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).