DP Chemistry (last assessment 2024)

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Question 21M.2.sl.TZ2.e(i)

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Date May 2021 Marks available [Maximum mark: 2] Reference code 21M.2.sl.TZ2.e(i)
Level sl Paper 2 Time zone TZ2
Command term Determine Question number e(i) Adapted from N/A
e(i).
[Maximum mark: 2]
21M.2.sl.TZ2.e(i)

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

Calcium hydroxide reacts with carbon dioxide to reform calcium carbonate.

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

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]

Markscheme

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

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