Calculate the percentage of CaCO3 present in the sample.

Question

a sample of CaCO3 of mass 2.8g was reacted with 75 cm3 of 1.0 mol/dm3 hydrochloride acid. The resulting solution was completely transferred into 250 cm3 of volumetric flask and the mixture made up to the mark with distilled water. 25 cm3 of this solution needed 20.05 cm3 of 1.0 mol/dm3 of NaOH to neutralize the excess acid. Calculate the percentage of CaCO3 present in the sample.

Answer 1

There seems to be a problem/error with the way the question is presented because according to the below calculations, there is more HCl left over (in excess) than there was originally added. You can certainly check my calculations, but also please check the precise wording of the question.

First reaction:

CaCO₃ + 2HCl ==> CaCl₂ + H₂CO₃ <--> CO₂ + H₂O

Second reaction:

HCl + NaOH = NaCl + H2O (this is titration of the excess HCl from the first reaction.

moles NaOH used = 0.02005 L x 1.0 mol/L = 0.02005 moles NaOH = 0.02005 moles HCl

Correcting for the aliquoting/dilution factor, you have...

0.02005 moles HCl/25 ml x 250 ml = 0.2005 moles HCl left over.

Moles HCl originally present for reaction 1 = 0.075 moles x 1.0 mol/L = 0.075 moles HCl

So, problem is there seems to be more HCl left over (0.2005 moles) than there was originally present (0.075 moles). How can this be?