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Theoretical Yield (g) of PbCrO4. Theoretical Yield (g) of KNO3

Theoretical Yield (g) of PbCrO4. Theoretical Yield (g) of KNO3

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter3: Chemical Reactions
Section: Chapter Questions
Problem 141QRT
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I need help on the Theoretical yield. It’s above question 6!! thank you!!!
Name Stoichiometry: Mass-Mass Experimentation has shown that there is usually a definite relationship between the number of moles of reactant and the number of moles of product in a chemical reaction. If one mole of A reacts with two moles of B, then it is reasonable to assume that one molecule of A can react with two molecules of B. The quantitative study of chemical reactions has led to the development of many important theories about the nature of matter. Mass-Mass experiments involve the mass relationships which accompany a chemical reaction. This experiment can yield very accurate results if care is taken in weighing, decanting, and filtering both reactants and products. The reaction involves: K2CrO4 (aq) Pb(NO3)(aq) → 2KNO3(aq) PbCrO4 (s) Potassium Chromate Lead Nitrate Potassium Nitrate Lead Chromate The PbCrO4 (s) is a yellow precipitate that can be collected on a filter paper and weighed. DATA 33.ul0s 1. First calculate the mass of exactly 0.005 moles of Potassium Chromate and Lead Nitrate Show work here: O.v05 194.19-0.971 .00Saえしs grams K,CrOa/calculated (this is what needs to be weighed) 0.97| grams Pb(NO;)»/calculated (this is what needs to be weighed)LCSS9 Mass of 250 ml beaker10.06g Mass of 250 ml flask 2.261 Mass of filter Paper Mass of filter paper and collected PbCrO4 (yellow stuff) Mass of flask with evaporated KNO, solution_ 973 Calculations (show the work): Mass of PbCrO4 (yellow stuff, this will require some subtraction)looMg ey eeam 1. /14- 110.06=394 1.679
Transcribed Image Text:Name Stoichiometry: Mass-Mass Experimentation has shown that there is usually a definite relationship between the number of moles of reactant and the number of moles of product in a chemical reaction. If one mole of A reacts with two moles of B, then it is reasonable to assume that one molecule of A can react with two molecules of B. The quantitative study of chemical reactions has led to the development of many important theories about the nature of matter. Mass-Mass experiments involve the mass relationships which accompany a chemical reaction. This experiment can yield very accurate results if care is taken in weighing, decanting, and filtering both reactants and products. The reaction involves: K2CrO4 (aq) Pb(NO3)(aq) → 2KNO3(aq) PbCrO4 (s) Potassium Chromate Lead Nitrate Potassium Nitrate Lead Chromate The PbCrO4 (s) is a yellow precipitate that can be collected on a filter paper and weighed. DATA 33.ul0s 1. First calculate the mass of exactly 0.005 moles of Potassium Chromate and Lead Nitrate Show work here: O.v05 194.19-0.971 .00Saえしs grams K,CrOa/calculated (this is what needs to be weighed) 0.97| grams Pb(NO;)»/calculated (this is what needs to be weighed)LCSS9 Mass of 250 ml beaker10.06g Mass of 250 ml flask 2.261 Mass of filter Paper Mass of filter paper and collected PbCrO4 (yellow stuff) Mass of flask with evaporated KNO, solution_ 973 Calculations (show the work): Mass of PbCrO4 (yellow stuff, this will require some subtraction)looMg ey eeam 1. /14- 110.06=394 1.679
2. Mass of KNO3 (this will require some subtraction) 0153 111.973-111.02 = 0.953 3. Calculate the moles of each product you obtained (use the masses you obtained): 114-2.619-110-06g=1.679 1,679 303.193 Moles of PbCrO4. - 0953 lo1.1032 Moles of KNO;,426~105 = 5,195 A103 9.426 a10 5.195 x 10-3 4. Compare the total mass of the reactants with the total mass of the products: 0.971+1.655 O.953+1.674 Total Mass of the Reactants d.626 2-632 vs. Total Mass of the Products Using Stoichiometry (show work) calculate the Theoretical Yield for both products starting with the masses that you weighed for the reagents. 5. Using Reagent K,cro md ,005mot L2Croul Tma I mul 0US mol Kacrou C) = .005 ml PoCro2 (322a=1.ul65g Pbcray I Imal Tmat Pocroa Tmal Theoretical Yield (g) of PbCrO4_l,lel65 Theoretical Yield (g) of KNO31.011 Using Reagent Pb(NO,) Theoretical Yield (g) of PbCrO4. Theoretical Yield (g) of KNO3 6. How does the theoretical yield (g) compare to the actual mass values you got???
Transcribed Image Text:2. Mass of KNO3 (this will require some subtraction) 0153 111.973-111.02 = 0.953 3. Calculate the moles of each product you obtained (use the masses you obtained): 114-2.619-110-06g=1.679 1,679 303.193 Moles of PbCrO4. - 0953 lo1.1032 Moles of KNO;,426~105 = 5,195 A103 9.426 a10 5.195 x 10-3 4. Compare the total mass of the reactants with the total mass of the products: 0.971+1.655 O.953+1.674 Total Mass of the Reactants d.626 2-632 vs. Total Mass of the Products Using Stoichiometry (show work) calculate the Theoretical Yield for both products starting with the masses that you weighed for the reagents. 5. Using Reagent K,cro md ,005mot L2Croul Tma I mul 0US mol Kacrou C) = .005 ml PoCro2 (322a=1.ul65g Pbcray I Imal Tmat Pocroa Tmal Theoretical Yield (g) of PbCrO4_l,lel65 Theoretical Yield (g) of KNO31.011 Using Reagent Pb(NO,) Theoretical Yield (g) of PbCrO4. Theoretical Yield (g) of KNO3 6. How does the theoretical yield (g) compare to the actual mass values you got???
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