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this is a lab report on stochiometry
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Lab report done by: Aaida Hossain (2132500)
Partner: Rosalie Meunier
Presented to: Matthew Hachey
202 - NYA-05, Section 03003
Date of the Experiment: 22- 09 - 2021
Date of the Lab Report Submission: 06- 10 - 2021
1.0 Introduction
Solution stoichiometry is the stoichiometry of chemical reaction carried out in solution. Classical
stoichiometric calculations involve the following steps:
product that are expected, thence the expected mass of product (theoretical yield).
yield) by calculating the percent yield.
For solution stoichiometry, in the first step, the number of moles of reactants is calculated by
using the definition of the molar concentration:
Here, ๐ is the number of moles of reactant, ๐ is the molarity of the solution and ๐ is the aliquot
of this solution being used.
In this experiment, aliquots of sodium carbonate, ๐๐ 2
3
, and calcium chloride solutions,
2
, are mixed and the reaction described by the following equation occurs:
2
3
2
3
The resulting precipitate is filtered, washed and dried and the mass of the precipitate obtained is
compared with the theoretical mass by calculating the %๐ฆ๐๐๐๐. (1)
2.0 Objective
The objective of this laboratory is to study the reaction between calcium chlorite and sodium
carbonate. Hence, the aim is to experimentally determine the mass of calcium carbonate
produced and to compare it to our theoretical expected value by calculating its percent yield. (1)
3.0 Material:
2
2
5.0 Results:
The table below shows the results of the experiment, and section 5.1 shows the sample
calculation used to find the values displayed in the table
Table 1: Lab results
Mass of the empty weighing dish. 2.7442 g
Mass of the weighing dish and the ๐ถ๐๐ถ๐
2
2
6.7480 g
Mass of the ๐ถ๐๐ถ๐
2
2
๐ 4.0 038 g
Moles of the ๐ถ๐๐ถ๐
2
2
๐ 0.02723 mol
2
2
3
Volume of the ๐ถ๐๐ถ๐
2
solution used. 10.00 mL
Volume of the ๐๐
2
3
solution used.
10.00 mL
Mass of the filter paper and the watch glass. 53.9009 g
Mass of the filter paper, the watch glass and
the dry final product.
54.1721g
Mass of the dry final product. 0.2712 g
Moles of ๐ถ๐๐ถ๐
2
used. 0.0027 mol
Moles of ๐๐
2
3
used.
0.0033 mol
Limiting Reagent. ๐ถ๐๐ถ๐
2
Calculated mass of excess reagent
remaining in the mixture after the reaction.
0.06 g of ๐๐
2
3
Theoretical yield 0.2700 g
Percent yield 99%
5.1 Sample Calculations
2
2
2
2
according to the correct ratios.
2
2
๐
๐๐
, where ๐ is the number of moles of the product, ๐ is the mass of the
product and ๐๐ is the molar mass of the product.
๐ก๐๐ก
2
๐
๐
, where ๐ is the molarity, ๐ is the number of moles and ๐ is the total
volume of the solution.
2
volume of the solution.
2
2
create. The reactant that creates the least amount of moles of final product is the limiting
reactant.
2
3
2
3
By observing this balance equation, we find out the number of moles of each reactant is equal to
the number of moles of the final product.
3
2
3
3
2
3
3
The limiting reactant in this reaction is: ๐ถ๐๐ถ๐
2
2
3
3
๐๐๐ก๐ข๐๐๐ฆ๐๐๐๐
๐กโ๐๐๐๐๐ก๐๐๐๐๐ฆ๐๐๐๐
3
8.0 Appendix
8. 1 Procedure
8.3 Post-Lab Questions
solution, thus reducing the number of moles of the final product that can be possibly
formed.
moisture that was present in the final product has been evaporated
b) It would make the percent yield invalid because the mass of the final product weighed
would not only be bigger than the theoretical mass, but the final product would not be
pure as it contains other substances
b)The final dry product would not be pure as it would contain not only the precipitate but
any excess reactants/products
substance would make the task of getting rid of any other substances easier.
b) The final dry product would not be pure as it would contain not only the precipitate but
any excess reactants/products
solution.
b)Yes, the theoretical yield would change because in this case the theoretical mass is
0.4900 g.
