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Stoichiometry and Limiting Reagent

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Stoichiometry and Limiting Reagents Theodore A. Bieniosek I. Purpose and Theory The purpose of the experiment is to study and apply the processes of stoichiometric calculation on a controlled chemical reaction. We will be adding variable amounts of reactants in a chemical reaction in order to demonstrate the effect of limiting reagents. Based on the volumes of the reactants, and their respective molarities, we can calculate the theoretical yield of the reaction and compare it to the amount of products experimentally yielded.

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From the molarities of the solutions, the amount of moles can be calculated, and in turn fed into the balanced chemical equation to determine the theoretical yield. II. Data and Calculations Data Test #1Test #2Test #3 Initial Volume of Sodium Hydroxide:42. 4 ml26. 6 ml30. 6 ml Final Volume of Sodium Hydroxide:44. 4 ml30. 6 ml35. 6 ml Volume of Sodium Hydroxide Delivered:2. 0 ml4. 0 ml5. 0 ml Initial Volume of Calcium Chloride:42. 6 ml39. 45 ml10. 7 ml Final Volume of Calcium Chloride:47. 6 ml44. 45 ml15. 7 ml Volume of Calcium Chloride Delivered:5.

0 ml5. 0 ml5. 0 ml

Excess Reagent Test with Sodium Hydroxide:positivepositivenegative with Calcium Chloride:negativenegativepositive Initial Mass of Filter Paper:. 89 g. 87 g. 76 g Final Mass of Filter Paper with precipitate:1. 14 g1. 27 g1. 15 g Mass of Precipitate. 25 g. 40 g. 39 g Calculations CaCl2(aq) + 2NaOH(aq) Ca(OH)2(s) + 2NaCl(aq) Moles = [volume (L)] * (Molarity) Molarity CaCl2: 1. 0 M Molarity NaOH: 2. 5 M Test #1 Volume CaCl2: 5. 0 mL = . 005 L Volume NaOH: 2. 0 mL = . 002 L Moles CaCl2 = (. 005 L) * (1. 0 mol/L) = . 005 moles Moles NaOH = (. 002 L) * (2. 5 mol/L) = . 05 moles Limiting Reactant: NaOH, because precipitate was formed when Sodium Hydroxide was added to filtered solution, CaCl2 was in excess, meaning the NaOH was completely used up and therefore limited the reaction. [ (2 mole NaOH) / (1 mole CaCl2) ]*[ . 005 mol CaCl2 ] = . 010 moles NaOH needed to completely react CaCl2, only . 005 moles available, therefore NaOH limits the reaction Theoretical Yield of Ca(OH)2: [ . 005 moles NaOH ] * [ (1 mole Ca(OH)2 ) / (2 moles NaOH) ] = . 0025 moles Ca(OH)2 formed [ . 0025 moles Ca(OH)2 ] * [ (74. 926 g Ca(OH)2 ) / ( 1 mole Ca(OH)2 ) ] = . 1852 grams Ca(OH)2 Percent Yield [ . 25 g / . 1852 g ] * 100 % = 134. 989 % Test #2 Volume CaCl2: 5. 0 mL = . 005 L Volume NaOH: 4. 0 mL = . 004 L Moles CaCl2 = (. 005 L) * (1. 0 mol/L) = . 005 moles Moles NaOH = (. 004 L) * (2. 5 mol/L) = . 010 moles Limiting Reactant: NaOH, because precipitate was formed when Sodium Hydroxide was added to filtered solution, CaCl2 was in excess, meaning the NaOH was completely used up and therefore limited the reaction. [ (2 mole NaOH) / (1 mole CaCl2) ]*[ . 05 mol CaCl2 ] = . 010 moles NaOH needed to completely react CaCl2, only . 010 moles available, theoretically all reagents are completely reacted. Measurement error has apparently left excess CaCl2 in filtered solution, and lack of NaOH limited the reaction. Theoretical Yield of Ca(OH)2: [ . 010 moles NaOH ] * [ (1 mole Ca(OH)2 ) / (2 moles NaOH) ] = . 005 moles Ca(OH)2 formed [ . 005 moles Ca(OH)2 ] * [ (74. 0926 g Ca(OH)2 ) / ( 1 mole Ca(OH)2 ) ] = . 3705 grams Ca(OH)2 Percent Yield [ actual yield / theoretical yield ] * 100% [ . 0 g / . 3705 g ] * 100% = 107. 962 % Test #3 Volume CaCl2: 5. 0 mL = . 005 L Volume NaOH: 5. 0 mL = . 005 L Moles CaCl2 = (. 005 L) * (1. 0 mol/L) = . 005 moles Moles NaOH = (. 005 L) * (2. 5 mol/L) = . 0125 moles Limiting Reactant: CaCl2, because precipitate was formed when Calcium Chloride was added to filtered solution, NaOH was in excess, meaning the CaCl2 was completely used up and therefore limited the reaction. [ (1 mole CaCl2) / (2 mole NaOH) ]*[ . 0125 mol NaOH ] = . 00625 moles CaCl2 needed to completely react the NaOH, only . 05 moles CaCl2 available, therefore CaCl2 limits the reaction Theoretical Yield of Ca(OH)2: [ . 005 moles CaCl2 ] * [ (1 mole Ca(OH)2 ) / (1 mole CaCl2) ] = . 005 moles Ca(OH)2 formed [ . 005 moles Ca(OH)2 ] * [ (74. 0926 g Ca(OH)2 ) / ( 1 mole Ca(OH)2 ) ] = . 3705 grams Ca(OH)2 Percent Yield [ . 39 g / . 3705 g ] * 100% = 105. 263% III. Results, Discussion, and Conclusion After concluding the experiment, I learned that the products of a chemical reaction can be created in precise amounts when the masses of the products can be measured to exact standards.

