“1.25 g of crushed limestone was added to 50.0 cm3 of 1.00 mol dm-3 hydrochloric acid (an excess). The mixture was left until all bubbling stopped and was then made up carefully to 250 cm3 with pure water. A 25.0 cm3 sample of this was pipetted into a conical flask and some methyl orange indicator was added. Sodium hydroxide solution of concentration 0.100 mol dm-3 was added from a burette. 30.0 cm3 were needed to reach the end-point of the indicator. We can then calculate the percentage of calcium carbonate in the sample.” What is the mass percentage of carbonate in the sample? Assume all carbonate is calcium carbonate, what is the mass percentage of the remainder ( or is this assumption not tenable) ?

Answer 1

The idea here is that the CaCO3 reacts with a known excess of HCl, the excess HCl is then titrated in a regular acid/base titn to determine how much of an excess was there.
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

1.25 g sample-->added 50 x 1 mol/dm^3 HCl. = 50 mmoles HCl-->diluted to 250 cc-->25 cc aliquot taken and titrated with 30 cc of 0.1 mole/dm^3 HCl.
I would start with the titration step.
30 cc x 0.1 mol/dm^3 = 3 mmoles NaOH.
3 mmoles NaOH = 3 mmoles excess HCl (in the 25 cc aliquot). Since that was a 25 cc aliquote from a 250 cc flask, then 3 mmoles x (250/25) = 30 mmoles in the entire flask. We started with 50 mmoles HCl; therefore 50-30=20 mmoles HCl were used by the CaCO3. Looking at the coefficients in the equation, 20 mmoles HCl means 10 mmoles CaCO3 reacted. I would then convert 10 mmoles to moles, grams CaCO3 = moles CaCO3 x molar mass CaCO3 and %CaCO3 = (mass CaCO3/mass sample)*100 = ??
mass of sample, of course, is 1.25 g.
I get an answer about 80% CaCO3 and it makes no difference which part we talk about. It's 80% in the original sample and 80% in the sample taken for titration.

Answer 2

Where did you get this question from