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IB Chemistry SL 2.1 Amount of Chemical Change Questions

Practise IB Chemistry SL 2.1 with questions that balance equations, calculate reacting quantities, identify limiting reactants and evaluate yield and atom economy.

Syllabus
First assessment 2025
Course
Chemistry SL
Level
SL

Exam points

  • Balance equations with state symbols, then use mole ratios to calculate masses, concentrations and gas volumes.
  • Identify limiting reactants and calculate theoretical, experimental and percentage yields from data.
  • Calculate atom economy and link a low value to waste in a green-chemistry judgement.

Covered syllabus points

2.1.1—Chemical equations2.1.2—Mole ratio2.1.3—Limiting reactant2.1.4—Percentage yield2.1.5—Atom economy

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Study 2.1 Amount of chemical change conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

Question 1

[Maximum number: 1]

Iron may be extracted from iron (II) sulfide, FeS.

The first step in the extraction of iron from iron (II) sulfide is to roast it in air to form iron (III) oxide and sulfur dioxide.

Write the equation for this reaction.

Question 2

[Maximum number: 2]

The reaction between ethanoic acid and ethanol is homogeneous and reversible.

C2H6O(l)+C2H4O2(l)⇌C4H8O2(l)+H2O(l)ΔHθr=−4 kJ mol−1\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(\mathrm{l})+\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}(\mathrm{l}) \rightleftharpoons \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}(\mathrm{l})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\theta}{ }_{\mathrm{r}}=-4 \mathrm{~kJ} \mathrm{~mol}^{-1}

0.6\quad 0.6 moles each of ethanol and ethanoic acid at 60∘C60^{\circ} \mathrm{C} reacted in the presence of an acid catalyst. The volume remained constant. At equilibrium 0.2 moles of ethanoic acid remained in the reaction mixture.

Calculate the amounts, in mol, of ethanol, ethyl ethanoate and water at equilibrium.

Question 3

[Maximum number: 3]

Chlorine undergoes many reactions.

Question (a)

(a)

2.67 g of manganese(IV) oxide was added to 200.0 cm3200.0 \mathrm{~cm}^{3} of 2.00moldmm−3HCl2.00 \mathrm{moldm} \mathrm{m}^{-3} \mathrm{HCl}.

MnO2( s)+4HCl(aq)→Cl2( g)+2H2O(l)+MnCl2(aq)\mathrm{MnO}_{2}(\mathrm{~s})+4 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{MnCl}_{2}(\mathrm{aq})
[ 3 ]

Question (i)

(i)

Determine the limiting reactant, showing your calculations.

[ 2 ]

Question (ii)

(ii)

Determine the excess amount, in mol , of the other reactant.

[ 1 ]
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