Chemistry • Year 11 • Module 2 • Lesson 15
Gas Stoichiometry
Lock in the molar volume values, the modified 4-step pathway and the conditions that determine which value to use before tackling harder questions.
1. Term–definition match
The definitions below are shuffled. Write the matching term from this list in the right-hand column: molar volume, STP, SATP, n = V ÷ Vm, V = n × Vm, Avogadro’s law, gas stoichiometry, mole ratio, RTP, ideal gas. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The volume occupied by one mole of any gas under stated temperature and pressure conditions. | |
| 1.2 | Standard Temperature and Pressure: 0 °C (273 K) and 100 kPa; molar volume = 22.71 L mol−1 (NESA standard). | |
| 1.3 | Standard Ambient Temperature and Pressure: 25 °C (298 K) and 100 kPa; molar volume = 24.8 L mol−1. | |
| 1.4 | The formula used to convert a gas volume into moles when temperature and pressure are specified. | |
| 1.5 | The formula used to convert moles of gas into a volume at specified conditions. | |
| 1.6 | The principle that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. | |
| 1.7 | A stoichiometry calculation in which at least one reactant or product is a gas, requiring an extra conversion step using molar volume. | |
| 1.8 | The ratio of moles of one species to moles of another, read directly from the balanced equation coefficients. | |
| 1.9 | Commonly called “room temperature and pressure” — 25 °C, 100 kPa — used in most NSW HSC calculations. | |
| 1.10 | A theoretical gas that follows Avogadro’s law exactly; one mole of any ideal gas occupies the same volume (the molar volume) under the same temperature and pressure conditions. |
2. True or false — with correction
Circle T or F. If false, write the corrected statement on the line below. 12 marks (1 T/F + 1 correction each)
2.1 The NESA-standard molar volume at STP (0 °C, 100 kPa) is 22.4 L mol−1. T / F
2.2 At SATP (25 °C, 100 kPa), one mole of CO2 gas occupies 24.8 L. T / F
2.3 When a gas volume is the input (given) in a stoichiometry problem, you should convert it to moles using n = V ÷ Vm before applying the mole ratio. T / F
2.4 The volume ratio shortcut (volume ratio = coefficient ratio) can be applied to any reaction, even when solid or liquid species are involved. T / F
2.5 In the reaction CaCO3(s) → CaO(s) + CO2(g), it is valid to say “1 L of CaCO3 produces 1 L of CO2”. T / F
2.6 A higher temperature means a larger molar volume for a gas, which is why 24.8 L mol−1 (SATP, 25 °C) is greater than 22.71 L mol−1 (STP, 0 °C). T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word or phrase is used once. 8 marks (1 per blank)
Word bank:
22.71 · 24.8 · coefficient · conditions · divide · molar volume · moles · multiply
Gas stoichiometry is the standard 4-step method with one extra conversion step. The key formula is n = V ÷ ___________, which converts a gas volume into ___________. To convert moles back to a gas volume, you ___________ by the molar volume. The value used for the molar volume depends on the stated ___________ in the question. At STP (0 °C, 100 kPa) use ___________ L mol−1; at SATP (25 °C, 100 kPa) use ___________ L mol−1. When all species are gases at the same temperature and pressure, the volume ratio equals the ___________ ratio directly, so you can ___________ by the relevant coefficient fraction without converting to moles first.
4. Function recall
Answer each question in 1–2 sentences using precise chemical language. 8 marks (2 each)
4.1 What does Avogadro’s law state, and why does it mean that all gases have the same molar volume at the same temperature and pressure?
4.2 What is the function of the “Step 0” conversion (n = V ÷ Vm) in a gas stoichiometry problem where a gas volume is given?
4.3 Why is it incorrect to use the “volume ratio shortcut” for the reaction 2KClO3(s) → 2KCl(s) + 3O2(g) to find the volume of O2 from a given mass of KClO3?
4.4 How does the molar volume value differ between STP and SATP, and what physical reason explains this difference?
5. Connect the concepts
Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “determines”, “converts to”, “equals at same T and P”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Terms: molar volume · moles of gas · gas volume · temperature & pressure · coefficient ratio · Avogadro’s law.
