Chemical Formulae, Equations and Calculations Part 2
6 marks · NormalA chemistry class investigates the reaction between solid calcium carbonate and dilute sulfuric acid. The reaction is:
Explain why, after a short time, the reaction between calcium carbonate and dilute sulfuric acid slows down, even though there is still acid present.
A student wants to calculate the concentration (in mol/dm³) of sulfuric acid if 5.0 g of CaCO3 is completely reacted using 250 cm³ of acid. (Ar: Ca = 40, C = 12, O = 16)
The same student collects the carbon dioxide produced. Explain how they can use the volume of CO₂ collected at rtp to check if the reaction went to completion.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) A layer of insoluble calcium sulfate forms on the calcium carbonate, preventing the acid from reaching it. · must show mentions CaSO4 forms as a solid coating, explains hinders acid access
(b) Concentration = 0.2 mol / 0.25 dm³ = 0.80 mol/dm³ · must show calculates moles of CaCO3 (0.05 mol), uses stoichiometry (CaCO3:H2SO4 ratio 1:1), calculates concentration with volume in dm³
(c) Compare measured volume of CO2 gas to the theoretical maximum. If they match, the reaction is complete; if less, the reaction is incomplete. · must show mentions comparing actual to theoretical, explains matching means completion
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Links slowing of reaction to buildup of insoluble calcium sulfate layer.
B2 (a) · 1 mark · Explains that the layer blocks acid from reacting with remaining CaCO3.
M1 (b) · 1 mark · Finds moles CaCO3: 5.0 g / 100 g/mol = 0.05 mol, uses 1:1 ratio for H2SO4.
A1 (b) · 1 mark · Correctly finds concentration: 0.05 mol / 0.25 dm³ = 0.20 mol/dm³ OR 0.80 mol/dm³ if all CaCO3 is used, accept correct method.
B3 (c) · 1 mark · Explains comparison of actual CO2 volume to theoretical value.
B4 (c) · 1 mark · Explains that matching volumes mean full reaction completion.
Worked solution
a) Calcium sulfate (CaSO4) formed in the reaction is insoluble and creates a layer over the CaCO3, preventing further acid from reaching it, so the reaction slows down even though some acid remains.
b) Mr of CaCO3 = 40 + 12 + (3 × 16) = 100.
Moles of CaCO3 = 5.0 g / 100 g/mol = 0.050 mol.
From the equation, 1 mol CaCO3 reacts with 1 mol H2SO4 ⇒ H2SO4 required = 0.050 mol
Concentration H2SO4 = moles / volume = 0.050 mol / 0.250 dm³ = 0.20 mol/dm³
c) Calculate theoretical amount of CO2 from moles CaCO3 (0.050 mol), so volume = 0.050 × 24,000 = 1,200 cm³ at rtp. Measure actual collected CO2 volume: if actual matches, the reaction is complete; if less, it is incomplete.