From the complete decomposition of 20g CaCO3 at STP the volume of CO2 obtained is:
Correct answer: C. 4.48dm3
- A. 2.24dm3
- B. 44.8dm3
- C. 4.48dm3
- D. 48.4dm3
Explanation
Balanced equation of decomposition of CaCO3: CaCO3 (s) → CaO (s) + CO2 (g) mass of CaCO3 : 20g molecular mass of CaCO3 : (1 x 20) + (1 x 12) + (3 x 16) = 100 no. of moles of CaCO3 : 20 / 100 = 0.2 moles stoichiometry ratio of CaCO3 : CO2 = 1 : 1 no. of moles of CO2 = 0.2 volume of 1 mol of CO2 at S.T.P = 22.4 dm3 volume of 0.2 mol of CO2 at S.T.P = 0.2 x 22.4 = 4.48 dm3 Thus option C is correct.
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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