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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