Question

You weigh out an antacid tablet and determine the mass to be 1.1990 g. After taking a 0.2455 g sample of the antacid tablet, you dissolve it in 25.00 mL of a 0.1006 M solution of hydrochloric acid in a 250.0 mL Erlenmeyer flask. After heating the solution to get rid of carbon dioxide, you titrate the leftover hydrochloric acid in the solution in the Erlenmeyer flask with a 0.09913 M solution of sodium hydroxide and determine that it takes 14.71 mL of the sodium hydroxide solution to react with the leftover hydrochloric acid in solution. Determine the milligrams of calcium carbonate in the sample of antacid tablet and then determine the milligrams of calcium carbonate in the entire antacid tablet.

Answer 1

CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

moles of HCl added initially = M x L = 0.002515 moles

That HCl is more than enough to dissolve all of the tablet and have some HCl left over.
moless NaOH needed to neutralize the excess HCl = M x L = 0.0014582 moles.
moles of HCl used for the tablet is
0.002515 - 0.0014582 = 0.0010568 moles
Convert moles of HCl used in the neutralization of the tablet to moles of CaCO3 using the coefficients in the balanaced equation. That will be moles of CaCO3 = (1/2 x moles HCl) = 0.0005284 moles

mass of CaCO3 = 0.05284g in 1.1990 g tablet

then in 0.2455 g tablet = 0.0065187 g = 6.5187 mg
Then convert to mg in the entire sample.