Analysis of a mixture of 90% pure limestone and sodium chloride

Analysis of a mixture of 90% pure limestone and sodium chloride
Aim:	To determine the composition of limestone and sodium chloride in a given mixture.

INTRODUCTION

Limestone is generally described as rock made from calcium carbonate, CaCO3. In this experiment, the mixture contains 90% pure limestone (CaCO3) and sodium chloride. The analysis will be carried out by reacting the mixture with hydrochloric acid and by measuring the amount of carbon dioxide evolved by making use of a gas syringe. A diagrammatic representation of the set up is given below;

Click to view experimental setup

Only the limestone will react with the excess acid, liberating carbon dioxide. The reaction proceeds as follows;

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

The sodium chloride will not have so much interaction with the acid, so the composition of the mixture may then be determined by monitoring the amount of carbon dioxide gas liberated.

PROCEDURE

IMPORTANT: Record all your observations as you work and record all data as required.

1)	Set up the apparatus as shown above.(Carefully place the test tube containing the mixture in the Erlenmeyer flask)

2)	Carefully shake the Erlenmeyer flask so that the mixture falls into it.

3)	Make use of a magnetic stirrer so as to make sure that all the solid mixture gets interacted with the acid.

4)	Monitor the amount of carbon dioxide liberated and when the gas syringe stops expanding, record the volume of gas evolved.

CALCULATIONS & ANALYSIS

Mass of sample = -- g

Volume of gas evolved = -- cm3

1)	Calculate the number of moles of carbon dioxide evolved from your results.

2)	From your results, deduce the mass of limestone that were present in the mixture.

3)	Deduce the number of moles of sodium chloride (Mr = 58.5) in the mixture.

Model Answer:

Mass of sample = 8.75 g [arbitrary]

Volume of gas evolved = 520 cm3 [arbitrary]

1)	Number of moles of carbon dioxide evolved;

At r.t.p 1 mole of gas occupies 24000 cm3

520 cm3 of gas will contain 0.0217 moles of carbon dioxide (0.953 g).

2)	Mass of limestone that were present;

From equation; 44 g of Carbon dioxide would be produced from 100 g of CaCO3.

0.953 g of Carbon dioxide would be produced from 2.17 g of CaCO3

3)	Number of moles of sodium chloride (Mr = 58.5) in the mixture;

Mass of sodium chloride in mixture	= (8.75 – 2.17) g = 6.58 g

No of moles of sodium chloride = 0.1125 moles

NOTE: ALL MEASUREMENTS WERE TAKEN AT ROOM TEMPERATURE AND PRESSURE

TUTORIALS

1)	Calculate the number of moles of Calcium Carbonate in the limestone.

2)	What mass of sodium chloride was present in the mixture if 1040 cm3 of carbon dioxide gas was evolved during the reaction?

3)	Which reagent was in excess in the experiment?

4) 	The sodium chloride remaining in the mixture was filtered and reacted with aqueous silver nitrate solution, and it was found that 6.75g of salt was formed;

(i)	Write a balanced equation for this reaction.

(ii)	Name the salt which was formed.

(iii)	Calculate the number of moles of sodium chloride which were present in the mixture.

5)	Give 5 uses of limestone.

6)	Describe briefly how you would analyze the purity of the calcium carbonate. Give the apparatus and reagents required.