Essay on Stoichiometry: Stoichiometry and Mol Caco 3== 0 , 2726 G Caco 3

Submitted By laurazepbed
Words: 368
Pages: 2

Laura Zeppetelli-Bédard
ID: 1438543
Steffie Ares

SOLUTION STOICHIOMETRY (EXP. #1)
Maria Ann Di Stefano

October 2nd 2014

INTRODUCTION:

In this experiment, there will be evaluation of the solution stoichiometry of the reaction between calcium chloride ( CaCl2) and sodium carbonate (Na2CO3) carried out in a aqueous solution.
Here, aliquots of sodium carbonate and calcium chloride solutions are mixed to gives us the following reaction equation:
Na2CO3 (aq) + CaCl2 (aq) CaCO3 (s) + 2NaCl (aq)
The theoretical mass of the precipitate CaCO3 (s) will be calculated with classical stoichiometric calculations involving the number of moles of reactants (using definition of molar concentration: n=MV, where V=aliquot of solution and M=molarity of solution), the determination of the limiting reactant and the mass of product.
The resulting precipitate obtained experimentally (after filtration and elimination of excess solution or moisture) will be weighed and compared to the theoretical mass by using the formula for the %yield.

PROCEDURE:

Please refer to pages 15-16 of the General Chemestry 202-NYA-05 lab manual.

DATA AND RESULTS:

Experimental data:

Mass (g)
Empty weighing dish
2,3784
Weighing dish
6,3823
CaCl2-H2O
4,0039
Filter paper + watch glass
40,4789
Filter paper + watch glass + dry product (final)
40,7358

-volume of CaCl2 solution used: 10,00 mL
-volume of Na2CO3 solution used: 10,00 mL

Calculated data:
-mass of dry product: 0,2526g
-moles of CaCl2 used: 0,0027233 mol
-moles of Na2CO3 used: 0,0027233 mol
-mass of excess reagent (Na2CO3) remaining: 0,0643g
-Theoretical yield: 0,2726g CaCO3
-% yield: 94,24%

SAMPLE CLACULATIONS:

equation:
Na2CO3