Chemistry · Year 11 · Module 2 · Lesson 15
HSC Exam Practice
Gas Stoichiometry
Short answer
1.Short answer
Define molar volume and state the two values used in NSW HSC Chemistry calculations. In your answer, identify the temperature and pressure conditions associated with each value.
State Avogadro’s law and explain how it justifies using the volume ratio shortcut (volume ratio = coefficient ratio) for all-gas reactions at constant temperature and pressure.
Calculate the volume of CO2 produced at SATP when 25.0 g of CaCO3 completely decomposes: CaCO3(s) → CaO(s) + CO2(g). Show all working. (Ca = 40.078, C = 12.011, O = 15.999)
What mass of zinc is required to produce 3.72 L of H2(g) at SATP? Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g). Show full working. (Zn = 65.38)
In the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), 9.00 L of H2 is completely consumed at constant temperature and pressure. Calculate the volume of NH3 produced and identify the method you used. Explain why this method is valid.
A student calculates the volume of H2 produced from 6.54 g of Zn at SATP but uses the STP molar volume (22.71 L mol−1) by mistake. Calculate the student’s incorrect answer and the correct answer, then determine the percentage error introduced. (Zn = 65.38)
Data response
2.Data response — hydrogen peroxide decomposition
A student decomposes hydrogen peroxide solution using manganese dioxide as a catalyst: 2H2O2(aq) → 2H2O(l) + O2(g). They measure the cumulative volume of O2 collected in a gas syringe at SATP. The table below shows the data.
| Time (s) | Volume O2 collected (mL) | Rate (mL s−1) |
|---|---|---|
| 0 | 0 | — |
| 30 | 18.4 | |
| 60 | 33.1 | |
| 90 | 44.6 | |
| 120 | 52.8 | |
| 150 | 57.9 | |
| 180 | 60.0 |
(a) Calculate the average rate of gas production (mL s−1) for each 30 s interval and complete the Rate column in the table. Describe how the rate changes over time and suggest a reason. (3 marks)
(b) Using the final volume of O2 collected (60.0 mL at SATP), calculate: (i) the moles of O2 produced; (ii) the moles of H2O2 that decomposed; (iii) the mass of H2O2 that decomposed. Show all working. (4 marks)
(c) If the same experiment were performed at STP instead of SATP with the same amount of H2O2, predict whether the final O2 volume would be greater, less or the same. Calculate the expected STP volume. (2 marks)
Extended response
3.Extended response
Evaluate the claim that “using the wrong molar volume in a gas stoichiometry calculation is only a minor mistake because the difference between 22.71 L mol−1 and 24.8 L mol−1 is small.” In your response you must:
- Calculate a specific worked example showing the correct and incorrect answers when the wrong molar volume is used.
- Quantify the percentage error and explain whether this is significant in an exam context.
- Explain, using particle theory and Avogadro’s law, why the two molar volumes differ.
- Describe a strategy a student should use before every gas stoichiometry calculation to avoid this error.
- Refer to the consequences of this error in an industrial gas measurement context (e.g. a natural gas processing plant).
Chemistry · Year 11 · Module 2 · Lesson 15
Answer Key & Marking Guidelines
Section 1 · Short answer · 3 marks · Band 3
Sample response. Molar volume (Vm) is the volume occupied by one mole of any ideal gas under specified conditions of temperature and pressure. In NSW HSC Chemistry, two values are used: (1) 22.71 L mol−1 at STP (Standard Temperature and Pressure: 0 °C / 273 K, 100 kPa); (2) 24.8 L mol−1 at SATP (Standard Ambient Temperature and Pressure, also called RTP: 25 °C / 298 K, 100 kPa). The SATP value is used in most NSW HSC calculations; STP is used when the question specifies 0 °C or “standard conditions”.
Marking notes. 1 mark for a correct definition of molar volume (volume of 1 mol of any gas at specified T and P). 1 mark for correctly stating both values with their associated temperatures (22.71 L mol−1 at 0 °C / STP; 24.8 L mol−1 at 25 °C / SATP). 1 mark for stating both pressures are 100 kPa and correctly identifying which is the NESA default.
