Year 11 Chemistry Module 2 Full Synthesis ⏱ ~45 min Lesson 19 of 20

Module 2
Synthesis & Exam Practice

Module 2 mastery means solving a 5-step problem without losing the thread. This lesson chains all three inquiry questions together in exam-style problems — the same kind you'll face in your assessment tasks.

🏗️
1

Understand the core concepts covered in this lesson.

2

Apply your knowledge to solve problems and explain phenomena.

3

Evaluate and analyse scientific information and data.

Inquiry Question 1

Chemical Reactions & Stoichiometry

  • Mole ratios
  • Limiting reagent
  • % yield & % purity
  • Gas & solution stoich
Inquiry Question 2

Mole Concept

  • n = m ÷ MM
  • N = n × Nₐ
  • V = n × molar vol
  • Empirical formula
Inquiry Question 3

Concentration & Analysis

  • c = n ÷ V
  • c₁V₁ = c₂V₂
  • Primary standards
  • Titration & gravimetric
MODULE 2 MASTER FORMULA WEB — n (moles) connects everything N = n × Nₐ n = N ÷ Nₐ m = n × MM n = m ÷ MM V = n × Vₘ n = V ÷ Vₘ n = c × V c = n ÷ V n(B) = n(A)×ratio ratio = coeff(B)÷coeff(A) ÷MM → simplest ratio → EF ×n → MF (if MM given) n MOLES N particles / atoms m mass (grams) V (gas) 22.71 L/mol STP c × V concentration n(B) stoich via mole ratio EF / MF empirical formula IQ2: N, m, V, EF IQ3: concentration c×V IQ1: stoichiometry n(A)→n(B)
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Think First

A chemist reacts a 92% pure sample of Na₂CO₃ with HCl to produce CO₂ gas. If she mixes up the order of operations — applying the stoichiometry first, then correcting for purity afterwards — what would happen to her final answer? And how many distinct "steps" do you think are involved in going from a concentration of HCl to a volume of gas produced?

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Chain Problem 1 — Concentration → Stoichiometry → Mass

IQ2 + IQ3 · 5 steps
A student mixes 35.0 mL of 0.400 mol/L H₂SO₄ with excess Zn metal. The reaction produces ZnSO₄ and H₂ gas. Zn + H₂SO₄ → ZnSO₄ + H₂.
  1. Part (a) — 1 mark

    Calculate n(H₂SO₄) in the solution.

  2. Part (b) — 1 mark

    Using the mole ratio, determine n(H₂) produced.

  3. Part (c) — 1 mark

    Calculate the volume of H₂ produced at RTP.

  4. Part (d) — 2 marks

    Calculate the mass of Zn consumed.

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Show Model Answer
(a) V = 35.0 mL = 0.0350 L; n(H₂SO₄) = 0.400 × 0.0350 = 0.01400 mol (b) Ratio 1:1; n(H₂) = 0.01400 mol (c) V(H₂) = 0.01400 × 24.8 = 0.347 L at RTP (d) MM(Zn) = 65.38; n(Zn) = 0.01400; m = 0.01400 × 65.38 = 0.915 g

Chain Problem 2 — Purity → Stoichiometry → % Yield

IQ3 Full Integration · 6 steps
An iron ore sample is 76.0% Fe₂O₃ by mass. 500 g of the ore is reduced in a blast furnace: Fe₂O₃ + 3CO → 2Fe + 3CO₂.
  1. Part (a) — 1 mark

    Calculate the mass of pure Fe₂O₃ in 500 g of ore.

  2. Part (b) — 2 marks

    Calculate the theoretical yield of Fe from this sample. (Fe = 55.845, O = 15.999)

  3. Part (c) — 2 marks

    The actual yield of Fe collected is 245 g. Calculate the percentage yield.

  4. Part (d) — 1 mark

    Give one reason why the actual yield is less than theoretical.

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(a) m(Fe₂O₃) = 500 × 0.760 = 380 g (b) MM(Fe₂O₃) = 2(55.845) + 3(15.999) = 159.69; n = 380 ÷ 159.69 = 2.379 mol Ratio 1:2; n(Fe) = 4.758 mol; MM(Fe) = 55.845; theoretical = 4.758 × 55.845 = 265.8 g (c) % yield = (245 ÷ 265.8) × 100 = 92.2% (d) Any valid: incomplete reduction of ore; some product lost during collection; side reactions forming other iron compounds; impurities reacting with CO.

Chain Problem 3 — Standard Solution → Titration Back-Calc → Purity

IQ2 + IQ3 Advanced · 7 steps
A chemist wants to determine the purity of a commercial NaOH sample. She dissolves 2.40 g of the sample in water and makes it up to 250 mL. 25.0 mL aliquots are titrated against 0.0980 mol/L HCl. Average concordant titre = 23.6 mL. HCl + NaOH → NaCl + H₂O. (Na = 22.990, O = 15.999, H = 1.008)
  1. Part (a) — 2 marks

    Find the concentration of NaOH in the prepared solution.

