Question

A 0.450-gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction ise excess HCl(aq) is titrated by 7.45 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO3(s) in the sample.

Answer 1

73.05%

Explanation

i)The reaction between HCl and NaOH is as follows

HCl(aq) + NaOH(aq) ------> NaCl(aq) + H2O(aq)

Stoichiometrically ,1mole of NaOH react with 1mole of HCl

Number of moles of NaOH consumed = (0.125mol/1000ml)×7.45ml = 0.00093125

0.00093125 moles of NaOH react with 0.00093125 moles of HCl

excess moles of HCl = 0.00093125mol

ii) Total moles of HCl = (0.150mol/1000ml)×50ml = 0.0075mol

moles of HCl reacted with CaCO3 = Total moles of HCl - excess moles of HCl

moles of HCl reacted with HCl = 0.0075mol - 0.00093125mol = 0.0065688mol

iii)The reaction between CaCO3 and HCl is as follows

CaCO3(s) + 2HCl(aq) -------> CaCl2(aq) + H2O(l) + CO2(g)

Stoichiometrically , 2moles of HCl react with 1mole of CaCO3

So,

0.0065688moles of HCl react with 0.0065688mol/2=0.0032844moles of CaCO3

Molar mass of CaCO3 = 100.089g/mol

Mass of CaCO3 = Number of moles of CaCO3 × molar mass of CaCO3

Mass of CaCO3 = 0.0032844mol × 100.089g/mol = 0.32873g

percent mass of CaCO3 in the sample = (0.32873g/0.450g)×100 = 73.05%