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JAMB Chemistry 2009 Objective — Question 4

Question 4 of 50 from the Joint Admissions and Matriculation Board (JAMB) Chemistry 2009 Objective paper, with the correct answer and a full explanation.

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0.0075 mole of calcium trioxocarbonate (IV) is added to 0.015 mole of a solution of hydrochloric acid. The volume of gas evolved at s.t.p. is

  • A. 224cm3
  • B. 168cm3
  • C. 112cm3Correct
  • D. 100cm3

Explanation

CaCO3(s)+2HCl(aq)->CaCl2(aq)+H2O(l)+CO2(g). 1 mole of CaCO3 requires 2 moles of HCl. Thus, 0.0075mol of CaCO3 will take 0.0075x2=0.015mol of HCl. Since there is no limiting reagent, we can therefore use either of the two reagents to determine the volume of CO2 produced. 1 mole of CaCO3 produces 1 mole of CO2. Thus, mole of CO2 = 0.0075mol. Volume of CO2 = mole x G.M.V = (0.0075 x 22.4)dm3 = 0.168dm3 = 168cm3.

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