Question

A 1.605g sample of dry calcium carbonate and calcium chloride mixture was dissolved in 35.00mL of 1.000mol-dm-3 hydrochloric acid solution. The only portion of the mixture that reacts with the hydrochloric acid solution is the calcium carbonate. What was the calcium chloride percentage in the original sample, if 31.51mL of 0.0950mol-dm-3 sodium hydroxide was used to titrate excess hydrochloric acid?

I don't know where to go from here:
CaCO3 + 2HCl -> CaCl2 + CO2 +H2O
NaOH +HCl -> H2O + NaCl

Please help!! :)

Answer 1

mmol = millimole
mmol HCl initially = mL x M = 35.00 x 1.00 = 35.00.
mmols NaOH used to titrate excess is 31.51 x 0.0950 = 3 approx but you need to do a more accurate calculation. All of the other numbers that follow are approximations, too, so you need to redo all of this more accurately.

mols HCl used in the reaction with CaCO3 = Initial - final = approx 35 - 3 = 32 or 0.032 mols HCl
Convert that to mols CaCO3 and that's 0.032 x (1 mol CaCO3/2 mols HCl) = 0.016 mols CaCO3.
Than g CaCO3 = mols CaCO3 x molar mass = g CaCO3
1.605 g mixture - grams CaCO3 = grams CaCl2.
%CaCl2 = (g CaCl2/1.605)*100 = ?
Post your work if you get stuck. I think the percent CaCl2 is a small number.
g CaCO3 = 0.032 x molar mass = approx 3.2

Answer 2

Thanks, this helped me figure out what I needed to do, I got 0.25% CaCl2

Answer 3

Ignore the last line. I don't know how it got posted; it shouldn't be there. The rest of the problem is ok.