The shape of the lesson
By the end of the lesson, learners can
- The molar gas volume: Set the amount of gas in the piston and read the volume at room temperature and pressure.
- Volume ratios from the equation: When every volume is measured at the same temperature and pressure, the coefficients in the equation give the volume ratio directly.
- Masses into volumes: When one substance is a solid or solution and the other a gas, go through moles as usual and finish with the molar volume.
- n = V ÷ 24.0 with V in dm3 at room conditions.
- cm3 to dm3: divide by 1000.
- Gas reactions: volume ratio = mole ratio.
- Combustion: water is liquid; NaOH removes CO2.
One picture every idea returns to
Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. So one mole of any gas occupies the same volume: 24.0 dm3 at room conditions, the Data Booklet value. Amount of gas n = V ÷ 24.0 with V in dm3 (or V ÷ 24 000 with V in cm3). For reactions between gases, the volume ratio equals the mole ratio in the equation.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, One mole, one volume.. At the same temperature and pressure, a mole of hydrogen fills the same space as a mole of carbon dioxide. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The molar gas volume | Set the amount of gas in the piston and read the volume at room temperature and pressure. Project tab 5 of the lab (Stoichiometry lab, gas volumes) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Volume ratios from the equation | When every volume is measured at the same temperature and pressure, the coefficients in the equation give the volume ratio directly. | Pairs |
| 21 to 30 min | Teach and check: Masses into volumes | When one substance is a solid or solution and the other a gas, go through moles as usual and finish with the molar volume. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Masses into volumes, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Watch the units: 1 dm 3 = 1000 cm 3, so use 24.0 with dm 3 and 24 000 with cm 3. | Water formed in a combustion reaction is a liquid at room temperature, so it does not count in the final gas volume. And the molar volume applies only to gases, never to solids, liquids or solutions. |
Volumes of gases: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
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)
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.
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.
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.
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)
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.
Answer key and marking notes
- 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.
- 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.
- 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.
- 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.
- n(ethanol) = 2.30 ÷ 46.0 = 0.0500 mol, so n(CO2) = 0.100 mol.
- V = 0.100 × 24.0 = 2.40 dm3.
- 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.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
What volume does 8.00 g of oxygen gas, O2, occupy at room conditions? (Ar: O 16.0; molar gas volume 24.0 dm3 mol−1)
- A6.00 dm3
- B12.0 dm3
- C192 dm3
- D6000 dm3
What is the mass of 360 cm3 of carbon dioxide at room conditions? (Ar: C 12.0, O 16.0; molar gas volume 24.0 dm3 mol−1)
- A0.660 g
- B660 g
- C0.707 g
- D0.0150 g
40 cm3 of hydrogen is mixed with 40 cm3 of oxygen and exploded: 2H2(g) + O2(g) → 2H2O(l). What is the total volume of gas left after cooling to room temperature and pressure?
- A20 cm3
- B60 cm3
- C0 cm3
- D40 cm3
0.460 g of sodium reacts completely with water: 2Na + 2H2O → 2NaOH + H2. What volume of hydrogen forms at room conditions? (Ar: Na 23.0; molar gas volume 24.0 dm3 mol−1)
- A240 cm3
- B480 cm3
- C120 cm3
- D0.240 cm3
Samples of four gases each have a volume of 1.00 dm3 at the same temperature and pressure. Which sample has the greatest mass? (Ar: H 1.0, C 12.0, O 16.0)
- Acarbon dioxide, CO2
- Boxygen, O2
- Cmethane, CH4
- Dall four have the same mass
10 cm3 of ethene, C2H4, is burned in 50 cm3 of oxygen: C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l). What is the total volume of gas after cooling to room temperature and pressure?
