Gas Stoichiometry

Q1 — Sample Band 6 response (9 marks), annotated

Calculations — both runs:

For Run A (SATP) and Run B (STP): All species in N₂(g) + 3H₂(g) ⇌ 2NH₃(g) are gases. By Avogadro’s law, at constant temperature and pressure, equal volumes contain equal moles, so the volume ratio equals the coefficient ratio directly — the volume ratio shortcut is valid for both runs [1 — method identified and justified].

H₂:NH₃ coefficient ratio = 3:2. V(NH₃) = 9.00 × (2 ÷ 3) = 6.00 L for both Run A and Run B [1 — correct answer for Run A; 1 — correct answer for Run B].

Difference between runs: The theoretical V(NH₃) is identical (6.00 L) in both cases because the volume ratio shortcut does not depend on the value of the molar volume — it depends only on the coefficient ratio [1]. Absolute difference = 0 L; percentage difference = 0%. The molar volume cancels out when all species are gases at the same T and P: the ratio V(H₂) / V(NH₃) = n(H₂) × V_m / n(NH₃) × V_m = n(H₂) / n(NH₃) regardless of V_m [1 — explanation with algebra or reasoning].

Where the wrong V_m causes an error: If a student mistakenly converts the 9.00 L of H₂ to moles using the wrong molar volume before applying the ratio, then converts back, errors compound. Example: using V_m = 22.71 at SATP → n(H₂) = 9.00 ÷ 22.71 = 0.3965 mol; n(NH₃) = 0.3965 × (2 ÷ 3) = 0.2643 mol; V(NH₃) = 0.2643 × 24.8 = 6.56 L — an error of 0.56 L (9.3%). The industrial consequence is a 9.3% mis-estimate of ammonia produced per batch; at industrial scale this could mean misallocation of storage tanks, incorrect billing of product, or safety risks from pressure build-up in downstream vessels [1 — consequence identified and quantified; 1 — industrial significance explained].

Avogadro’s law explanation: Avogadro’s law states that at constant T and P, equal volumes of all gases contain equal numbers of molecules. Therefore the mole ratio between H₂ and NH₃ is exactly reflected in their volume ratio (3 L H₂ produces 2 L NH₃). The molar volume value (22.71 or 24.8 L mol⁻¹) sets the absolute scale but cancels in the ratio; the ratio is condition-independent as long as all species are at the same conditions [1 — Avogadro’s law correctly referenced].

Marking criteria summary (9 marks): 1 = identifies and justifies volume ratio shortcut. 1 = correct V(NH₃) Run A = 6.00 L. 1 = correct V(NH₃) Run B = 6.00 L. 1 = explains the two answers are equal with mathematical/logical reasoning. 1 = explains how molar volume cancels in the ratio (or equivalent algebraic reasoning). 1 = demonstrates a concrete worked example of the error caused by using the wrong V_m. 1 = quantifies the percentage error (9.3% or equivalent). 1 = identifies at least one industrial consequence. 1 = uses precise chemical vocabulary throughout (Avogadro’s law, molar volume, coefficient ratio, SATP, STP).

Q2 — Sample Band 6 response (8 marks), annotated

Hypothesis: If the molar volume of CO₂ at SATP is 24.8 L mol⁻¹, then when a known mass of CaCO₃ is dissolved in excess HCl at 25 °C and 100 kPa, the ratio V(CO₂) ÷ n(CO₂) will equal 24.8 L mol⁻¹. IV = mass of CaCO₃; DV = volume of CO₂ collected (mL); controlled: concentration of HCl (excess; 1.0 mol L⁻¹), temperature (25 °C; measured with thermometer), pressure (100 kPa; atmospheric) [1 — hypothesis + IV, DV + 2 controlled variables].

Recommended sample mass calculation: n(CO₂) = 0.0500 L ÷ 24.8 L mol⁻¹ = 0.002016 mol; ratio 1:1 → n(CaCO₃) = 0.002016 mol; m(CaCO₃) = 0.002016 × 100.09 = 0.202 g [1 — correct mass with working].

Procedure: (1) Measure and record the laboratory temperature (aim 25 °C) and atmospheric pressure. (2) Accurately weigh 0.202 g of marble chips (CaCO₃) on a balance (±0.01 g). (3) Add 50 mL of 1.0 mol L⁻¹ HCl to the conical flask; connect the gas syringe via the rubber stopper and delivery tube; ensure all joints are airtight. (4) Add the marble chips to the flask and reseal immediately; record the syringe volume every 30 s until no further gas is produced. (5) Read the total volume of CO₂ collected (V, mL) from the syringe when volume is stable. Calculate n(CaCO₃) = m ÷ 100.09; since ratio is 1:1, n(CO₂) = n(CaCO₃); calculate experimental V_m = V(L) ÷ n(CO₂) and compare to 24.8 L mol⁻¹ [1 — five steps; 1 — calculation method described].

Error 1 — CO₂ dissolving in water: CO₂ is slightly soluble in water; some gas dissolves in the HCl solution rather than reaching the syringe. Result: V(CO₂) collected is less than the true value → calculated V_m = V ÷ n is underestimated compared to 24.8 L mol⁻¹ [1 — source identified; 1 — direction of error explained].

Error 2 — Reaction is exothermic; temperature above 25 °C: The reaction of CaCO₃ with HCl releases heat; the solution temperature rises above 25 °C during the reaction. At higher temperatures, gas molecules have more kinetic energy and occupy more volume — V(CO₂) is measured while hot, then the gas cools in the syringe after collection, giving an inflated volume reading. Result: V(CO₂) measured is slightly greater than at true 25 °C → V_m is overestimated [1 — source identified; 1 — direction of error explained].

Marking criteria summary (8 marks): 1 = hypothesis + IV, DV, 2 controlled. 1 = correct mass calculation (0.202 g). 1 = procedure in at least 5 numbered steps. 1 = calculation of V_m described. 1 = source of error 1 (CO₂ dissolution) identified. 1 = direction of error 1 on V_m (underestimate). 1 = source of error 2 (exothermic heating) identified. 1 = direction of error 2 on V_m (overestimate).