Stoichiometry: A Precipitation Reaction and Limiting Reagent Calculations Name: Lab Partners: Show the equations used fo

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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 CaCl, 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 KIO, 126.90 + 3x16= 174.900g/mol 6. Molar mass of Ca(10₁)2 40.88 + 16 x 6+126.9 x 2 = 389.88g/mol 2. Mass of KIO, 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 KIO, used 1.192/214.001 = 0.0056mol 10. Moles of Ca(103); produced 0.706/389.88 = 0.0018mol
11. Theoretical yield of Ca(10), 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), based on KIO, 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 KC1/2mol 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 21. Moles of limiting reagent per mole of Ca(103) = 0.0056mol/0.0018mol = 3.11g Total mass of excess-mass of excess consumed Given total mass of excess=1.076g mol of excess consumed = mol of limiting reactantx molar ratio -0.0056mol KIO3 x [1mol Ca02/2mol K303) = 0.0028mal Mass of excess consumed-mol x molar mass= 0.0028mal 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
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.