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2011 topic 01 lecture 3 - limiting reactant and percent yield

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  • 1. IB Chemistry Power Points Topic 1 Quantitative Chemistrywww.pedagogics.ca Lecture 3 Limiting Reactant Percent Yield
  • 2. Consider the following reaction 2 H2 + O2  2 H2O
  • 3. Reactants are combined in perfect proportions 3 molecules6 molecules 6 molecules
  • 4. In reality this never happens 3 molecules6 molecules 6 molecules
  • 5. Consider 3 molecules4 molecules 4 molecules + leftover oxygen
  • 6. Consider EXCESSLIMITING REACTANT Amount ofREACTANT PRODUCT is determined by limiting reactant
  • 7. Consider 2 molecules6 molecules 4 molecules + leftover hydrogen
  • 8. Consider LIMITINGEXCESS REACTANT Amount ofREACTANT PRODUCT is determined by limiting reactant
  • 9. Given 24 grams of O2 and 5.0 grams of H2determine the mass of H2O produced. 2 H2 + O2  2 H2Othe mass of H2O produced will be determined by thelimiting reactant - do TWO calculations
  • 10. calculation for 24 grams of O2 24 g O2 2 H2O 18.0 g mol-1 = 27 g of H2O32.0 g mol-1 1 O2
  • 11. calculation for 24 grams of O2 24 g O2 2 H2O 18.0 g mol-1 = 27 g of H2O32.0 g mol-1 1 O2calculation for 5.0 grams of H2 5 g H2 2 H2O 18.0 g mol-1 = 45 g of H2O 2.0 g mol-1 2 H2
  • 12. calculation for 24 grams of O2 24 g O2 2 H2O 18.0 g mol-1 = 27 g of H2O32.0 g mol-1 1 O2O2 is the LIMITING REACTANT and determines theamount of productcalculation for 5.0 grams of H2 5 g H2 2 H2O 18.0 g mol-1 = 45 g of H2O 2.0 g mol-1 2 H2H2 is the EXCESS REACTANT (some would be leftover)
  • 13. How much hydrogen gas would be left over?To calculate, first determine how much reactswith all of the oxygen
  • 14. How much hydrogen gas would be left over?To calculate, first determine how much reactswith all of the oxygengiven 24 grams of O2 24 g O2 2 H2 2.0 g mol-1 = 3.0 g of H232.0 g mol-1 1 O23.0 g of H2 reacts so
  • 15. How much hydrogen gas would be left over?To calculate, first determine how much reactswith all of the oxygengiven 24 grams of O2 24 g O2 2 H2 2.0 g mol-1 = 3.0 g of H232.0 g mol-1 1 O23.0 g of H2 reacts so5.0 g – 3.0 g = 2.0 g of hydrogen remains
  • 16. Percent YieldEnoch the Red, an alchemist,wants to try to turn lead intogold (which you can’t dochemically). He finds thatmixing lead with anunidentified compound (goldIII chloride) actually producessmall amounts of gold. Thereaction is as follows: 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
  • 17. Percent YieldEnoch reacts 14.0 g of gold IIIchloride with excess leadmetal. What would be themaximum, THEORETICAL yieldof this reaction? 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
  • 18. Percent YieldEnoch reacts 14.0 g of gold IIIchloride with excess leadmetal. What would be themaximum, THEORETICAL yieldof this reaction? 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
  • 19. Percent YieldEnoch reacts 14.0 g of gold IIIchloride with excess leadmetal. What would be themaximum, THEORETICAL yieldof this reaction? 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au given 14.0 g of AuCl3 14.0 g AuCl3 2 AuCl3 196.97 g mol-1 = 9.09 g Au 303.5 g mol-1 2 Au
  • 20. Percent YieldEnoch recovers only 1.05 g ofgold from the reaction. Thiscould be for many differentreasons some product was lost in the recovery process the reaction did not go to completion the AuCl3 is not pure
  • 21. Percent Yieldthe percentage yield expressesthe proportion of the expectedproduct that was actuallyobtained. actual % yield= ×100% theoretical 1.05 % yield= ×100%=11.6% 9.09

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