DP Chemistry (last assessment 2024)
Question 21M.2.hl.TZ2.d(i)
| Date | May 2021 | Marks available | [Maximum mark: 2] | Reference code | 21M.2.hl.TZ2.d(i) |
| Level | hl | Paper | 2 | Time zone | TZ2 |
| Command term | Determine | Question number | d(i) | Adapted from | N/A |
d(i).
[Maximum mark: 2]
21M.2.hl.TZ2.d(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 ==» 0.0330 «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».
