volume ratio 1:3:2 when all three gas volumes are compared under the same conditions.

B. Limiting-gas workflow

  1. Write the balanced equation.
  2. Divide each available gas volume by its coefficient.
  3. The smaller result identifies the limiting gas.
  4. Use that gas and the equation ratio to find product volume.
  5. Calculate any excess volume from supplied minus reacted volume.

C. When the shortcut is invalid

Gas volume changes with temperature and pressure. Volumes measured under different conditions cannot be compared directly as amounts. Do not introduce pV = nRT into an H1 answer unless the relationship is explicitly supplied; instead use only the information and relationships given in the question.

Worked Examples

Modelled example 1

Use a Direct Gas-volume Ratio

Core

Problem

For 2H₂(g) + O₂(g) → 2H₂O(g), what volume of oxygen reacts with 120 cm³ of hydrogen when all gas volumes are measured at the same temperature and pressure?

Study the worked solution

  1. Read the gaseous ratio

    Method

    Use the coefficients of hydrogen and oxygen as a 2:1 volume ratio.

    Reason

    At the same temperature and pressure, gas volume is proportional to amount.

    Working

    H₂:O₂ = 2:1

  2. Scale the ratio

    Method

    Multiply the hydrogen volume by 1/2.

    Reason

    One oxygen volume reacts with every two hydrogen volumes.

    Working

    V(O₂) = 120(1/2) = 60 cm³

Guided practice 2

Find Product and Residual Gas Volumes

About 6 min

Problem

30 cm³ of methane reacts completely with 80 cm³ of oxygen: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) Find the carbon dioxide volume and the unused oxygen volume at the original temperature and pressure after the water has condensed.

Try this before viewing the solution

Unit: cm³
Unit: cm³

Hints

Hint 1: test the methane requirement
Each methane volume requires two oxygen volumes.
Hint 2: separate product from excess
Use 1:1 for methane to carbon dioxide, then subtract consumed oxygen from supplied oxygen.
View solution step by step

  1. Identify the limiting gas

    Method

    Calculate the oxygen required by 30 cm³ methane.

    Reason

    The 1:2 ratio requires 60 cm³ oxygen, which is less than the 80 cm³ supplied.

    Working

    V(O₂\ consumed) = 30(2) = 60 cm³

  2. Find carbon dioxide

    Reason

    Methane and carbon dioxide have equal coefficients.

    Working

    V(CO₂) = 30 cm³

  3. Find residual oxygen

    Method

    Subtract the consumed oxygen from the initial oxygen.

    Reason

    Oxygen is the excess gas and water is not part of the cooled gas mixture.

    Working

    V(O₂\ unused) = 80-60 = 20 cm³

Common misconception 3

Correct a Smaller-volume Limiting Claim

Find and correct the mistake

Learner claim

For 2CO(g) + O₂(g) → 2CO₂(g), 50 cm³ carbon monoxide is mixed with 40 cm³ oxygen. A learner says oxygen is limiting because its initial volume is smaller. Identify the first error, find the limiting gas, and calculate the product and residual gas volumes.

Find the error before calculating

Limiting gas

View solution step by step

  1. Test the oxygen requirement

    Method

    Apply the 2:1 carbon monoxide-to-oxygen ratio.

    Reason

    50 cm³ carbon monoxide needs only 25 cm³ oxygen, so oxygen is in excess.

    Working

    V(O₂\ required) = 50(1/2) = 25 cm³

  2. Calculate product and excess

    Method

    Use the 2:2 product ratio and subtract the oxygen consumed.

    Reason

    Carbon monoxide is limiting and is converted to the same gas volume of carbon dioxide.

    Working

    V(CO₂) = 50 cm³; V(O₂\ unused) = 40-25 = 15 cm³.

Examiner practice 4

Determine a Final Gas Mixture

4 marks

Problem

50 cm³ nitrogen monoxide is mixed with 40 cm³ oxygen at the same temperature and pressure: 2NO(g) + O₂(g) → 2NO₂(g) Determine the limiting reactant, the nitrogen dioxide volume, the unused gas volume and the final total gas volume. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Calculate oxygen required

    1 mark

    Method

    Halve the nitrogen monoxide volume.

    Reason

    The 2:1 ratio means 50 cm³ nitrogen monoxide requires 25 cm³ oxygen.

    Working

    V(O₂\ required) = 25 cm³, so NO is limiting.

  2. Find nitrogen dioxide

    1 mark

    Reason

    The 2:2 ratio gives equal nitrogen monoxide and nitrogen dioxide volumes.

    Working

    V(NO₂) = 50 cm³

  3. Find unused oxygen

    1 mark

    Reason

    Oxygen supplied minus oxygen consumed gives the residual gas.

    Working

    V(O₂\ unused) = 40-25 = 15 cm³

  4. Find the final total

    1 mark

    Method

    Add the product and residual gas volumes.

    Reason

    These are the only gases remaining after complete consumption of nitrogen monoxide.

    Working

    V_final = 50 + 15 = 65 cm³

Challenge 5

Deduce the Equation from Gas-volume Evidence

Minimal support

Problem

At the same temperature and pressure, 30 cm³ nitrogen reacts completely with 90 cm³ hydrogen to form 60 cm³ ammonia. Deduce the smallest whole-number equation, then predict the ammonia volume formed from 45 cm³ hydrogen with excess nitrogen.

Try this before viewing the solution

Hints

Hint 1: simplify the observations
Reduce the volume ratio 30:90:60 to its smallest whole numbers.
Hint 2: reuse the hydrogen ratio
Once the equation is known, compare the hydrogen and ammonia coefficients.
View solution step by step

  1. Infer the coefficients

    Method

    Divide all three measured volumes by 30.

    Reason

    Gas volumes at the same conditions are proportional to reacting amounts.

    Working

    30:90:60 = 1:3:2

  2. Write the equation

    Reason

    The simplified volume ratio supplies the smallest gaseous coefficients.

    Working

    N₂(g) + 3H₂(g) → 2NH₃(g)

  3. Predict from hydrogen

    Method

    Multiply the hydrogen volume by 2/3.

    Reason

    Three hydrogen volumes form two ammonia volumes when nitrogen is in excess.

    Working

    V(NH₃) = 45(2/3) = 30 cm³

Mind Stretchers

Attempt each unfamiliar application before opening the hint, then compare your reasoning chain with the solution.

Mind stretcher 1: Competing reactant volumesExtension

Question. 40 cm³ N₂ reacts with 100 cm³ H₂ according to N₂ + 3H₂ → 2NH₃. Find the residual reactant and ammonia volumes.

Show Hint

Compare 40/1 with 100/3 to identify the limiting gas.

Show Answer

Hydrogen is limiting. It reacts with 100/3 = 33.3 cm³ nitrogen, leaving 6.7 cm³ nitrogen. The H₂:NH₃ ratio is 3:2, so 66.7 cm³ ammonia forms.

Mind stretcher 2: Deducing a gas-mixture compositionExtension

Question. A 40.0 cm³ mixture of methane and ethene consumes exactly 100 cm³ oxygen on complete combustion at common conditions. Find each hydrocarbon volume.

Show Hint

Let methane be x. Methane needs twice its volume of oxygen; ethene needs three times its volume.

Show Answer

Ethene volume is 40.0-x. Oxygen used is 2x + 3(40.0-x) = 100, so x = 20.0. The mixture contains 20.0 cm³ methane and 20.0 cm³ ethene.