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Date May 2021 Marks available 1 Reference code 21M.2.sl.TZ2.1
Level SL Paper 2 Time zone TZ2
Command term Calculate Question number 1 Adapted from N/A

Question

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

The second step of the lime cycle produces calcium hydroxide, Ca(OH)2.

Calcium hydroxide reacts with carbon dioxide to reform calcium carbonate.

Ca(OH)2 (aq) + CO2 (g) → CaCO3 (s) + H2O (l)

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]
a.

Thermodynamic data for the decomposition of calcium carbonate is given.

Calculate the enthalpy change of reaction, ΔH, in kJ, for the decomposition of calcium carbonate.

[2]
b.

The potential energy profile for a reaction is shown. Sketch a dotted line labelled “Catalysed” to indicate the effect of a catalyst.

[1]
c(i).

Outline why a catalyst has such an effect.

[1]
c(ii).

Write the equation for the reaction of Ca(OH)2 (aq) with hydrochloric acid, HCl (aq).

[1]
d(i).

Determine the volume, in dm3, of 0.015 mol dm−3 calcium hydroxide solution needed to neutralize 35.0 cm3 of 0.025 mol dm−3 HCl (aq).

[2]
d(ii).

Saturated calcium hydroxide solution is used to test for carbon dioxide. Calculate the pH of a 2.33 × 10−2 mol dm−3 solution of calcium hydroxide, a strong base.

[2]
d(iii).

Determine the mass, in g, of CaCO3 (s) produced by reacting 2.41 dm3 of 2.33 × 10−2 mol dm−3 of Ca(OH)2 (aq) with 0.750 dm3 of CO2 (g) at STP.

[2]
e(i).

2.85 g of CaCO3 was collected in the experiment in e(i). Calculate the percentage yield of CaCO3.

(If you did not obtain an answer to e(i), use 4.00 g, but this is not the correct value.)

[1]
e(ii).

Outline how one calcium compound in the lime cycle can reduce a problem caused by acid deposition.

[1]
f.

Markscheme

«nCaCO3555g11.09gmol-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 to obtain the volume with p as either 100 kPa (126 dm3) or 101.3 kPa (125 dm3).

Do not penalize use of 22.4 dm3 mol–1 to obtain the volume (124 dm3).

a.

«ΔH =» (−635 «kJ» – 393.5 «kJ») – (−1207 «kJ») ✓

«ΔH = + » 179 «kJ» ✓


Award [2] for correct final answer.

Award [1 max] for −179 kJ.

Ignore an extra step to determine total enthalpy change in kJ: 179 kJ mol−1 x 5.55 mol = 993 kJ.

Award [2] for an answer in the range 990 - 993« kJ».

b.

lower activation energy curve between same reactant and product levels ✓


Accept curve with or without an intermediate.

Accept a horizontal straight line below current line with the activation energy with catalyst/Ecat clearly labelled.

c(i).

provides an alternative «reaction» pathway/mechanism ✓


Do not accept “lower activation energy” only.

c(ii).

Ca(OH)2 (aq) + 2HCl (aq) → 2H2O (l) + CaCl2 (aq) ✓

d(i).

«nHCl = 0.0350 dm3 × 0.025 mol dm−3 =» 0.00088 «mol»
OR
nCa(OH)212 nHCl/0.00044 «mol» ✓

«V12×0.00088mol0.015moldm-3 =» 0.029 «dm3» ✓


Award [2] for correct final answer.

Award [1 max] for 0.058 «dm3».

d(ii).

Alternative 1:

[OH] = « 2 × 2.33 × 10−2 mol dm−3 =» 0.0466 «mol dm−3» ✓

«[H+] = 1.00×10-140.0466 = 2.15 × 10−13 mol dm−3»
pH = « −log(2.15 × 10−13) =» 12.668 ✓

 

Alternative 2:

[OH] =« 2 × 2.33 × 10−2 mol dm−3 =» 0.0466 «mol dm−3» ✓

«pOH = −log (0.0466) = 1.332»

pH = «14.000 – pOH = 14.000 – 1.332 =» 12.668 ✓

 

Award [2] for correct final answer.

Award [1 max] for pH =12.367.

d(iii).

«nCa(OH)2 = 2.41 dm3 × 2.33 × 10−2 mol dm−3 =» 0.0562 «mol» AND
«nCO2 =0.750dm322.7moldm-3=» 0.0330 «mol» ✓

«CO2 is the limiting reactant»

«mCaCO3 = 0.0330 mol × 100.09 g mol−1 =» 3.30 «g» ✓


Only award ECF for M2 if limiting reagent is used.

Accept answers in the range 3.30 - 3.35 «g».

e(i).

«2.853.30 × 100 =» 86.4 «%» ✓


Accept answers in the range 86.1-86.4 «%».

Accept “71.3 %” for using the incorrect given value of 4.00 g.

e(ii).

«add» Ca(OH)2/CaCO3/CaO AND to «acidic» water/river/lake/soil
OR
«use» Ca(OH)2/CaCO3/CaO in scrubbers «to prevent release of acidic pollution» ✓


Accept any correct name for any of the calcium compounds listed.

f.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c(i).
[N/A]
c(ii).
[N/A]
d(i).
[N/A]
d(ii).
[N/A]
d(iii).
[N/A]
e(i).
[N/A]
e(ii).
[N/A]
f.

Syllabus sections

Core » Topic 1: Stoichiometric relationships » 1.3 Reacting masses and volumes
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