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Purpose: To calculate the yield percentage of a chemical reaction between sodium carbonate and calcium chloride. Na2C03 Used CaCI2 Used Mass of empty filter paper Mass of filter paper and dry calcium carbonate precipitate Analyze and Interpret Questions 1. a) What is the solid precipitate that formed in this reaction? The solid precipitate that formed in this reaction is Sodium Chloride (NACI). b) Which reactant was the limiting reactant, and which was the excess reactant? CaC12 was the limiting reactant, Na2CO3 was the excess reactant. Na2CO3(aq) + CaCI2 (aq) →> 2NaCI (aq) + CaCO3(aq) Sodium Carbonate c) Calculate the theoretical yield of the precipitate, in grams, using the actual recorded masses. 4.77 g (22.99 X 2) + (16 X 3) + 12.01 Moles = 4.77/105.99 Moles 0.045 mol = Observations Calcium Chloride 4.4392 g (35.45 X 2) + 40.08 Moles = 4.4392/110.98 Moles 0.04 mol 75ml 100ml 1.75g 4.9g =

Purpose: To calculate the yield percentage of a chemical reaction between sodium carbonate and calcium chloride. Na2C03 Used CaCI2 Used Mass of empty filter paper Mass of filter paper and dry calcium carbonate precipitate Analyze and Interpret Questions 1. a) What is the solid precipitate that formed in this reaction? The solid precipitate that formed in this reaction is Sodium Chloride (NACI). b) Which reactant was the limiting reactant, and which was the excess reactant? CaC12 was the limiting reactant, Na2CO3 was the excess reactant. Na2CO3(aq) + CaCI2 (aq) →> 2NaCI (aq) + CaCO3(aq) Sodium Carbonate c) Calculate the theoretical yield of the precipitate, in grams, using the actual recorded masses. 4.77 g (22.99 X 2) + (16 X 3) + 12.01 Moles = 4.77/105.99 Moles 0.045 mol = Observations Calcium Chloride 4.4392 g (35.45 X 2) + 40.08 Moles = 4.4392/110.98 Moles 0.04 mol 75ml 100ml 1.75g 4.9g =

Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Mark S. Cracolice, Ed Peters
Chapter16: Solutions
Section: Chapter Questions
Problem 148E: An analytical procedure for finding the chloride ion concentration in a solution involves the...
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  1. d) Compare the mass of the precipitate you collected (the actual yield) with the expected theoretical yield. Describe any differences.


2. Calculate the percentage yield of this reaction.

Purpose: To calculate the yield percentage of a chemical reaction between sodium carbonate and calcium chloride. Na2C03 Used CaCI2 Used Mass of empty filter paper Mass of filter paper and dry calcium carbonate precipitate Analyze and Interpret Questions 1. a) What is the solid precipitate that formed in this reaction? The solid precipitate that formed in this reaction is Sodium Chloride (NACI). b) Which reactant was the limiting reactant, and which was the excess reactant? CaC12 was the limiting reactant, Na2CO3 was the excess reactant. Na2CO3(aq) + CaCI2 (aq) →> 2NaCI (aq) + CaCO3(aq) Sodium Carbonate c) Calculate the theoretical yield of the precipitate, in grams, using the actual recorded masses. 4.77 g (22.99 X 2) + (16 X 3) + 12.01 Moles = 4.77/105.99 Moles 0.045 mol = Observations Calcium Chloride 4.4392 g (35.45 X 2) + 40.08 Moles = 4.4392/110.98 Moles 0.04 mol 75ml 100ml 1.75g 4.9g =
Transcribed Image Text:Purpose: To calculate the yield percentage of a chemical reaction between sodium carbonate and calcium chloride. Na2C03 Used CaCI2 Used Mass of empty filter paper Mass of filter paper and dry calcium carbonate precipitate Analyze and Interpret Questions 1. a) What is the solid precipitate that formed in this reaction? The solid precipitate that formed in this reaction is Sodium Chloride (NACI). b) Which reactant was the limiting reactant, and which was the excess reactant? CaC12 was the limiting reactant, Na2CO3 was the excess reactant. Na2CO3(aq) + CaCI2 (aq) →> 2NaCI (aq) + CaCO3(aq) Sodium Carbonate c) Calculate the theoretical yield of the precipitate, in grams, using the actual recorded masses. 4.77 g (22.99 X 2) + (16 X 3) + 12.01 Moles = 4.77/105.99 Moles 0.045 mol = Observations Calcium Chloride 4.4392 g (35.45 X 2) + 40.08 Moles = 4.4392/110.98 Moles 0.04 mol 75ml 100ml 1.75g 4.9g =
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