Stoichiometry: A Precipitation Reaction and Limiting Reagent Data and Calculations Name: Lab Partners: DATA A. Mass of b
Posted: Wed Jul 06, 2022 10:55 am
questions, would prefer done in a googledoc
Stoichiometry: A Precipitation Reaction and Limiting Reagent Data and Calculations Name: Lab Partners: DATA A. Mass of beaker 1 B. Mass of beaker 1 and CaCl₂ C. Mass of beaker 2 D. Mass of beaker 2 and KIO3 E. Mass of watch glass and filter paper 57.167g F. Mass of watch glass and filter paper and Ca(103)2 after 1rst heating 57.869g H. Mass of beaker 2 and residue after 1rst heating 1. Mass of beaker 2 and residue after 2nd heating G. Mass of watch glass and filter paper and Ca(103)2 after 2nd heating 57.873g 2. Mass of KIO3 used 3. Mass of Ca(103)2 produced 4. Molar mass of CaCl₂ CALCULATIONS (Do not forget to use units and significant figures.) 1. Mass of CaCl₂ used 1.076g 1.192g 5. Molar mass of KIO3 6. Molar mass of Ca(103)2 106.031g 7. Molar mass of KCI 8. Moles of CaCl₂ used 127.984g 104.955g 127.725g 127.723g 0.706g 236.792g 110.980g/mol 174.900g/mol 74.55g/mol 0.0097mol 389.88g/mol
9. Moles of KIO3 used 10. Moles of Ca(103)2 produced 11. Theoretical yield of Ca(103)2 based on CaCl₂ used 12. Theoretical yield of Ca(103)2 based on KIO3 used 13. Limiting reagent 14. Reagent in excess 15. Percent yield of Ca(103)2 16. 17. 18. 19. 0.0056mol 20. KIO3 Mass of excess reactant (#14 above) that reacted Mass of excess reactant (#14 above) that remains unreacted 0.765g Theoretical yield of KCI based on limiting reagent Mass of residue in beaker 2 Sum of #17 plus #18 above 21. Moles of limiting reagent per mole of Ca(103)2 Cacl2 0.0018mol 1.706g 0.596g 118.5% 1.076g 0.417g 127.723g 1.182g 3.11g
Stoichiometry: A Precipitation Reaction and Limiting Reagent Calculations Name: Lab Partners: Show the equations used for each of the calculations that follow. You must use units after all numbers and express your answers using significant digits in order to receive full credit for your work. You may use a pencil to complete this page. CALCULATIONS Balanced chemical equation for reaction: 1. Mass of CaCl₂ used (B-A) (106.031 104.955)g = 1.076g 4. Molar mass of CaCl₂ 40.08 + 2x35.45 = 110.980g/mol 5. Molar mass of KIO3 126.90 + 3x16= 174.900g/mol 6. Molar mass of Ca(103)2 40.88 + 16 x 6 + 126.9 x 2 = 389.88g/mol 2. Mass of KIO3 used (127.90 +126.792)g = 1.192g 3. Mass of Ca(103)2 produced (57.873-57.167)g = 0.706g 7. Molar mass of KCI 39.1 + 35.45 = 74.55g/mol 8. Moles of CaCl₂ used 1.076/110.98 = 0.0097mol 9. Moles of KIO3 used 1.192/214.001 = 0.0056mol 10. Moles of Ca(103)2 produced 0.706/389.88 = 0.0018mol
11. Theoretical yield of Ca(103)2 based on CaCl₂ used Theoretical yield of Ca(103)2 based on Cacl2 used. CaCl2 + 2KIO3 -- 2KCI + Ca (103)2 1 mol of Ca (103)2 produced from 1 mol of Cacl2 Theoretical yield = 1/1 x 1.706g = 1.706g 12. Theoretical yield of Ca(103)2 based on KIO3 used Theoretical yield of Ca(103)2 based on KIO3 used. 1 mol of Ca(103)2 produced from 2 mol of KIO3 Theoretical yield = 1/2 x 1.192g = 0.596g 13. Limiting reagent KIO3 14. Reagent in excess CaCl2 15. Percent yield of Ca(103)2 Actual yield/Theoretical yield x 100 = 0.706/0.596 x 100 = 118.5% 16. Mass of excess reactant (#14 above) that reacted 1.076g 17. Mass of excess reactant (#14 above) that remains unreacted 18. Theoretical yield of KCI based on limiting, reagent mol of KCI producéd = mol of limiting reactant x molar ratio = 0.0056mol x [2mol KCI/2mol KIO3] = 0.0056mol Theoretical yield of KCI = mol x molar mass = 0.0056mol x 74.559g/mol = 0.417g 19. Mass of residue in beaker 2 127.723g Total mass of excess - mass of excess consumed Given total mass of excess = 1.076g mol of excess consumed = mol of limiting reactant x molar ratio 21. Moles of limiting reagent per mole of Ca(103)2 = 0.0056mol/0.0018mol = 3.11g = 0.0056mol KIO3 x [1mol CaC12/2mol KIO3] = 0.0028mol Mass of excess consumed = mol x molar mass= 0.0028mol x 110.989g/mol = 0.3107g ; mass of excees remain = 1.076-0.3107 = 0.765g 20. Sum of #17 plus #18 above 0.765g + 0.417g = 1.182g
Stoichiometry: A Precipitation Reaction and Limiting Reagent Lab Questions Name: Lab Partners: 1. Does your data support the Law of Conservation of Matter? Use your measurements to validate your response. # 2. According to your data, what are the coefficients for the limiting reagent and the precipitate (refer to #21 in calculations)? Are these in agreement with the balanced chemical equation? If yes, explain why this is to be expected. If not, identify some possible sources of error in your experimental procedure. 3. What substances were present in the residue in beaker 2? Explain. 4. Which measurements should be equal to the answer in #20? Explain.