Further, I learned more about the molarity of aqueous solutions, and how that quantity, along with the volume of the solution, can be manipulated to find the exact number of moles in a given volume. In this case, we were given aqueous solutions of NaOH and CaCl2 in known molarities and then had them react with one another to yield a precipitate of Ca(OH)2. The precipitate was filtered out of the remaining aqueous solution of NaCl, dried, and finally weighed. We compared this measured value against the theoretical yield of product, which we calculated using simple principles of stoichiometry.

In our case, all three tests yielded more mass than should have occurred. We attribute this excess to inaccuracy in measurement, both of the volumes of reactants and of the masses of the products filtered. It is possible that the water had not completely evaporated from the filtered precipitate, which would add excess mass. By dealing with such small quantities of reagents, any small inaccuracy in measurement creates a large difference in actual yield from theoretical yield.

Through simple molar calculations, using the coefficients in the balanced chemical equation (2NaOH(aq) + CaCl2(aq) 2NaCl(aq) + Ca(OH)2(s)), the limiting reagent could be determined from the volumes of reactants used. The first and the third tests turned out as expected, NaOH and CaCl2 respectively being the limiting reagents. In the second test, according to our calculations, the products should have completely reacted, leaving pure water after filtration. In the test, the addition of NaOH to the filtered solution yielded more precipitate, which should not have occurred if the filtrate was pure water.

We hypothesize that either the products were not measured in exact quantities, or in the test tube the products did not completely react with one another, leaving some CaCl2 in solution. In conclusion, I found that by measuring reactants accurately, the products of a chemical reaction can be created to precise standards. In our third trial, we were able to achieve a yield within six percent of the theoretical yield. Only through accurate calculations and measurements can the yield of products be determined before a reaction occurs.

Cite this Stoichiometry and Limiting Reagent

Stoichiometry and Limiting Reagent. (2019, May 01). Retrieved from https://graduateway.com/stoichiometry-and-limiting-reagent/

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