Question

A 0.450 g sample of impure CaCO₃ is dissolved in 50.0mL of 0.150M HCl.

CaCO₃ + 2HCl ---> CaCl₂ + H₂0 + CO₂

the excess HCl is titrated by 7.45mL of 0.125M NaOH

Calculate the mass percentage of CaCo3 in the sample?

Answer 1

Answer – We are given, mass of sample = 0.450 g , volume of HCl = 50.0 mL

[HCl] = 0.150 M HCl

Moles of HCl = 0.150 M * 0.050 L

                      = 0.0075 moles

The excess HCl was titrated with 7.45mL of 0.125M NaOH, so

Moles of NaOH = 0.125 M * 0.00745 L

                          = 0.000931 moles

We know , 1 moles of HCl = 1 moles of NaOH

So, excess moles of HCl = 0.000931 moles

Moles of HCl reacted with CaCO₃ = 0.075 – 0.000931

                                                     = 0.00657 moles

From the reaction –

CaCO₃ + 2HCl -----> CaCl₂ + H₂O + CO₂

2 moles of HCl = 1 moles of CaCO₃

So, 0.00657 moles of HCl = ?

= 0.00328 moles of CaCO₃

So, mass of CaCO₃ = 0.00328 moles * 100.087 g/mol

                                = 0.329 g

So, mass percent of CaCO₃ = 0.329 g / 0.450 g *100 %

                                             = 73.0 %