IB Questionbank

User interface language: English | Español

Date	May 2021	Marks available	1	Reference code	21M.2.sl.TZ2.1
Level	SL	Paper	2	Time zone	TZ2
Command term	Write	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]

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]

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 × 10⁻²mol 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]

Markscheme

«n_CaCO3 = $\frac{555 g}{11.09 g {mol}^{- 1}}$ =» 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).

«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».

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/E_cat 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).

«n_HCl = 0.0350 dm³ × 0.025 mol dm⁻³ =» 0.00088 «mol»
OR
n_Ca(OH)2 = $\frac{1}{2}$ n_HCl/0.00044 «mol» ✓

«V = $\frac{\frac{1}{2} \times 0.00088 mol}{0.015 mol {dm}^{- 3}}$ =» 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⁺] = $\frac{1.00 \times 10^{- 14}}{0.0466}$ = 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).

«n_Ca(OH)2 = 2.41 dm³ × 2.33 × 10⁻²mol dm⁻³ =» 0.0562 «mol» AND
«n_CO2 = $\frac{0.750 {dm}^{3}}{22.7 mol {dm}^{- 3}}$ =» 0.0330 «mol» ✓

«CO₂ is the limiting reactant»

«m_CaCO3 = 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).

« $\frac{2.85}{3.30}$ × 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.

Examiners report

[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]

Syllabus sections

Core » Topic 1: Stoichiometric relationships » 1.1 Introduction to the particulate nature of matter and chemical change

Show 65 related questions

Hide related questions

Core » Topic 1: Stoichiometric relationships

Core

Question

Markscheme

Examiners report

Syllabus sections

View options