IB Chemistry SL 2.1 Amount of Chemical Change Questions

Practise IB Chemistry SL 2.1 by balancing equations, converting reacting amounts with mole ratios, identifying limiting reagents and calculating yield and atom economy.

Syllabus
First assessment 2025
Course
Chemistry SL
Level
SL

Exam points

  • Balance equations from named reactants and products, including molecular formulas and state symbols.
  • Use equation coefficients to convert masses, concentrations or gas volumes into reacting amounts.
  • Identify the limiting reagent from mole amounts and calculate theoretical product and excess remaining.
  • Calculate percentage yield from actual and theoretical product, and explain measured losses.
  • Calculate atom economy from the desired product and total reactants, then compare waste.

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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