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AS · Practice questions · Volumes of gases

Read the volumes.

Six original Cambridge-style questions on gas volumes: molar volume, gas and mass calculations, volume ratios, Mr from a gas sample and combustion analysis.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Same T and p, same volume per mole.

01
Calculation
[2 marks]

What mass of magnesium reacts with excess dilute hydrochloric acid to produce 2.40 dm3 of hydrogen at room conditions? Mg + 2HCl → MgCl2 + H2. (Ar: Mg 24.3)

  • n(H2) = 2.40 ÷ 24.0 = 0.100 mol, so n(Mg) = 0.100 mol. ✓
  • Mass of Mg = 0.100 × 24.3 = 2.43 g. ✓
02
Calculation
[2 marks]

50 cm3 of carbon monoxide is mixed with 50 cm3 of oxygen and ignited. 2CO(g) + O2(g) → 2CO2(g). Calculate the total volume of gas at the end, all volumes at the same temperature and pressure.

  • 50 cm3 of CO reacts with 25 cm3 of O2 to form 50 cm3 of CO2. ✓
  • 25 cm3 of O2 is left over, so the final volume is 50 + 25 = 75 cm3. ✓
03
Analysis
[4 marks]

15 cm3 of a gaseous hydrocarbon is burned in 120 cm3 of oxygen (an excess). After cooling to room temperature the gas volume is 105 cm3. After passing through aqueous sodium hydroxide, 75 cm3 remains. Deduce the molecular formula of the hydrocarbon.

  • The NaOH absorbs CO2: volume of CO2 = 105 − 75 = 30 cm3. The 75 cm3 left is unreacted oxygen. ✓
  • CO2 : hydrocarbon = 30 : 15 = 2, so x = 2. ✓
  • Oxygen used = 120 − 75 = 45 cm3, so O2 : hydrocarbon = 3 = x + y/4, giving y/4 = 1 and y = 4. ✓
  • The hydrocarbon is C2H4. ✓
04
Analysis
[2 marks]

A 0.640 g sample of a gas occupies 240 cm3 at room conditions. Calculate its Mr and suggest which of CH4, O2 and SO2 it is.

  • n = 240 ÷ 24 000 = 0.0100 mol; M = 0.640 ÷ 0.0100 = 64.0 g mol−1. ✓
  • Mr = 64.0, which matches SO2 (Mr 64.1). CH4 is 16.0 and O2 is 32.0. ✓
05
Calculation
[2 marks]

Calculate the volume of carbon dioxide, at room conditions, produced by the complete combustion of 2.30 g of ethanol. C2H5OH + 3O2 → 2CO2 + 3H2O. (Mr: C2H5OH 46.0)

  • n(ethanol) = 2.30 ÷ 46.0 = 0.0500 mol, so n(CO2) = 0.100 mol. ✓
  • V = 0.100 × 24.0 = 2.40 dm3. ✓
06
Explain
[2 marks]

In combustion analysis of a hydrocarbon, explain why the water formed does not contribute to the final gas volume, and why aqueous sodium hydroxide is used.

  • The volumes are measured at room temperature, where water is a liquid; its volume is negligible compared with the gases. ✓
  • Carbon dioxide is an acidic gas and reacts with sodium hydroxide, so the decrease in volume on passing through NaOH(aq) equals the volume of CO2. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.