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2011 topic 01 lecture 3 - limiting reactant and percent yield

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1) The document discusses quantitative chemistry and limiting/excess reactants. It provides examples of calculations to determine the limiting reactant and amount of product formed based on the limiting reactant. 2) It also discusses percent yield calculations. For the example reaction of gold (III) chloride and lead, the theoretical yield of gold is calculated to be 9.09g based on the given amount of gold (III) chloride. 3) The actual yield obtained in the example is 1.05g gold. The percentage yield is calculated to be 11.6%, expressing the proportion of theoretical yield that was actually obtained.

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IB Chemistry Power Points Topic 1 Quantitative Chemistry www.pedagogics.ca Lecture 3 Limiting Reactant Percent Yield
Consider the following reaction 2 H2 + O2  2 H2O
Reactants are combined in perfect proportions 3 molecules 6 molecules 6 molecules
In reality this never happens 3 molecules 6 molecules 6 molecules
Consider 3 molecules 4 molecules 4 molecules + leftover oxygen
Consider EXCESS LIMITING REACTANT Amount of REACTANT PRODUCT is determined by limiting reactant
Consider 2 molecules 6 molecules 4 molecules + leftover hydrogen
Consider LIMITING EXCESS REACTANT Amount of REACTANT PRODUCT is determined by limiting reactant
Given 24 grams of O2 and 5.0 grams of H2 determine the mass of H2O produced. 2 H2 + O2  2 H2O the mass of H2O produced will be determined by the limiting reactant - do TWO calculations
calculation for 24 grams of O2 24 g O2 2 H2O 18.0 g mol-1 = 27 g of H2O 32.0 g mol-1 1 O2
calculation for 24 grams of O2 24 g O2 2 H2O 18.0 g mol-1 = 27 g of H2O 32.0 g mol-1 1 O2 calculation 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
calculation for 24 grams of O2 24 g O2 2 H2O 18.0 g mol-1 = 27 g of H2O 32.0 g mol-1 1 O2 O2 is the LIMITING REACTANT and determines the amount of product calculation 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 H2 is the EXCESS REACTANT (some would be left over)
How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen
How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen given 24 grams of O2 24 g O2 2 H2 2.0 g mol-1 = 3.0 g of H2 32.0 g mol-1 1 O2 3.0 g of H2 reacts so
How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen given 24 grams of O2 24 g O2 2 H2 2.0 g mol-1 = 3.0 g of H2 32.0 g mol-1 1 O2 3.0 g of H2 reacts so 5.0 g – 3.0 g = 2.0 g of hydrogen remains
Percent Yield Enoch the Red, an alchemist, wants to try to turn lead into gold (which you can’t do chemically). He finds that mixing lead with an unidentified compound (gold III chloride) actually produces small amounts of gold. The reaction is as follows: 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
Percent Yield Enoch reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction? 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
Percent Yield Enoch reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction? 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
Percent Yield Enoch reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of 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
Percent Yield Enoch recovers only 1.05 g of gold from the reaction. This could be for many different reasons some product was lost in the recovery process the reaction did not go to completion the AuCl3 is not pure
Percent Yield the percentage yield expresses the proportion of the expected product that was actually obtained. actual % yield= ×100% theoretical 1.05 % yield= ×100%=11.6% 9.09

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2011 topic 01 lecture 3 - limiting reactant and percent yield

  • 1. IB Chemistry Power Points Topic 1 Quantitative Chemistry www.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 molecules 6 molecules 6 molecules
  • 4. In reality this never happens 3 molecules 6 molecules 6 molecules
  • 5. Consider 3 molecules 4 molecules 4 molecules + leftover oxygen
  • 6. Consider EXCESS LIMITING REACTANT Amount of REACTANT PRODUCT is determined by limiting reactant
  • 7. Consider 2 molecules 6 molecules 4 molecules + leftover hydrogen
  • 8. Consider LIMITING EXCESS REACTANT Amount of REACTANT PRODUCT is determined by limiting reactant
  • 9. Given 24 grams of O2 and 5.0 grams of H2 determine the mass of H2O produced. 2 H2 + O2  2 H2O the mass of H2O produced will be determined by the limiting reactant - do TWO calculations
  • 10. calculation for 24 grams of O2 24 g O2 2 H2O 18.0 g mol-1 = 27 g of H2O 32.0 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 H2O 32.0 g mol-1 1 O2 calculation 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 H2O 32.0 g mol-1 1 O2 O2 is the LIMITING REACTANT and determines the amount of product calculation 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 H2 is the EXCESS REACTANT (some would be left over)
  • 13. How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen
  • 14. How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen given 24 grams of O2 24 g O2 2 H2 2.0 g mol-1 = 3.0 g of H2 32.0 g mol-1 1 O2 3.0 g of H2 reacts so
  • 15. How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen given 24 grams of O2 24 g O2 2 H2 2.0 g mol-1 = 3.0 g of H2 32.0 g mol-1 1 O2 3.0 g of H2 reacts so 5.0 g – 3.0 g = 2.0 g of hydrogen remains
  • 16. Percent Yield Enoch the Red, an alchemist, wants to try to turn lead into gold (which you can’t do chemically). He finds that mixing lead with an unidentified compound (gold III chloride) actually produces small amounts of gold. The reaction is as follows: 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
  • 17. Percent Yield Enoch reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction? 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
  • 18. Percent Yield Enoch reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction? 2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
  • 19. Percent Yield Enoch reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of 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 Yield Enoch recovers only 1.05 g of gold from the reaction. This could be for many different reasons some product was lost in the recovery process the reaction did not go to completion the AuCl3 is not pure
  • 21. Percent Yield the percentage yield expresses the proportion of the expected product that was actually obtained. actual % yield= ×100% theoretical 1.05 % yield= ×100%=11.6% 9.09