Lab 8 Limiting Reactant PDF

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Lab 8 Limiting Reactant Name: Tashfia Hasan Lab Partner: Hayley Karr, Natalie Sondhi, Tina Takla, Becca, Som Keshav Course: E01, Tuesdays 6 PM - 8:45 PM Instructor Name: Obiajulu Nwanze Laboratory Assistant Name: Camila Gonzalez Date Experiment was Performed: September 22, 2020

Prelab Questions: 1) The limiting reactant is determined in this experiment. a. What are the reactants (and their molar masses) in the experiment? CaCl2•2H2O(aq) : 147.01 g/mol K2C2O4•H2O(aq) : 184.23 g/mol b. How is the limiting reactant determined in the experiment? The limiting reactant is determined by calculating the amount of product that can be formed by each reactant; the one that produces less product is the limiting reagent. In Part A of this experiment, the solid reactant salts CaCl2•2H2O and K2C2O4•H2O forms a heterogeneous mixture of unknown composition. Water is added to the reactants and insoluble CaC2O4•H2O forms and the precipitate is collected via gravity filtration and dried, and its mass is measured.

2) Experimental Procedure, Part A.2.What is the procedure and purpose of“digesting the precipitate”? The procedure to “digesting the precipitate” requires heat to be applied to a covering beaker with a watch glass and warm the solution on a hot plate to a maximum temperature of 75°C for ~15 minutes. During the heating process it is important to periodically stir the solution. After ~15 minutes, remove the heat and allow the precipitate to settle; the solution does not n eed to cool to room temperature.This allows the equilibrium between the solid and its ions in solution to form larger solid particles in order to make their filtration more efficient. 3) Two special steps in the Experimental Procedure are incorporated to reduce the loss of the calcium oxalate precipitate. Identify the steps in the procedure and the reason for each step. The two special steps are the digestion of calcium oxalate which is reprecipitated onto large particles making the filtering process more efficient. Also the step using the filter paper reduced the chance of ppt going through. 4) a. A sample of a CaCl2•2H2O/K2C2O4•H2O solid salt mixture is dissolved in ~150 mL of deionized water previously adjusted to a pH that is basic. The precipitate, after having been filtered, was air-dried and weighed. Data for Trial 1 were obtained as shown. Complete the following table. (See Report Sheet . ) Record calculated values with the correct number of significant figures. A. Precipitation of CaC2O4•H2O from the Salt Mixture 3. Mass of salt mixture (g) -> 0.879 g

4. Mass of filter paper (g) -> 1.896 g 5. Mass of filter paper and CaC2O4•H2O (g) -> 2.180 g 6. Mass of air-dried CaC2O4•H2O (g) ->0.284 g B. Determination of Limiting Reactant 1. Limiting reactant in salt mixture -> CaCl2•2H2O 2. Excess reactant in salt mixture -> K₂C₂O₄·H₂O Data Analysis: 1. Moles of CaC2O4•H2O precipitated (mol) -> .001943 mole CaC₂O₄·H₂O 2. Moles of limiting reactant in salt mixture (mol ) -> .001943 mole CaCl₂ 3. Mass of limiting reactant in salt mixture (g ) -> .286 g CaCl₂ 4. Mass of excess reactant in salt mixture (g ) -> .593 g K₂C₂O₄·H₂O 5. Percent limiting reactant in salt mixture (%) -> 32.5% 6. Percent excess reactant in salt mixture (%) -> 67.5% Calculations for Data Analysis: 4A) 6. Mass of filter paper and CaC2O4•H2O (g) - Mass of filter paper (g) = 2.180 g - 1.896 g = .284 g CaC2O4•H2O (g) Finding the limiting reactant: If the limiting reactant was CaCl₂·2H₂O, than the excess must be K₂C₂O₄·H₂O.

1. Mass of CaC2O4•H2O (g) (moles) = .284 g CaC₂O₄·H₂ (1 mol/ 146.13g) = .001943 2. .001943 mole CaC₂O₄·H₂O (1mole CaCl₂·2H₂O/1 mole CaC₂O₄·H₂O) = .001943 mole CaCl₂ 3. .001943 mol CaCl₂ (147.02 g/ 1 mole) = .2855 4. Mass of salt mixture (g) - Mass of limiting reactant in salt mixture (g ) = .879 g- .286 g = .593 g 5. Mass of limiting reactant in salt mixture (g ) /Mass of salt mixture (g) * 100 =.286 g/.879 g= .3254 * 100 = 32.5% 6. Mass of limiting reactant/Mass of salt mixture (g) * 100= .593 g/.879 g = .6746 * 100 = 67.5%

Data: A. Precipitation of CaC2O4•H2O from the Salt Mixture Unknown number: Questions