Section 1 · Short answer · 3 marks · Band 3
Sample response. Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. Therefore, at constant T and P, one mole of any gas occupies the same volume. As a result, the ratio of volumes of two gases in a reaction equals the ratio of their moles, which in turn equals the ratio of their stoichiometric coefficients from the balanced equation. This means that for any all-gas reaction at constant T and P, VA / VB = coeffA / coeffB, and no molar volume conversion is needed — the volume ratio can be applied directly.
Marking notes. 1 mark for a correct statement of Avogadro’s law (equal volumes, same T and P, equal molecules). 1 mark for the logical link: mole ratio = coefficient ratio; and at constant T/P, moles ∝ volume, so volume ratio = coefficient ratio. 1 mark for explicitly stating the condition that all species must be gases at the same T and P for the shortcut to be valid.
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. MM(CaCO3) = 40.078 + 12.011 + 3 × 15.999 = 100.09 g mol−1. n(CaCO3) = 25.0 ÷ 100.09 = 0.2498 mol [Step 2]. Mole ratio CaCO3:CO2 = 1:1; n(CO2) = 0.2498 mol [Step 3]. V(CO2) = 0.2498 × 24.8 = 6.19 L at SATP [Step 5].
Marking notes. 1 mark for correct n(CaCO3) = 0.2498 mol. 1 mark for correct mole ratio and n(CO2) = 0.2498 mol. 1 mark for correct V = 0.2498 × 24.8 = 6.19 L with SATP molar volume stated. Deduct 1 mark if 22.71 used instead of 24.8 but all other working is correct (consequential error).
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. Step 0: n(H2) = 3.72 ÷ 24.8 = 0.1500 mol [gas volume to moles]. Mole ratio Zn:H2 = 1:1; n(Zn) = 0.1500 mol. m(Zn) = 0.1500 × 65.38 = 9.81 g.
Marking notes. 1 mark for correct n(H2) = 0.1500 mol (using 24.8 L mol−1). 1 mark for correct n(Zn) = 0.1500 mol via 1:1 ratio. 1 mark for correct m(Zn) = 9.81 g.
Section 1 · Short answer · 3 marks · Band 4
Sample response. Method: volume ratio shortcut (all species are gases at constant T and P). Ratio H2:NH3 = 3:2; V(NH3) = 9.00 × (2 ÷ 3) = 6.00 L. This is valid because all three species (N2, H2, NH3) are gases at the same temperature and pressure; by Avogadro’s law, equal volumes contain equal moles, so the volume ratio equals the coefficient ratio directly. No molar volume conversion is needed.
Marking notes. 1 mark for correctly identifying the volume ratio shortcut as the appropriate method. 1 mark for correct answer V(NH3) = 6.00 L. 1 mark for valid justification referencing Avogadro’s law and the condition that all species are gases at the same T and P.
Section 1 · Short answer · 4 marks · Band 4
Sample response. n(Zn) = 6.54 ÷ 65.38 = 0.1000 mol; ratio 1:1; n(H2) = 0.1000 mol. Correct SATP answer: V = 0.1000 × 24.8 = 2.48 L. Student’s incorrect STP answer: V = 0.1000 × 22.71 = 2.27 L. Percentage error = (2.48 − 2.27) ÷ 2.48 × 100 = 8.5%.
Marking notes. 1 mark for correct n(Zn) = n(H2) = 0.1000 mol. 1 mark for correct SATP answer 2.48 L. 1 mark for correct incorrect STP answer 2.27 L. 1 mark for correct % error = 8.5% (accept 8.4–8.5%).
Section 2 · Data response · 9 marks · Band 4–5
Sample response (a) — rates and trend. Rates: 0–30 s = 18.4 ÷ 30 = 0.613 mL s−1; 30–60 s = 14.7 ÷ 30 = 0.490; 60–90 s = 11.5 ÷ 30 = 0.383; 90–120 s = 8.2 ÷ 30 = 0.273; 120–150 s = 5.1 ÷ 30 = 0.170; 150–180 s = 2.1 ÷ 30 = 0.070 mL s−1. The rate decreases steadily over time [1]. This is because as the reaction proceeds, the concentration of H2O2 decreases; fewer reactant particles are available per unit volume, so the frequency of successful collisions (and thus the rate) decreases [1].