  2. Part (b) — 2 marks

    Find the mass of pure NaOH in the full 250 mL solution.

  3. Part (c) — 1 mark

    Calculate the percentage purity of the commercial NaOH sample.

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(a) n(HCl) = 0.0980 × 0.0236 = 2.313×10⁻³; ratio 1:1; n(NaOH) = 2.313×10⁻³ c(NaOH) = 2.313×10⁻³ ÷ 0.0250 = 0.09250 mol/L (b) n(NaOH total) = 0.09250 × 0.250 = 0.02313 mol MM(NaOH) = 22.990 + 15.999 + 1.008 = 39.997; m(NaOH) = 0.02313 × 39.997 = 0.9250 g (c) % purity = (0.9250 ÷ 2.40) × 100 = 38.5% Note: The low purity (38.5%) indicates the commercial sample contains significant impurities — this is realistic for technical-grade NaOH.
Key Terms
MoleThe SI unit for amount of substance; contains exactly 6.022 × 10²³ particles.
Avogadro's Number6.022 × 10²³ — the number of particles in one mole of a substance.
Molar MassThe mass of one mole of a substance, measured in g/mol.
Limiting ReagentThe reactant that is completely consumed first, limiting the amount of product formed.
Empirical FormulaThe simplest whole-number ratio of atoms in a compound.
Molecular FormulaThe actual number of atoms of each element in a molecule of a compound.

Misconceptions to Fix

Wrong: The mole is a measure of mass.

Right: The mole is a measure of amount of substance; one mole contains Avogadro's number of particles.

Key Point

Pay close attention to reaction conditions and reagents. These determine the products formed.

Module 2 Competency Checklist

Tick each item when you can do it without referring to notes. Be honest — this is your own study tool.

Inquiry Question 2 — Mole Concept

Calculate moles from mass using n = m ÷ MM
Find number of particles using N = n × Nₐ
Derive empirical formula from % composition
Calculate gas volumes at STP and RTP

Inquiry Question 3 — Concentration & Analysis

Calculate concentration using c = n ÷ V (V in litres)
Dilute a solution using c₁V₁ = c₂V₂
Describe how to prepare a standard solution
Find [unknown] from titration data (back-calc)
Find [unknown] from precipitate mass (gravimetric)

Inquiry Question 1 — Chemical Reactions & Stoichiometry

Balance equations and extract mole ratios
Perform mass–mass stoichiometry (4-step method)
Identify limiting reagent and calculate theoretical yield
Calculate % yield and % purity
Perform gas stoichiometry using molar volumes
Chain concentration + stoichiometry in one problem

Working Scientifically

List properties of a primary standard
Describe systematic vs random error with examples
Identify how a specific error affects the result (higher/lower)
Calculate concordant average titre and explain why rough is excluded

📐 Always Write the Equation

Even if a question seems straightforward, write and balance the equation first. The mole ratio is hidden in the coefficients — you can't access it without the equation.

🔢 Show All Steps

In NSW exams, working is marked. A correct answer with no working gets no part-marks if wrong. Write: given → formula → substitution → answer with units.

📏 Units at Every Step

Write units after every number throughout your calculation. Mol, g/mol, g, L, mol/L. Units tell you when you've made an error — mass ÷ mass gives a dimensionless ratio, not moles.

🔄 Sanity Check

After calculating, ask: is this a reasonable quantity? If you get 1200 g of product from 0.01 mol of reactant, something is wrong. Order-of-magnitude checks catch arithmetic errors.

⚗️ Purity Comes First

If a reactant is impure, correct for purity BEFORE calculating moles. m(pure) = m(sample) × (% purity ÷ 100). Never apply purity after the stoichiometry — it's already too late.

💧 mL → L Every Time

Concentration calculations require volumes in litres. The habit of immediately writing "V = ___ mL = ___ L" before any calculation will eliminate one of the most common error sources.

📝 How are you completing this lesson?

Interactive: Module 2 Mastery Marathon
Revisit — Think First

At the start of this lesson, you thought about what happens when purity is applied in the wrong order, and how many steps are in a chained stoichiometry problem.

Purity must be applied before stoichiometry — you need the mass of pure reactant before you can calculate moles. A full chained problem typically involves 5–7 steps: convert concentration to moles, apply purity if needed, use the mole ratio, convert moles to the required quantity (mass, volume, or concentration). Keeping track of what you know and what you need at each step is the key to solving these problems without losing the thread.

Reflect: how did your initial thinking compare to what you've learned?

