IB Chemistry SL Reactivity 2.1 Amount of Chemical Change

Practise using balanced equations as quantitative mole-ratio maps. Convert masses, concentrations or gas volumes to amount, identify the limiting reactant, calculate theoretical…

Syllabus
First assessment 2025
Topic
2.1
Level
SL

Exam points

  • convert each given quantity to moles before applying the balanced-equation ratio
  • identify the limiting reactant and use it to calculate the theoretical product amount
  • calculate percentage yield or atom economy and distinguish what each efficiency measures

Covered syllabus points

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

2.1 Amount of chemical change 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.

2.1 Amount of chemical change 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.

2.1 Amount of chemical change 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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