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JAMB Chemistry 1987 Question 3

Question 3

Mole Concept and Stoichiometry

The number of molecules of carbon (IV) oxide produced when 10.0 g of CaCO3 is treated with 0.2 dm3 of 1 M HCl in the equation CaCO3 + 2HCl → CaCl2 + H2O + CO2 is [Ca = 40, O = 16, C = 12, NA = 6.02 × 1023, H = 1, Cl = 35.5]

  1. A.1.00 × 1023
  2. B.6.02 × 1023
  3. C.6.02 × 1022
  4. D.6.02 × 1024
Show the answer and explanation

Answer: C (6.02 × 1022)

Explanation

10.0 g of CaCO3 (M = 100) is 0.1 mol, and 0.2 dm3 of 1 M HCl is 0.2 mol, exactly enough. So 0.1 mol of CO2 forms: 0.1 × 6.02 × 1023 = 6.02 × 1022 molecules.