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2. A 0.2719 g sample containing CaCO3 reacted with 20.00 mL of 0.2254 M HCl. Given that HCI was excess. The excess HCl required exactly 20.00 mL of 0.1041 M NaOH to reach the end-point using phenolphthalein indicator. Determine percentage purity of CaCO3 in the sample. The reraction involved is CaCO3(s) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(l) The titration reaction is HC(aq) + NaOH(aq) NaCl(aq) + H₂O(1)

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The percentage purity of the calcium carbonate is  44%. The percentage purity gives the amount of pure CaCO3 in the sample.

What is excess titration?

In excess titration or back titration, we neutralize the excess titrand left in a system.

We have the reaction;

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

Number of moles of HCl = 0.2254 M * 20/1000 L = 0.0045 moles

The reaction of the excess acid is according to the reaction;

HCl + NaOH ----> NaCl + H2O

Number of moles of NaOH = 0.1041 M * 20/1000 = 0.0021 moles

Since the reaction is 1:1, 0.0021 moles of HCl reacted also

Number of moles of HCl that reacted with CaCO3  =  0.0045 moles - 0.0021 moles = 0.0024 moles

If 2 moles of HCl reacts with 1 mole of CaCO3

0.0024 moles of HCl reacts with 0.0024 moles * 1/2

= 0.0012 moles

Mass of pure CaCO3 present = 0.0012 moles * 100 g/mol = 0.12g

Percent purity of the sample =  0.12g/0.2719 g * 100/1

= 44%

Learn more about percent purity: https://brainly.com/question/10962305?