6. Conditions checklist — choose the correct molar volume
For each question stem below, write the correct molar volume value (22.71 L mol−1 or 24.8 L mol−1) and the keyword that told you which one to use. 6 marks (1 per row)
| # | Condition stated in the question | Molar volume to use | Keyword / phrase |
|---|---|---|---|
| 6.1 | “Calculate the volume of gas produced at room temperature and pressure.” | ||
| 6.2 | “All measurements taken at 0 °C and 100 kPa.” | ||
| 6.3 | “The gas is collected at standard conditions.” | ||
| 6.4 | “Assume SATP throughout.” | ||
| 6.5 | “The experiment is conducted at 25 °C, 100 kPa.” | ||
| 6.6 | “STP conditions apply.” |
Q1 — Term–definition match
1.1 molar volume • 1.2 STP • 1.3 SATP • 1.4 n = V ÷ Vm • 1.5 V = n × Vm • 1.6 Avogadro’s law • 1.7 gas stoichiometry • 1.8 mole ratio • 1.9 RTP • 1.10 ideal gas.
Marking note: accept RTP or SATP for 1.9 as they are used interchangeably in the NSW HSC context (both 25 °C, 100 kPa).
Q2 — True / false with correction
2.1 False. The NESA-standard molar volume at STP (0 °C, 100 kPa) is 22.71 L mol−1, not 22.4 L mol−1. The 22.4 value corresponds to the older standard (0 °C, 101.325 kPa) and is no longer used by NESA.
2.2 True. At SATP, any ideal gas occupies 24.8 L mol−1; CO2 is no exception.
2.3 True. Converting volume to moles via n = V ÷ Vm is Step 0 — it must be done before applying the mole ratio (Step 3).
2.4 False. The volume ratio shortcut applies only when all species being compared are gases at the same temperature and pressure. It cannot be applied when solids or liquids are involved.
2.5 False. CaCO3 and CaO are solids, not gases; they do not have gas volumes in the Avogadro’s law sense. The shortcut is invalid here. You must use the full 4-step method (mass → moles → ratio → moles of CO2 → volume).
2.6 True. At higher temperature, gas molecules move faster and occupy more space, giving a larger molar volume. Hence 24.8 L mol−1 (25 °C) > 22.71 L mol−1 (0 °C) at the same pressure.
Q3 — Cloze paragraph
In order: molar volume / moles / multiply / conditions / 22.71 / 24.8 / coefficient / divide (the last blank — “divide by the relevant coefficient fraction” means multiplying by the ratio; accept “multiply by” if the student writes it as “multiply by (coefficient of product / coefficient of reactant)”; the intended fill is “divide” matching the word bank word).
Marking note: the eighth blank could reasonably be answered “multiply” since you multiply by the volume ratio; award the mark if the student’s fill is consistent with the word bank and their explanation is correct.
Q4.1 — Avogadro’s law and molar volume
Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. Because every mole of any gas contains the same number of particles (6.022 × 1023), and because at a given T and P each molecule occupies the same average volume, one mole of any gas must occupy the same volume — the molar volume.
Q4.2 — Function of Step 0
Step 0 converts the measured gas volume into moles so that the mole ratio from the balanced equation can be correctly applied. The mole ratio compares moles, not litres; Step 0 bridges the gap between the measurable volume and the chemically meaningful mole count.
Q4.3 — Why the volume ratio shortcut fails for KClO3
The volume ratio shortcut applies only when all species being compared are gases at the same T and P. KClO3 and KCl are solids; they do not have gas volumes in the Avogadro’s law sense. To find V(O2) from a mass of KClO3, you must use the full method: m(KClO3) → n(KClO3) → n(O2) via mole ratio → V(O2).
Q4.4 — STP vs SATP molar volume difference
At STP (0 °C, 100 kPa) the molar volume is 22.71 L mol−1; at SATP (25 °C, 100 kPa) it is 24.8 L mol−1. The higher temperature at SATP gives gas molecules greater average kinetic energy, so they move faster and spread further apart, requiring a larger volume per mole at the same pressure.
Q5 — Sample concept map
Award 1 mark per valid labelled arrow, up to 6. Sample arrows:
- gas volume — divided by → molar volume → gives → moles of gas
- temperature & pressure — determines → molar volume
- Avogadro’s law — explains why → molar volume (is same for all gases)
- Avogadro’s law — justifies use of → coefficient ratio (when all species are gases)
- moles of gas — multiplied by molar volume gives → gas volume
- coefficient ratio — equals volume ratio when → temperature & pressure (are constant for all gas species)
Q6 — Conditions checklist
6.1 24.8 L mol−1 • keyword: “room temperature and pressure”.
6.2 22.71 L mol−1 • keyword: “0 °C and 100 kPa”.
6.3 22.71 L mol−1 • keyword: “standard conditions” (STP).
6.4 24.8 L mol−1 • keyword: “SATP”.
6.5 24.8 L mol−1 • keyword: “25 °C, 100 kPa”.
6.6 22.71 L mol−1 • keyword: “STP”.