I only need the lab Stoichiometry: A Precipitation Reaction and Limiting Reagent Data and Calculations Name: Lab Partners: DATA A. Mass of beaker 1 B. Mass of beaker 1 and CaCl₂ C. Mass of beaker 2 D. Mass of beaker 2 and KIO3 E. Mass of watch glass and filter paper 57.167g F. Mass of watch glass and filter paper and Ca(103)2 after 1rst heating 57.869g H. Mass of beaker 2 and residue after 1rst heating 1. Mass of beaker 2 and residue after 2nd heating G. Mass of watch glass and filter paper and Ca(103)2 after 2nd heating 57.873g 2. Mass of KIO3 used 3. Mass of Ca(103)2 produced 4. Molar mass of CaCl₂ CALCULATIONS (Do not forget to use units and significant figures.) 1. Mass of CaCl₂ used 1.076g 1.192g 5. Molar mass of KIO3 6. Molar mass of Ca(103)2 106.031g 7. Molar mass of KCI 8. Moles of CaCl₂ used 127.984g 104.955g 127.725g 127.723g 0.706g 236.792g 110.980g/mol 174.900g/mol 74.55g/mol 0.0097mol 389.88g/mol
9. Moles of KIO3 used 10. Moles of Ca(103)2 produced 11. Theoretical yield of Ca(103)2 based on CaCl₂ used 12. Theoretical yield of Ca(103)2 based on KIO3 used 13. Limiting reagent 14. Reagent in excess 15. Percent yield of Ca(103)2 16. 17. 18. 19. 0.0056mol 20. KIO3 Mass of excess reactant (#14 above) that reacted Mass of excess reactant (#14 above) that remains unreacted 0.765g Theoretical yield of KCI based on limiting reagent Mass of residue in beaker 2 Sum of #17 plus #18 above 21. Moles of limiting reagent per mole of Ca(103)2 Cacl2 0.0018mol 1.706g 0.596g 118.5% 1.076g 0.417g 127.723g 1.182g 3.11g
Stoichiometry: A Precipitation Reaction and Limiting Reagent Calculations Name: Lab Partners: Show the equations used for each of the calculations that follow. You must use units after all numbers and express your answers using significant digits in order to receive full credit for your work. You may use a pencil to complete this page. CALCULATIONS Balanced chemical equation for reaction: 1. Mass of CaCl₂ used (B-A) (106.031 104.955)g = 1.076g 4. Molar mass of CaCl₂ 40.08 + 2x35.45 = 110.980g/mol 5. Molar mass of KIO3 126.90 + 3x16= 174.900g/mol 6. Molar mass of Ca(103)2 40.88 + 16 x 6 + 126.9 x 2 = 389.88g/mol 2. Mass of KIO3 used (127.90 +126.792)g = 1.192g 3. Mass of Ca(103)2 produced (57.873-57.167)g = 0.706g 7. Molar mass of KCI 39.1 + 35.45 = 74.55g/mol 8. Moles of CaCl₂ used 1.076/110.98 = 0.0097mol 9. Moles of KIO3 used 1.192/214.001 = 0.0056mol 10. Moles of Ca(103)2 produced 0.706/389.88 = 0.0018mol
11. Theoretical yield of Ca(103)2 based on CaCl₂ used Theoretical yield of Ca(103)2 based on Cacl2 used. CaCl2 + 2KIO3 -- 2KCI + Ca (103)2 1 mol of Ca (103)2 produced from 1 mol of Cacl2 Theoretical yield = 1/1 x 1.706g = 1.706g 12. Theoretical yield of Ca(103)2 based on KIO3 used Theoretical yield of Ca(103)2 based on KIO3 used. 1 mol of Ca(103)2 produced from 2 mol of KIO3 Theoretical yield = 1/2 x 1.192g = 0.596g 13. Limiting reagent KIO3 14. Reagent in excess CaCl2 15. Percent yield of Ca(103)2 Actual yield/Theoretical yield x 100 = 0.706/0.596 x 100 = 118.5% 16. Mass of excess reactant (#14 above) that reacted 1.076g 17. Mass of excess reactant (#14 above) that remains unreacted 18. Theoretical yield of KCI based on limiting, reagent mol of KCI producéd = mol of limiting reactant x molar ratio = 0.0056mol x [2mol KCI/2mol KIO3] = 0.0056mol Theoretical yield of KCI = mol x molar mass = 0.0056mol x 74.559g/mol = 0.417g 19. Mass of residue in beaker 2 127.723g Total mass of excess - mass of excess consumed Given total mass of excess = 1.076g mol of excess consumed = mol of limiting reactant x molar ratio 21. Moles of limiting reagent per mole of Ca(103)2 = 0.0056mol/0.0018mol = 3.11g = 0.0056mol KIO3 x [1mol CaC12/2mol KIO3] = 0.0028mol Mass of excess consumed = mol x molar mass= 0.0028mol x 110.989g/mol = 0.3107g ; mass of excees remain = 1.076-0.3107 = 0.765g 20. Sum of #17 plus #18 above 0.765g + 0.417g = 1.182g
Stoichiometry: A Precipitation Reaction and Limiting Reagent Lab Questions Name: Lab Partners: 1. Does your data support the Law of Conservation of Matter? Use your measurements to validate your response. # 2. According to your data, what are the coefficients for the limiting reagent and the precipitate (refer to #21 in calculations)? Are these in agreement with the balanced chemical equation? If yes, explain why this is to be expected. If not, identify some possible sources of error in your experimental procedure. 3. What substances were present in the residue in beaker 2? Explain. 4. Which measurements should be equal to the answer in #20? Explain.