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.

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.

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

Outline why a catalyst has such an effect.

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

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).

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.

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.

2.85 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.)

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

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

provides an alternative «reaction» pathway/mechanism ✓

Do not accept “lower activation energy” only.

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

«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 = $\frac{{\displaystyle \frac{1}{2}\times 0.00088 \mathrm{mol}}}{0.015 \mathrm{mol} {\mathrm{dm}}^{-3}}$ =» 0.029 «dm³» ✓

Award [2] for correct final answer.

Award [1 max] for 0.058 «dm³».

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.

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

«$\frac{2.85}{3.30}$ × 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.

«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.