IB DP Chemistry Topic 1.3 Reacting masses and volumes HL Paper 2

Question

2.67 g of manganese(IV) oxide was added to 200.0 cm3 of 2.00 mol dm3 HCl.

MnO2 (s) + 4HCl (aq) → Cl2 (g) + 2H2O (l) + MnCl2 (aq)

(i) Calculate the amount, in mol, of manganese(IV) oxide added.                                   [1]

(ii) Determine the limiting reactant, showing your calculations.                                      [2]

(iii) Determine the excess amount, in mol, of the other reactant.                                    [1]

(iv) Calculate the volume of chlorine, in dm3, produced if the reaction is conducted at standard temperature and pressure (STP). Use section 2 of the data booklet. [1]

Answer/Explanation

Ans

 i « \frac{2.69g}{86.94mol^-^1} » 0.0307 «mol»   

ii

«nHCl = 2.00 mol dm-3 x 0.2000 dm3» = 0.400 mol 

« \frac{0.400}{4} =» 0.100 mol AND MnO2 is the limiting reactant 

Accept other valid methods of determining the limiting reactant in M2.

iii

«0.0307 mol × 4 = 0.123 mol»

«0.400 mol – 0.123 mol =» 0.277 «mol» 

iv «0.0307 mol × 22.7 dm3 mol−1 =» 0.697 «dm3»  Accept methods employing pV = nRT.

Question

Limestone can be converted into a variety of useful commercial products through the lime cycle. Limestone contains high percentages of calcium carbonate, CaCO3.

Calcium carbonate is heated to produce calcium oxide, CaO.

CaCO3 (s) → CaO (s) + CO2 (g)

Calculate the volume of carbon dioxide produced at STP when 555 g of calcium carbonate decomposes. Use sections 2 and 6 of the data booklet. [2]

Answer/Explanation

Ans:

 «nCaCO3 = \tfrac{555g}{100.09mol^-^1} =» 5.55 «mol» ✓ «V = 5.55 mol × 22.7 dm3 mol−1 =» 126 «dm3»

✓Award [2] for correct final answer.Accept method using pV = nRT toobtain the volume with p as either 100kPa (126 dm3) or 101.3 kPa (125 dm3).

Do not penalize use of 22.4 dm3 mol-1 toobtain the volume (124 dm3)

Question

The percentage of iron(II) ions, \({\text{F}}{{\text{e}}^{2 + }}\), in a vitamin tablet can be estimated by dissolving the tablet in dilute sulfuric acid and titrating with standard potassium manganate(VII) solution, \({\text{KMn}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\). During the process iron(II) is oxidized to iron(III) and the manganate(VII) ion is reduced to the manganese(II) ion, \({\text{M}}{{\text{n}}^{2 + }}{\text{(aq)}}\). It was found that one tablet with a mass of 1.43 g required \({\text{11.6 c}}{{\text{m}}^{\text{3}}}\) of \(2.00 \times {10^{ – 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) \({\text{KMn}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\) to reach the end-point.

State the half-equation for the oxidation of the iron(II) ions.

[1]
a.i.

State the half-equation for the reduction of the \({\text{MnO}}_4^ – \) ions in acidic solution.

[1]
a.ii.

Deduce the overall redox equation for the reaction.

[1]
a.iii.

Calculate the amount, in moles, of \({\text{MnO}}_4^ – \) ions present in \({\text{11.6 c}}{{\text{m}}^{\text{3}}}\) of \(2.00 \times {10^{ – 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) \({\text{KMn}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\).

[1]
b.i.

Calculate the amount, in moles, of \({\text{F}}{{\text{e}}^{2 + }}\) ions present in the vitamin tablet.

[1]
b.ii.

Determine the percentage by mass of \({\text{F}}{{\text{e}}^{2 + }}\) ions present in the vitamin tablet.

[2]
b.iii.
Answer/Explanation

Markscheme

\({\text{F}}{{\text{e}}^{2 + }} \to {\text{F}}{{\text{e}}^{3 + }} + {{\text{e}}^ – }\);

a.i.

\({\text{MnO}}_4^ –  + {\text{8}}{{\text{H}}^ + } + {\text{5}}{{\text{e}}^ – } \to {\text{M}}{{\text{n}}^{2 + }} + {\text{4}}{{\text{H}}_2}{\text{O}}\);

a.ii.

\({\text{MnO}}_4^ –  + {\text{5F}}{{\text{e}}^{2 + }} + {\text{8}}{{\text{H}}^ + } \to {\text{M}}{{\text{n}}^{2 + }} + {\text{5F}}{{\text{e}}^{3 + }} + {\text{4}}{{\text{H}}_2}{\text{O}}\);

Accept e instead of e.

a.iii.

\({\text{amount of MnO}}_4^ –  = \frac{{11.6}}{{1000}} \times 0.0200 = 2.32 \times {10^{ – 4}}{\text{ mol}}\);

b.i.

\({\text{amount of F}}{{\text{e}}^{2 + }} = 5 \times 2.32 \times {10^{ – 4}} = 1.16 \times {10^{ – 3}}{\text{ mol}}\);

b.ii.

\({\text{mass of F}}{{\text{e}}^{2 + }} = 55.85 \times 1.16 \times {10^{ – 3}} = 6.48 \times {10^{ – 2}}{\text{ g}}\);

\({\text{percentage of F}}{{\text{e}}^{2 + }}{\text{ in tablet}} = \frac{{6.48 \times {{10}^{ – 2}}}}{{1.43}} = 100 = 4.53\% \);

b.iii.

Examiners report

This question was generally well answered. A common mistake with writing half-equations was the failure to realise that only single arrows should be used if oxidation and reduction are specifically asked for. Candidates were only penalized once for this error.

a.i.

Given that the half-equation involving \({\text{MnO}}_4^ – \) ions is provided in the Data Booklet, it was surprising that several candidates could not correctly write the equation for their reduction in acidic solution.

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