Chemistry • Year 11 • Module 2 • Lesson 19

Module 2 Synthesis & Exam Practice

Build HSC Band 5–6 extended-response technique on chained stoichiometry, error analysis, and experimental design in quantitative chemistry.

Master · Extended Response

1. Data + scenario: blast-furnace iron production (Band 5–6)

8 marks   Band 5–6

Scenario. Iron ore from Western Australia is charged into a blast furnace with excess coke (carbon). The dominant iron-producing reaction is:

Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

A shipment of 2.50 × 103 kg of ore arrives at the furnace. Assay data show the ore is 81.2% Fe2O3 by mass. The actual mass of iron produced from this charge is 1.02 × 103 kg. Molar masses: Fe = 55.845 g mol−1; O = 15.999 g mol−1; MM(Fe2O3) = 159.69 g mol−1.

Q1. Analyse and evaluate the blast-furnace data above. In your response you must:

  • Calculate the mass of pure Fe2O3 in the 2.50 × 103 kg ore charge.
  • Calculate the theoretical mass of iron that should be produced (show full working).
  • Calculate the percentage yield of iron and interpret what this means for the industrial process.
  • Identify two reasons why the actual yield is less than theoretical; classify each as relating to reaction completeness or product recovery.
  • Explain whether it would ever be possible to achieve a percentage yield greater than 100%, and why not.
Plan: (1) m(Fe2O3) = 2500 kg × 0.812; (2) n(Fe2O3) = m ÷ 159.69; ratio 1:2 → n(Fe) = 2×n(Fe2O3); m(Fe) theoretical = n(Fe)×55.845; (3) % yield = (1020 ÷ theoretical)×100%; (4) reaction completeness vs product recovery; (5) yield >100% impossible because theoretical = maximum if all Fe2O3 reacts.

2. Experimental design — determining the purity of a commercial HCl sample (Band 5–6)

7 marks   Band 5–6

Research question. A supplier claims their hydrochloric acid has a concentration of exactly 1.000 mol L−1. Design a titration-based investigation to verify this claim and calculate the actual concentration.

Constraints: You have access to: anhydrous Na2CO3 (primary standard; MM = 105.99 g mol−1), a 250 mL volumetric flask, a 25 mL pipette, a 50 mL burette, methyl orange indicator, and standard Year 11 glassware. The investigation must be completed in a single two-hour laboratory session.

Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

Q2. Design the investigation and present it using the structure below.

  • State your hypothesis (a testable prediction linking the stated concentration to a measured titre volume).
  • Identify the independent variable, dependent variable, and at least two controlled variables.
  • Describe the procedure in at least five numbered steps, including the preparation of the Na2CO3 standard solution and how you will determine the average concordant titre.
  • Show how you would use the average titre to calculate c(HCl), including the formula steps.
  • State two limitations of your design and suggest one improvement to increase reliability.
Plan: hypothesis (if c = 1.000 mol L−1, 25.0 mL Na2CO3 of known concentration should require a predictable titre of HCl); prep standard (dissolve exact mass in 250 mL flask); ratio 1:2 (Na2CO3 : HCl); average concordant ≤ 0.10 mL; limitations: indicator error, CO2 bubbling affecting endpoint clarity.
Answers — Do not peek before attempting

Q1 — Sample Band 6 response (8 marks), annotated

Mass of pure Fe2O3: m(Fe2O3) = 2.50 × 103 × 0.812 = 2030 kg = 2.030 × 106 g. [1 mark]

Theoretical mass of iron: n(Fe2O3) = 2.030 × 106 ÷ 159.69 = 1.271 × 104 mol. Ratio Fe2O3 : Fe = 1 : 2; n(Fe) = 2.541 × 104 mol. m(Fe) = 2.541 × 104 × 55.845 = 1.419 × 106 g = 1419 kg. [2 marks: 1 for n(Fe2O3) and ratio; 1 for correct theoretical mass]

Percentage yield: % yield = (1020 ÷ 1419) × 100 = 71.9%. This means only about 72% of the theoretical iron was recovered, which is typical for a large-scale industrial blast furnace where reaction conditions are not perfectly optimised and product recovery is imperfect. [1 mark for % yield; 1 mark for interpretation]

