Calcium carbonate CaCO 3 is the active ingredient in many antacid medicines. One
ID: 788170 • Letter: C
Question
Calcium carbonate CaCO3 is the active ingredient in many antacid medicines. One student decides to know how many grams of CaCO3 are present in one tablet. He crushes one tablet, dissolves it in distilled water and titrates the solution with 49.83 mL of 0.200 M HCl. Calculate the mass of calcium carbonate present in this tablet.
The molecular equation is: CaCO3(aq) + 2 HCl(aq) ->H2O(l) + CO2(g) + CaCl2(aq)
From the data collected by the student, there are______x 10^_____g of calcium carbonate in the tablet of anti-acid medicine. Please EXPLAIN.
Explanation / Answer
Millimoles of HCL used = 49.83 * 0.2 = 9.966
Hence millimoles of CaCO3 as per stoichiometry is 9.966/2 = 4.983 millimoles
Mass of CaCO3 hence will be 4.983 * 10^(-3)*100 g = 4.983*10^(-1) g