Answered

Find the best answers to your questions with the help of IDNLearn.com's expert contributors. Get step-by-step guidance for all your technical questions from our dedicated community members.

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)

Sagot :

Pstnonsonjokuidn

The calculated percentage purity of the original sample of calcium carbonate is 44%.

What is excess titration?

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

We have the reaction equation;

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

To get the moles of HCl = 0.2254 M * 20/1000 L = 0.0045 moles

The reaction involving the excess acid occurs thus;

HCl + NaOH ----> NaCl + H2O

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

We have a  1:1 hence 0.0021 moles of HCl reacted also

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

Given the stoichiometry of the reaction,  0.0012 moles of CaCO3 reacted.

Hence;

Mass of pure CaCO3 in the sample= 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?

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for choosing IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more solutions.