Trial 1

1. Mass of beaker (g)

133.000 g

2. Mass of beaker and salt mixture (g)

134.001 g

3. Mass of salt mixture (g)

1.001 g

4. Mass of filter paper (g)

0.456 g

5. Mass of filter paper and product after oven-dried (g)

0.652 g

6. Mass of product (g)

0.196 g

7. Formula of product

CaC2O4

B. Determination of Limiting React

Excess reactant in salt mixture (formula)

Ca2+

Limiting reactant in salt mixture (formula)

C2O42-

Data Analysis: Questions: 1. Moles of CaC2O4•H2O (or CaC2O4) precipitated (mol) 2. Moles of limiting reactant in salt mixture (mol) Formula of limiting hydrate

Trial 1 1.53 * 10^-3 mol 1.53 * 10^-3 mol

3. Mass of limiting reactant in salt mixture (g)

0.135 g

4. Mass of excess reactant in salt mixture (g)

0.866 g

5. Percent limiting reactant in salt mixture (%)

13.5%

6. Percent excess reactant in salt mixture (%)

86.5%

7. Mass of excess reactant that reacted (g)

0.196 g

8. Mass of excess reactant, unreacted (g)

0.67 g

Calculations: Data Analysis: 1. Moles of CaC2O4•H2O (or CaC2O4) precipitated (mol) CaC2O4= 128.0994g/ mol Moles (n)= 0.196 g/ 128.0994g/mol = 1.53 * 10^-3 mol

2. Moles of limiting reactant in salt mixture = Moles of CaC2O4•H2O (or CaC2O4)

precipitated = 1.5300 * 10^-3 mol as there is a 1:1 ratio in the reaction.

3. Mass of limiting reactant in salt mixture Moles of limiting reactant * molar mass of C2O4 1.5300 * 10^-3 mol * 88.02 g/mol= 1.3467 * 10^-1 mol -> 1.35 * 10^-1 mol

4. Mass of excess reactant in salt mixture (g) Mass of salt mixture(g) - Mass of limiting reactant in salt mixture (g) = 1.001 g - .135 g = 0.866 g

5. Percent limiting reactant in salt mixture (%) Mass of limiting reactant in salt mixture (g ) /Mass of salt mixture (g) * 100 0.135 g/ 1.001 g * 100 = 13.5%

6. Percent excess reactant in salt mixture (%) Mass of excess reactant/Mass of salt mixture (g) * 100 0.866 g/ (1.001 g) * 100 = 86.5%

7. Mass of excess reactant that reacted (g) Moles of CaC2O4*H2O * molar mass of CaC2O4*H2O 1.53 x 10^-3 mol CaC2O4*H2O* ( 128.0994 g CaC2O4*H2O/ mol /1 mol  CaC2O4*H2O) = 0.196 g

8. Mass of excess reactant that was unreacted (g) Total excess reactant (g) - Mass of excess reactant that reacted (g) 0.866 g - 0.196 g = 0.67 g Results: Based on the data and calculations of this experiment there is an observation that was apparent. When the precipitate is being digested, if the correct temperature, duration of heating or stirring isn’t done correctly, then the mass of the precipitate CaC2O4•H2O will be too high. When a substance is being filtered with the filter paper, not all of the supernatant fluid is able to be removed which causes some of the excess reactant to be unreacted. This is apparent in the large amount of the excess reactant that is unreacted in the trial where approximately 77.4% of the Ca is unreacted. This occurs because the reactant is using the quantity of the limiting reactant in its entirety. Heating the compound causes bonds in the compound to break and release energy and ultimately mass as a result. Brief Conclusion In this experiment, the objective was to determine t he limiting reactant in a mixture of two soluble salts and the percent composition of each substance in a salt mixture. These calculators were based on the grams of the salt mixture as well as the mass of the precipitate. To find the limiting reactant of this compound, the theoretical yield is essential to find so the molar ratio enables one to find the moles of each reactant. Then, the mass of each of the reactants can be calculated based on the molar amount. In turn, the percent composition can be calculated based on the quantity of each reactant in relation to the total mass of the salt mixture.

Post Lab Questions:

1) If the step for digesting the precipitate were omitted, will the reported “percent limiting reactant” in the salt mixture be too high, too low, or unaffected? Explain. The quantity will be too low because the purpose of digesting the precipitate is to make the filtering process easier to do because it essentially reduces the quantity of the precipitate. The product would be too low which would interfere with the amount of the limiting reactant. 3) Because of the porosity of the filter paper some of the CaC2O4•H2O precipitate passes through the filter paper. Will the reported percent of the limiting reactant in the original salt mixture be reported too high or too low? Explain. The quantity would be too low because it means that the precipitate which is essential to determine the limiting and excess reactant would be lost and a result in a decrease in the limiting reactant of the original salt. 5) The CaC2O4•H2O precipitate is not completely air-dried when its mass is determined.Will the reported mass of the limiting reactant in the original salt mixture be reported too high or too low? Explain. The precipitate would be too high because the precipitate hasn't evaporated completely meaning that the water would have a greater mass in comparision to the original salt mixture....