Question

if 21.21mL of 0.491M HCl are required to titrate CaCO₃ to a pale yellow endpoint, what is the mass of CaCO₃?

Answer 1

Balanced chemical equation is:

2 HCl + CaCO3 ---> CaCl2 + H2O + CO2

lets calculate the mol of HCl

volume , V = 21.21 mL

= 2.121*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.491*2.121*10^-2

= 1.041*10^-2 mol

According to balanced equation

mol of CaCO3 reacted = (1/2)* moles of HCl

= (1/2)*1.041*10^-2

= 5.207*10^-3 mol

This is number of moles of CaCO3

Molar mass of CaCO3,

MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

use:

mass of CaCO3,

m = number of mol * molar mass

= 5.207*10^-3 mol * 1.001*10^2 g/mol

= 0.5212 g

Answer: 0.521 g