Question

4(c) Calcium carbonate of mass 1.0g was heated until there was no further change.
(i) Write an equation for the reaction which took place.
(ii) Calculate the mass of the residue.
(iii) Calculate the volume of the gas evolved at s.t.p.
(iv) What would be the volume of the gas measured at 15°C and 760mmHg? [C=12.0, O=16.0, volume of a gas at s.t.p.=22.4dm³] [10 marks]

Accepted Answer

(i) CaCO3(s) →(Δ)→ CaO(s) + CO2(g)
(ii) Moles of CaCO3 = 1.0/100 = 0.01 mol. Since mole ratio CaCO3:CaO = 1:1, moles of CaO (residue) = 0.01 mol. Molar mass CaO = 56 g/mol, so mass of residue = 0.01×56 = 0.56 g.
(iii) Since mole ratio CaCO3:CO2 = 1:1, moles of CO2 = 0.01 mol. Volume at s.t.p. = moles × 22.4 dm³ = 0.01×22.4 = 0.224 dm³.
(iv) Using Charles'/combined gas law with P constant: V1/T1 = V2/T2, where V1=0.224 dm³, T1=273K, T2=(15+273)=288K.
  V2 = V1×T2/T1 = 0.224×288/273 = 0.236 dm³.

WAEC Chemistry 2019 Theory — Question 21

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