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R2.1 Exam Practice

Test your knowledge on Amount of Change

Section B: Data Analysis (Paper 1B Style)

Calculator and Data Booklet permitted. Show all working clearly.

Question 1: Iron & Sulphur Reaction Determine

5 marks

A student reacts iron powder with sulphur powder to synthesise iron(II) sulphide: Fe(s) + S(s) → FeS(s). The following data was recorded:

MeasurementValue
Mass of Fe(s)5.00 g
Mass of S(s)5.00 g
Actual mass of FeS collected7.20 g

(a) Determine the limiting reagent for this reaction. Show your working. [2]

(b) Calculate the percentage yield of iron(II) sulphide. [2]

(c) Explain one reason why the percentage yield is less than 100%. [1]

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(a)

  • n(Fe) = 5.00/55.85 = 0.0895 mol AND n(S) = 5.00/32.07 = 0.156 mol [1]
  • Ratio is 1:1, 0.0895 < 0.156, so Fe is limiting [1]

(b)

  • Theoretical yield = 0.0895 × 87.91 = 7.87 g [1]
  • \% yield = 7.20/7.87 × 100 = 91.5\% [1]

(c) Transfer losses / incomplete reaction / side reactions / impure reagents [1]

Examiner tip: Always use atomic masses from the IB Data Booklet to two decimal places (Fe = 55.85, S = 32.07). Premature rounding causes cascading errors.

Question 2: Atom Economy & Substitution Calculate

5 marks

1-bromopropane (C_3H_7Br) can be synthesised via a substitution reaction: C_3H_8 + Br_2 → C_3H_7Br + HBr

ReagentInitial Amount
Propane (C_3H_8)2.00 mol
Bromine (Br_2)1.50 mol
Actual yield of C_3H_7Br110.00 g

(a) Deduce which reactant is the limiting reagent. [1]

(b) Calculate the percentage yield of 1-bromopropane to three significant figures. [2]

(c) Calculate the atom economy for this reaction and explain why substitution reactions always have an atom economy below 100%. [2]

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(a) Bromine (Br_2) is limiting; 1:1 ratio means 1.50 mol < 2.00 mol [1]

(b)

  • Theoretical mass = 1.50 × 123.00 = 184.5 g [1]
  • \% yield = 110.00/184.5 × 100 = 59.6\% [1]

(c)

  • Atom economy = 123.00/203.90 × 100 = 60.3\% [1]
  • Substitution reactions produce waste by-products (HBr), so not all reactant atoms end up in the desired product [1]
Examiner tip: Do not confuse percentage yield (lab execution efficiency) with atom economy (inherent pathway efficiency from the balanced equation).
Links to: R2.1.3 Atom Economy & Green Chemistry

Section C: Structured Questions (Paper 2 Style)

Show all working. State answers with appropriate significant figures and units.

Question 3: Stoichiometry and the Mole Concept State

4 marks

(a) State the conditions of temperature and pressure for standard temperature and pressure (STP). [1]

(b) Calculate the volume, in dm³, occupied by 0.250 mol of an ideal gas at STP. [1]

(c) A student prepares 250.0 cm³ of a sodium hydroxide (NaOH) solution with a concentration of 0.200 mol dm⁻³. Determine the mass of NaOH required. [2]

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(a) 273.15 K (0 °C) and 100 kPa [1] (Both values required)

(b) V = 0.250 × 22.7 = 5.68 dm³ [1]

(c)

  • n = c × V = 0.200 × 0.2500 = 0.0500 mol [1]
  • m = n × M = 0.0500 × 40.00 = 2.00 g [1]
Examiner tip: Remember to convert cm³ to dm³ by dividing by 1000 before substituting into n = cV. This is one of the most common errors in IB exams.
Links to: R2.1.1 Amounts & Stoichiometry

Question 4: Green Chemistry and Atom Economy Define

4 marks

(a) Define the term atom economy. [1]

(b) Explain why addition reactions always have an atom economy of 100%. [2]

(c) A process has a 100% percentage yield but an atom economy of only 35%. Suggest why this process is not considered sustainable. [1]

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(a) A measure of how efficiently atoms in the reactants are incorporated into the desired product [1]

(b)

  • All reactant atoms join into a single product molecule [1]
  • No by-products are formed [1]

(c) 65% of reactant mass is inherently wasted as by-products, generating chemical waste / increasing environmental harm [1]

Examiner tip: Atom economy is calculated from the balanced equation alone; it tells you nothing about the actual laboratory execution. A reaction can have 100% yield but terrible atom economy.
Links to: R2.1.5 Atom Economy

Question 5: Percentage Yield and Limiting Reagents Deduce

5 marks

Calcium carbonate reacts with hydrochloric acid: CaCO_3(s) + 2HCl(aq) → CaCl_2(aq) + H_2O(l) + CO_2(g)

(a) 10.0 g of CaCO₃ reacts with 200.0 cm³ of 1.50 mol dm⁻³ HCl. Deduce which reagent is limiting. [2]

(b) Calculate the maximum mass of CO₂ that could be produced. [2]

(c) The student collects 3.52 g of CO₂. Calculate the percentage yield. [1]

Show Mark Scheme

(a)

  • n(CaCO_3) = 10.0/100.09 = 0.0999 mol; n(HCl) = 1.50 × 0.2000 = 0.300 mol [1]
  • Ratio is 1:2; CaCO₃ needs 0.200 mol HCl, 0.300 mol available → CaCO₃ is limiting [1]

(b)

  • 1:1 ratio → n(CO_2) = 0.0999 mol [1]
  • m = 0.0999 × 44.01 = 4.40 g [1]

(c) \% yield = 3.52/4.40 × 100 = 80.0\% [1]

Examiner tip: When identifying the limiting reagent with unequal stoichiometry, divide moles by the coefficient; the smallest value identifies the limiting reagent.
Links to: R2.1.2 Limiting & Excess Reagents, R2.1.4 Percentage Yield
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