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Multi-Step Stoichiometry: The Full Path Given data m, V, c Purity × purity (decimal) Moles n = m/M or cV Ratio balanced eqn Target mass / gas / c Mass stoichiometry m → n → ratio → n → m Solution stoichiometry c × V → n → ratio → target Gas stoichiometry m → n → ratio → n → V (gas)
MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

📝

Full Exam Practice

ApplyBand 3

Q1. A student dissolves 1.06 g of impure Na₂CO₃ (92.0% pure, MM = 105.99) in water and reacts it with excess HCl: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (a) Find the mass of pure Na₂CO₃ in the sample. (b) Calculate the theoretical volume of CO₂ at STP. 4 MARKS

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AnalyseBand 4

Q2. 30.0 mL of 0.300 mol/L Ca(OH)₂ is mixed with 50.0 mL of 0.250 mol/L HCl: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. (a) Identify the limiting reagent. (b) Calculate the mass of CaCl₂ formed. (Ca = 40.078, Cl = 35.453) 5 MARKS

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AnalyseBand 4

Q3. An ore sample (85.0% Fe₂O₃ by mass) is reacted with excess CO in a blast furnace: Fe₂O₃ + 3CO → 2Fe + 3CO₂. The actual mass of Fe collected from 300.0 g of the ore is 148 g. (a) Calculate the theoretical yield of Fe. (b) Calculate the percentage yield. (Fe = 55.845, O = 15.999) 5 MARKS

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EvaluateBand 5

Q4. A chemist calculates that 0.0250 mol of NaOH should react with 0.0250 mol of HCl (1:1 ratio). After mixing 25.0 mL of 1.00 mol/L NaOH with 25.0 mL of 1.00 mol/L HCl, she evaporates the solution and finds 1.38 g of NaCl instead of the theoretical 1.46 g. (a) Calculate the percentage yield of NaCl. (b) The chemist suggests two possible explanations: (i) some NaCl was lost when evaporating the solution, or (ii) the HCl solution was slightly less concentrated than 1.00 mol/L. For each explanation, identify whether it represents a random or systematic error and predict the effect on all future trials. (Na = 22.990, Cl = 35.453) 5 MARKS

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✅ Model Answers

Q1 (4 marks)

(a) m(pure Na₂CO₃) = 1.06 × 0.920 = 0.9752 g (b) n(Na₂CO₃) = 0.9752 ÷ 105.99 = 9.201×10⁻³ mol; ratio 1:1; n(CO₂) = 9.201×10⁻³ V(CO₂) = 9.201×10⁻³ × 22.4 = 0.206 L at STP

Q2 (5 marks)

(a) n(Ca(OH)₂) = 0.300 × 0.0300 = 9.00×10⁻³; ÷ 1 = 9.00×10⁻³ n(HCl) = 0.250 × 0.0500 = 1.25×10⁻²; ÷ 2 = 6.25×10⁻³

HCl has smaller n÷coeff value (6.25×10⁻³ < 9.00×10⁻³) → HCl is the limiting reagent

(b) Ratio HCl:CaCl₂ = 2:1; n(CaCl₂) = 1.25×10⁻² ÷ 2 = 6.25×10⁻³ mol MM(CaCl₂) = 110.98; m = 6.25×10⁻³ × 110.98 = 0.694 g

Q3 (5 marks)

(a) m(Fe₂O₃) = 300.0 × 0.850 = 255.0 g MM(Fe₂O₃) = 2(55.845) + 3(15.999) = 159.69; n(Fe₂O₃) = 255.0 ÷ 159.69 = 1.597 mol Ratio Fe₂O₃:Fe = 1:2; n(Fe) = 1.597 × 2 = 3.193 mol Theoretical m(Fe) = 3.193 × 55.845 = 178.3 g (b) % yield = (148 ÷ 178.3) × 100 = 83.0%

Q4 (5 marks)

(a) MM(NaCl) = 22.990 + 35.453 = 58.443; theoretical m = 0.0250 × 58.443 = 1.461 g % yield = (1.38 ÷ 1.461) × 100 = 94.5%

(b) Explanation (i) — NaCl lost during evaporation: this is a random error if the amount lost varies between trials. It could also be systematic if the student consistently loses the same amount (e.g., always stops heating too early). It would generally give a yield below 100% in every trial, with the magnitude varying randomly.

Explanation (ii) — HCl less concentrated than 1.00 mol/L: this is a systematic error. Every trial uses the same solution; every trial would have the same deficit in HCl moles. The theoretical yield calculated from 0.0250 mol would be wrong in the same way every time — the % yield would be consistently below the true value in a predictable, reproducible way.

Mark lesson as complete

Tick when you've finished all chain problems and the checklist.