- A40 cm3
- B60 cm3
- C20 cm3
- D70 cm3
2.45 g of potassium chlorate(V), KClO3 (Mr 122.6), is heated: 2KClO3(s) → 2KCl(s) + 3O2(g). 540 cm3 of oxygen is collected at room conditions. What is the percentage yield of oxygen? (molar gas volume 24.0 dm3 mol−1)
- A75.1%
- B113%
- C169%
- D80.4%
10 cm3 of a gaseous hydride of nitrogen decomposes completely into its elements, giving 10 cm3 of nitrogen, N2, and 20 cm3 of hydrogen, H2, all volumes measured at the same temperature and pressure. What is the formula of the hydride?
- AN2H4
- BNH2
- CNH3
- DN2H2
Quiz key
n = 8.00 ÷ 32.0 = 0.250 mol, so V = 0.250 × 24.0 = 6.00 dm3. 12.0 dm3 uses 16.0 as the molar mass, treating oxygen as single atoms. 192 dm3 multiplies the mass in grams by 24.0 without finding moles. 6000 dm3 uses 24 000, which is the molar volume in cm3, with a dm3 unit.
360 cm3 = 0.360 dm3; n = 0.360 ÷ 24.0 = 0.0150 mol; mass = 0.0150 × 44.0 = 0.660 g. 660 g divides 360 by 24.0 without converting cm3 to dm3, a factor of 1000 too large. 0.707 g uses 22.4 dm3 mol−1, the molar volume at 0 °C, not at room conditions. 0.0150 g is the amount in moles written as a mass.
40 cm3 of H2 needs only 20 cm3 of O2, so hydrogen is used up and 20 cm3 of oxygen is left. The water condenses to a liquid on cooling, with negligible volume, so 20 cm3 of gas remains. 60 cm3 counts 40 cm3 of steam as gas as well. 0 cm3 assumes both gases react completely. 40 cm3 assumes hydrogen and oxygen react 1 : 1 to give 40 cm3 of water, then counts that water as a gas.
n(Na) = 0.460 ÷ 23.0 = 0.0200 mol. Na : H2 = 2 : 1, so n(H2) = 0.0100 mol and V = 0.0100 × 24 000 = 240 cm3. 480 cm3 ignores the 2 : 1 ratio. 120 cm3 applies it twice (halving 0.0100 again). 0.240 cm3 is the correct number in dm3 written with the wrong unit.
By Avogadro's law, each sample holds the same amount in moles, so the heaviest molecules give the greatest mass: CO2 (Mr 44.0) beats O2 (32.0) and CH4 (16.0). Equal volume means equal numbers of molecules, not equal masses, so the last option confuses the two.
10 cm3 of ethene uses 30 cm3 of oxygen and forms 20 cm3 of CO2. Oxygen left = 50 − 30 = 20 cm3. Water is liquid, so the final gas is 20 + 20 = 40 cm3. 60 cm3 also counts 20 cm3 of water as a gas. 20 cm3 counts only one of the two gases, either the CO2 or the leftover oxygen. 70 cm3 adds the CO2 to all 50 cm3 of oxygen, forgetting that 30 cm3 was used up.
n(KClO3) = 2.45 ÷ 122.6 = 0.0200 mol; n(O2) = 0.0200 × 3/2 = 0.0300 mol; theoretical V = 0.0300 × 24 000 = 719 cm3. Yield = 540 ÷ 719 × 100 = 75.1%. 113% uses a 1 : 1 ratio (theoretical 480 cm3), and a yield over 100% should signal an error. 169% uses the ratio upside down (2/3 instead of 3/2). 80.4% uses 22.4 dm3 mol−1, which applies at 0 °C, not room conditions.
The volume ratio hydride : N2 : H2 = 10 : 10 : 20 = 1 : 1 : 2, so one molecule gives one N2 and two H2: 2 N atoms and 4 H atoms, N2H4 (hydrazine). NH2 has the right ratio of atoms but is only the empirical formula; it would give half a volume of N2 per volume of hydride. NH3 would give N2 and H2 in the ratio 1 : 3. N2H2 counts each volume of H2 as one H atom, forgetting that hydrogen is diatomic.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.