Chem homework help

Question

A sample of impure CaCO3(s) is dissolved in 50mL of 0.150M HCl(aq)

The equation for the reaction is

CaCO3 (s) +2HCl (aq)=CaCl2 (aq) + H2O (I) +CO2 (g)

The excess HCl(aq) is titrated by 7.30 mL of  0.125 M NaOH(aq) Calculate the mass percentage of CaCO3(s) in the sample.

Answer 1

To find the mass % of CaCO3, we first need to find the mass of PURE CaCO3. We do this from the balanced equation and the moles of HCl needed to react with the CaCO3.

Reaction 1: CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

moles HCl used = 50 ml x 1 L/1000 ml x 0.150 mol/L = 0.0075 mol HCl but some of this is in excess as indicated by the fact that a second reaction took place to neutralize the excess. This second reaction is

HCl + NaOH ==> NaCl + H2O

mol NaOH needed = 7.30 ml x 1 L/1000 ml x 0.125 mol/L = 0.0009125 moles and this is equal to the moles of HCl in excess in reaction 1 because NaOH and HCl react in a 1:1 mol ratio.

moles HCl actually used in reaction 1: 0.0075 mol - 0.0009125 mol = 0.006588 moles HCl

moles CaCO3 in original sample = 0.006588 mol HCl x 1 mol CaCO3 / 2 mol HCl = 0.003294 moles CaCO3

mass of CaCO3 in original sample = 0.003294 moles x 100.0 g/mol = 0.3204 g

mass % = 0.3204 g / original mass of sample (x100) = mass %

NOTE: The problem did not give the mass of the original impure sample, so the calculation cannot be completed until you get that value.