Question
Limestone can be converted into a variety of useful commercial products through the lime cycle. Limestone contains high percentages of calcium carbonate, CaCO3.
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]
▶️Answer/Explanation
Ans:
«nCaCO3 = =» 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 toobtain the volume with p as either 100kPa (126 dm3) or 101.3 kPa (125 dm3).
Do not penalize use of 22.4 dm3 mol-1 toobtain the volume (124 dm3)
Detailed Solution:
CaCO3 (s) → CaO (s) + CO2 (g)
555 g of calcium carbonate contains 555/100 = 5.55 moles
It produces 5.55 moles CO2 gas.
Volume of 1 mole gas at STP = 22.7 dm3
Volume of CO2 produced = 5.55*22.7 = 126 dm3 .
Question
Limestone can be converted into a variety of useful commercial products through the lime cycle. Limestone contains high percentages of calcium carbonate, CaCO3.
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 3 moles of calcium carbonate decomposes. Use sections 2 and 6 of the data booklet. [2]
▶️Answer/Explanation
Ans:
«nCaCO3 = 3 «mol» ✓ «V = 3 mol × 22.7 dm3 mol−1 =» 68.1 «dm3»
✓Award [2] for correct final answer.Accept method using pV = nRT to obtain the volume with p as either 100kPa or 101.3 kPa.
Do not penalize use of 22.4 dm3 mol-1 to obtain the volume (67.4 dm3)
Detailed Solution:
CaCO3 (s) → CaO (s) + CO2 (g)
1 mole of CaCO3 produces 1 mol CO2.
So, 3 moles CaCO3 will produce 3 moles CO2.
At STP, Volume of 1 mole gas = 22.7 dm3
Volume of CO2 produced = 3*22.7 = 68.1 dm3 .
Question
(a) The second step of the lime cycle produces calcium hydroxide, Ca(OH)2.
(i) Write the equation for the reaction of Ca(OH)2 (aq) with hydrochloric acid, HCl (aq). [1]
(ii) 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]
(b) Calcium hydroxide reacts with carbon dioxide to reform calcium carbonate.
Ca(OH)2 (aq) + CO2 (g) → CaCO3 (s) + H2O (l)
(i) 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]
(ii) 2.85 g of CaCO3 was collected in the experiment in b(i). Calculate the percentage yield of CaCO3. (If you did not obtain an answer to b(i), use 4.00 g, but this is not the correct value.) [1]
▶️Answer/Explanation
Ans:
a i Ca(OH)2 (aq) + 2HCl (aq) → 2H2O (l) + CaCl2 (aq)
a ii
«nHCl = 0.0350 dm3 × 0.025 mol dm−3 =» 0.00088 «mol»
OR nCa(OH)2 = 1 2 nHCl/0.00044 «mol»
nCa(OH)2 = = nHCl/0.00044 «mol» «V =
=» 0.029 «dm3»
b i
«nCa(OH)2 = 2.41 .dm3 × 2.33 × 10−2 mol dm−3 =» 0.0562 «mol» AND «nCO2 =
«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».
b ii
« × 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.
Detailed solution:
a.
i. It is an acid base reaction. The reaction between calcium hydroxide (Ca(OH)₂) and hydrochloric acid (HCl) can be represented by the following balanced chemical equation: Ca(OH)₂(aq) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(l)
ii. 35.0 cm3 of 0.025 mol dm–3 HCl (aq) contains 0.035*0.025 = .000875 moles
2 moles of HCl require 1 mole Ca(OH)₂
Moles of Ca(OH)₂ required = .000875/2 = .0004375 moles
V*0.015mol dm–3 = .0004375 moles
V = 0.029 dm3 .
b.
i. Ca(OH)2 (aq) + CO2 (g) → CaCO3 (s) + H2O (l)
Number of moles of Ca(OH)2 = 2.41 dm3 *2.33 × 10–2 mol dm–3 = 0.056 moles
Assuming 1 mole gas has 22.7 dm3 volume at STP,
no of moles of CO2 = .75/22.7 = 0.033 moles.
Hence, CO2 is the limiting reagent here.
No of moles of CaCO3 formed = 0.033
Mass of CaCO3 formed = 0.033*100 = 3.3 g.
ii. Percentage yield of CaCO3 = 100*(Mass collected/Theoretical mass) = 100*(2.85/3.3) = 86.36%.