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.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Key facts
- The four mole formulas (n = m/MM, N = n×Nₐ, V = n×Vₘ, c = n/V)
- Purity comes BEFORE stoichiometry; yield comes AFTER
- STP molar volume = 22.71 L/mol (NESA); RTP = 24.8 L/mol
- The mole (n) is the universal bridge for every calculation
Concepts
- Why the same calculation pathway works across all three IQs
- How to track 5–7 step problems without losing the thread
- When to apply purity vs % yield in chained problems
Skills
- Chain concentration → stoichiometry → mass or gas volume in one problem
- Apply purity before stoichiometry to find theoretical yield
- Solve standard solution → titration back-calc → purity problems
Module 2 is built around three inquiry questions. Each one has its own toolkit, but they share a common centre: the mole. Every formula and every calculation eventually passes through n.
- Chemical Reactions & Stoichiometry
- Mole ratios
- Limiting reagent
- % yield & % purity
- Gas & solution stoich
- Mole Concept
- n = m ÷ MM
- N = n × Nₐ
- V = n × molar vol
- Empirical formula
- Concentration & Analysis
- c = n ÷ V
- c₁V₁ = c₂V₂
- Primary standards
- Titration & gravimetric
Module 2 has three inquiry questions: IQ1 (stoichiometry + yield/purity), IQ2 (mole formulas), IQ3 (concentration + analysis). The mole (n) is the universal bridge — every calculation passes through n. Four core formulas: n = m ÷ MM; N = n × NA; V = n × Vm; n = c × V. In chained problems: convert known quantity → n → apply ratio → n(target) → convert back.
Pause — copy the highlighted summary into your book before moving on.
Did you get this? True or false: every Module 2 calculation passes through the moles (n) of at least one substance as an intermediate step.
We just saw the complete map of Module 2 formulas, with n as the central hub. That raises a question: when purity and yield are both involved in the same problem, does the order you apply them matter? This card answers it → purity must come before stoichiometry; yield comes after.
In chained problems, the order in which you apply purity and yield decides whether your answer is right or off by a fixed factor. The rule is simple:
So a complete chained problem with an impure reactant looks like: m(sample) → ×purity → m(pure) → n → ratio → n(product) → m(theoretical) → compare with m(actual) → % yield.
Purity is applied BEFORE stoichiometry: m(pure) = m(sample) × (% purity ÷ 100) — use m(pure) in the moles step. % yield is applied AFTER calculating theoretical yield: % yield = (actual ÷ theoretical) × 100. Applying them in the wrong order overestimates or underestimates theoretical yield by a constant factor. Also: convert mL → L before c = n ÷ V; use Vm = 24.8 L mol⁻¹ at SATP (NESA default).
Pause — copy the highlighted rules into your book before moving on.
Quick check: A student applies purity AFTER calculating the theoretical yield from the impure sample mass. What happens to their final answer for theoretical yield?
Fill the blanks: drag each token into the matching blank.
Apply ___ before converting to ___. Apply ___ at the end, after you have calculated the ___ yield.
Chain all Module 2 skills in exam-style problems.
Worked examples · chained exam-style problems
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. (a) Calculate n(H₂SO₄). (b) Find n(H₂). (c) Calculate V(H₂) at RTP. (d) Calculate the mass of Zn consumed.
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. (a) Calculate the mass of pure Fe₂O₃. (b) Calculate the theoretical yield of Fe. (Fe = 55.845, O = 15.999). (c) Actual yield = 245 g. Calculate % yield. (d) Give one reason actual yield is less than theoretical.
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). (a) Find c(NaOH). (b) Find m(pure NaOH) in 250 mL. (c) Calculate % purity.
Match each Module 2 quantity to the formula or relationship that defines it.
- Moles from mass
- Moles from gas volume at STP (0 °C, 100 kPa)
- Moles from solution concentration
- Percentage yield
- n = c × V (V in litres)
- n = m ÷ MM
- (actual ÷ theoretical) × 100
- n = V ÷ 22.71 (V in litres)
Sort the steps — given an impure reactant mass and asked for the actual mass of product (factoring in % purity and % yield), put the chain in order. Click two steps to swap them.
- n(reactant) = m(pure reactant) ÷ MM(reactant)
- Write the balanced equation and note the mole ratio
- Apply % yield: actual m(product) = theoretical m(product) × (% yield ÷ 100)
- Apply % purity FIRST: m(pure reactant) = m(sample) × (% purity ÷ 100)
- Use mole ratio → n(product), then m(product) = n × MM (theoretical yield)
Common errors · the 3 traps that cost marks
Applying purity AFTER stoichiometry
If you use the impure sample mass to calculate moles, your n is too large — and so is your theoretical yield. The whole chain is shifted upward.
Fix: Always m(pure) = m(sample) × (purity ÷ 100) FIRST, then convert to moles.
Forgetting mL → L in concentration calculations
n = c × V requires V in litres. Using 35 instead of 0.035 gives an answer 1000× too large. This happens every year in HSC exams.
Fix: Write "V = ___ mL = ___ L" before any calculation. Force the conversion to be visible.
Skipping the balanced equation
The mole ratio lives in the coefficients. Without writing the equation, you can't extract it — and 1:1 isn't always right (H₂SO₄ + 2NaOH, Fe₂O₃ → 2Fe, Na₂CO₃ + 2HCl are all non-1:1).
Fix: Write the balanced equation at the top of every chained problem. Circle the coefficients you'll use.
Quick-fire practice · 5 reps +2 XP per reveal
An ore sample is 65.0% FeS₂ by mass. Find the mass of pure FeS₂ in 200.0 g of ore.
50.0 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. Identify the limiting reagent.
1.06 g of impure Na₂CO₃ (92.0% pure, MM = 105.99) reacts with excess HCl: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. Calculate V(CO₂) at STP (22.71 L/mol).
The actual mass of Fe collected is 148 g, when the theoretical yield was 178.3 g. Calculate % yield.
0.0250 mol of NaOH reacts with 0.0250 mol of HCl (1:1). Theoretical m(NaCl) = 1.46 g; actual m(NaCl) = 1.38 g. Calculate % yield.
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.
Pick your answer, then rate your confidence — that tells the system what to drill next.
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.
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)
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)
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)
📖 Comprehensive answers (click to reveal)
Multiple choice — drill bank
MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.
Short answer 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 × 22.71 = 0.209 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 (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) = 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.
Five timed questions chaining all three inquiry questions. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaClimb platforms, hit checkpoints, and answer chained Module 2 questions. Quick recall, lighter than the boss.
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