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Limiting Reactant Percent Yield

Limiting Reactant Percent Yield. Consider the following reaction 2 H 2 + O 2  2 H 2 O. Reactants are combined in perfect proportions. 3 molecules. 6 molecules. 6 molecules. This is a theoretic reaction. 3 molecules. 6 molecules. 6 molecules. In reality this never happens.

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Limiting Reactant Percent Yield

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  1. Limiting Reactant Percent Yield

  2. Consider the following reaction2 H2 + O2 2 H2O

  3. Reactants are combined in perfect proportions 3 molecules 6 molecules 6 molecules

  4. This is a theoretic reaction 3 molecules 6 molecules 6 molecules

  5. In reality this never happens 3 molecules 6 molecules 6 molecules

  6. Consider 3 molecules 4 molecules 4 molecules + leftover oxygen

  7. Consider EXCESS REACTANT Amount of PRODUCT is determined by limiting reactant LIMITING REACTANT

  8. Consider 2 molecules 6 molecules 4 molecules + leftover hydrogen

  9. Consider LIMITING REACTANT Amount of PRODUCT is determined by limiting reactant EXCESS REACTANT

  10. 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

  11. calculation for 24 grams of O2 = 27 g of H2O

  12. calculation for 24 grams of O2 = 27 g of H2O calculation for 5.0 grams of H2 = 45 g of H2O

  13. calculation for 24 grams of O2 = 27 g of H2O O2 is the LIMITING REACTANT and determines the amount of product calculation for 5.0 grams of H2 = 45 g of H2O H2 is the EXCESS REACTANT (some would be left over)

  14. How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen

  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 = 3.0 g of H2 3.0 g of H2 reacts so

  16. 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 = 3.0 g of H2 3.0 g of H2 reacts so 5.0 g – 3.0 g = 2.0 g of hydrogen remains

  17. Percent Yield 2 AuCl3 +3 Pb 3 PbCl2 + 2 Au Dougie, 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:

  18. Percent Yield 2 AuCl3 +3 Pb 3 PbCl2 + 2 Au Dougiereacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction?

  19. Percent Yield 2 AuCl3 +3 Pb 3 PbCl2 + 2 Au Dougiereacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction?

  20. Percent Yield 2 AuCl3 +3 Pb 3 PbCl2 + 2 Au Dougiereacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction? given 14.0 g of AuCl3 = 9.09 g Au

  21. Percent Yield Dougierecovers 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

  22. Percent Yield The percentage yield expresses the proportion of the expected product that was actually obtained.

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