Two reasons yield < 100%: (1) Reaction completeness: The reaction of Fe2O3 with CO may not proceed to completion; some ore may pass through unreduced due to insufficient contact time or temperature variation. (2) Product recovery: Molten iron may be lost as slag inclusions, spilled during tapping, or solidify in the wrong location and not be fully collected. [1 mark each, correctly classified]

Why yield cannot exceed 100%: The theoretical yield represents the maximum mass of product obtainable if all of the limiting reagent is converted to product with no losses. The actual yield can never exceed this because no new atoms are created — conservation of mass guarantees that n(Fe) formed cannot exceed n(Fe) available from the ore. A reported yield > 100% would indicate an error in measurement, calculation or impurity accounting. [1 mark]

Marking criteria (8 marks): 1 = correct m(Fe2O3) with purity applied; 1 = correct n(Fe2O3) and mole ratio applied; 1 = correct theoretical mass of Fe; 1 = correct % yield; 1 = meaningful industrial interpretation of % yield; 1 = valid reaction-completeness reason; 1 = valid product-recovery reason (both correctly classified); 1 = clear explanation of why yield > 100% is impossible, referencing conservation of mass or the definition of theoretical yield.

Q2 — Sample Band 6 response (7 marks), annotated

Hypothesis: If the HCl solution has a concentration of exactly 1.000 mol L−1, then titrating 25.0 mL of a 0.0500 mol L−1 Na2CO3 standard solution (prepared from 1.325 g in 250 mL) against the HCl should require a titre of approximately 25.0 mL. Independent variable: volume of HCl used per titration. Dependent variable: calculated concentration of HCl. Controlled variables: volume of Na2CO3 pipetted per titration (25.0 mL); same indicator batch (methyl orange, 3 drops); same burette throughout. [1 mark]

Procedure:

  1. Accurately weigh 1.325 g of anhydrous Na2CO3 (MM = 105.99; n = 0.01250 mol). Dissolve in ~100 mL distilled water in a 250 mL volumetric flask. Make up to the 250 mL mark with distilled water. Invert 20 times to mix. Label the standard solution: c(Na2CO3) = 0.05000 mol L−1.
  2. Rinse the burette with distilled water then with the HCl solution. Fill the burette with HCl and remove any air bubbles. Record the initial reading.
  3. Pipette 25.0 mL of the Na2CO3 standard solution into a conical flask. Add 3 drops of methyl orange indicator. Perform a rough titration, swirling continuously, until the indicator changes from yellow to orange-red. Record the rough titre.
  4. Refill the burette. Perform two or more careful titrations, adding HCl dropwise near the endpoint. Record all burette readings. Identify concordant titres (within 0.10 mL); average them to obtain V(HCl).
  5. Calculate: n(Na2CO3) = 0.05000 × 0.02500 = 1.250 × 10−3 mol. Ratio 1 : 2; n(HCl) = 2.500 × 10−3 mol. c(HCl) = n ÷ V(HCl in litres).

[1 mark for five clear steps including standard prep and concordant average] [1 mark for showing the full calculation pathway]

Limitations: (1) Methyl orange endpoint with a CO2-producing reaction: CO2 bubbles near the equivalence point make the colour change harder to judge precisely, introducing endpoint error. (2) Weighing Na2CO3 on a balance with ±0.01 g uncertainty introduces a small systematic error in the calculated concentration of the standard. [1 mark each]

Improvement: Repeat the titration at least three times and use only the two closest concordant titres for the average. Alternatively, heat the flask gently near the endpoint to expel CO2 before the final drop, making the endpoint colour change sharper. [1 mark]

Marking criteria (7 marks): 1 = testable hypothesis with predicted titre from stated concentration; 1 = IV, DV and at least two controlled variables correctly identified; 1 = five-step procedure including standard preparation; 1 = correct calculation pathway (n Na2CO3 → ratio → n HCl → c HCl); 1 = one valid limitation; 1 = second valid limitation; 1 = one specific, actionable improvement to reliability.

HSCScience • Chemistry • Year 11 • Module 2 • Lesson 19 • Worksheet 3 (Master)