Sample response (b) — moles and mass. (i) n(O2) = V ÷ Vm = 0.0600 L ÷ 24.8 = 0.002419 mol [1]. (ii) Ratio H2O2:O2 = 2:1; n(H2O2) = 0.002419 × 2 = 0.004839 mol [1]. (iii) m(H2O2) = 0.004839 × 34.015 = 0.1646 g ≈ 0.165 g [1].
Sample response (c) — STP prediction. The STP volume would be less [1]. Same n(O2) = 0.002419 mol; V(STP) = 0.002419 × 22.71 = 0.05494 L = 54.9 mL [1]. Lower than SATP because 0 °C < 25 °C; gas molecules have less kinetic energy at lower temperature and occupy a smaller molar volume.
Marking notes. Part (a): 1 mark for all six rates correctly calculated; 1 mark for describing decreasing rate trend; 1 mark for correct reason (decreasing H2O2 concentration → fewer collisions). Part (b): 1 mark each for n(O2), n(H2O2), m(H2O2); 1 mark for correct mole ratio application. Part (c): 1 mark for correct prediction (less) with correct V = 54.9 mL; 1 mark for explanation referencing lower temperature → smaller molar volume.
Section 3 · Extended response · 7 marks · Band 5–6
Sample response. The claim is incorrect. While the absolute difference between the two molar volumes appears small (24.8 − 22.71 = 2.09 L mol−1), the relative difference is (2.09 ÷ 22.71) × 100 = 9.2%, which is significant in both exam and industrial contexts [1 — % error quantified]. Worked example: Consider producing H2 from 6.54 g Zn: n(Zn) = 0.1000 mol; n(H2) = 0.1000 mol. Correct SATP answer: V = 0.1000 × 24.8 = 2.48 L. Using STP by mistake: V = 0.1000 × 22.71 = 2.27 L. The student loses the final mark (wrong numerical answer) even if all steps before are correct [1 — worked example showing exam consequence]. In a 3-mark calculation, losing the answer mark for a systematic error that carries through is a meaningful penalty [1 — exam significance explained]. Particle theory explanation: At a higher temperature (25 °C vs 0 °C), gas molecules have greater average kinetic energy; they move faster, collide with container walls more forcefully, and require more space per mole to maintain the same pressure (100 kPa in both cases). By Avogadro’s law, one mole of any gas at the same T and P occupies the same volume; but since SATP is at a higher temperature than STP, the molar volume at SATP is larger [1 — particle theory + Avogadro’s law correctly referenced]. Prevention strategy: Before writing any working, underline the temperature condition stated in the question; write “SATP → 24.8 L mol−1” or “STP → 22.71 L mol−1” at the top of the working space. This forces the correct selection before calculations begin and prevents the error from propagating through multiple steps [1 — specific and actionable strategy]. Industrial context: At a natural gas processing plant, gas volumes are measured to calculate billing, storage requirements, and safety pressure limits. A 9.2% error in a gas volume measurement could mean under-ordering storage tanks (safety hazard if tanks overfill), over-billing customers for gas delivered, or calibrating pressure-relief valves incorrectly — all with real safety and financial consequences at the scale of millions of cubic metres processed per day [1 — industrial consequence named and explained]. Ultimately, the claim is false: a 9.2% error is neither minor in an exam nor in industry. The correct approach is to always read the conditions before selecting a molar volume [1 — clear evaluative conclusion].
Marking criteria (7 marks). 1 = quantifies % difference as approximately 9.2% and states this is significant. 1 = correct worked example with both correct and incorrect answers shown. 1 = explains exam consequence (mark lost for wrong numerical answer, even with correct method). 1 = explains difference in molar volumes using particle theory at different temperatures and Avogadro’s law. 1 = identifies a specific, actionable prevention strategy. 1 = identifies a plausible industrial consequence and explains why it matters. 1 = reaches an explicit evaluative judgement (claim is incorrect / error is not minor) integrated across both exam and industrial contexts.