Question

A 1.00 g sample of limestone is allowed to react with 100 cm^3 of 0.200 mol/dm^3 HCl. The excess acid required 24.8 cm^3 of 0.100 mol/dm^3 NaOH solution. Calculate the percentage of calcium carbonate in the limestone

Answer 1

You need to understand what's happening here. You add an excess of HCl to CaCO3 and it reacts completely but not all of the HCl is used. The NaOH is to determine how much of the HCl excess is there after all of the CaCO3 is gone.
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

millimols HCl added initially = mL x M = 100 x 0.2 = 20
mmols NaOH used for excess = mL x M = 24.8 x 0.1 = 2.48
mmols HCl used = 20-2.48 = 17.52 but you should confirm these numbers.
How many mmols CaCO3 must have been there? That's 17.52 mmols HCl x (1 mol CaCO3/2 mols HCl) = 17.52/2 = 8.76 or 0.00876 mols CaCO3.
g CaCO3 = mols CaCO3 x molar mass CaCO3.
%CaCO3 = (grams CaCO3/g of sample)*100 = ?%

Answer 2

87.7%

Answer 3

Are these answers correct?

Answer 4

Ibrahim