1 / 9

Percentage Yield

Percentage Yield. Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data. Include: theoretical yield , experimental yield Additional KEY Terms Actual yield.

nevina
Télécharger la présentation

Percentage Yield

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Percentage Yield

  2. Identify the limiting reactantand calculate the mass of a product, given the reaction equation and reactant data. • Include: theoretical yield, experimental yield • Additional KEY Terms • Actual yield

  3. Theoretical yield: amounts of product calculated from the limiting reagent. Actual yield: amount produced during the conducted experiment. Percentage yield: ratio of actual to theoretical yield expressed in percentage. actual yield x 100 percent yield = theoretical yield

  4. Reactions are influenced by "external" factors which effect the amount of yield produced: • Concentration • State (s, l, g, aq) • Temperature • Pressure (of gas reactants) • Human error Poor recovery Impurities in reactants Miscalculation A side reaction occurs

  5. 2 KClO3 2 KCl + 3 O2 5.00 g of KClO3 is heated and decomposes to yield 1.78 g of O2. What is the % yield? 5.00 g KClO3 1 mole 3 mole O2 32.0 g O2 1 mole O2 2 mole KClO3 122.5 g = 1.96 g O2 actual yield 1.78 g x 100 90.8 % percent yield = 1.96 g theoretical yield

  6. How much O2 would be produced if the percentage yield was 78.5%? Ac. yield 1.96 g x 100 78.5 % = % yield = Th. yield 1.53 g = Ac. yield

  7. What is the % yield if 58 g H2O are produced by combining 60.0 g O2 and 7.0 g H2? O2(g) + 2 H2 (g) 2 H2O (g) 60.0 g O2 1 mole O2 2 mole H2 2.0 g H2 1 mole O2 1 mole H2 32.0 g O2 = 7.5 g H2 H2 - limiting reactant, O2 - excess reactant

  8. 7.0 g H2 1 mole H2 2 mole H2O 18.0 g 2 mole H2 1 mole 2.0 g H2 = 63 g H2O produced actual yield 58 g x 100 92 % percent yield = theoretical yield 63 g

  9. CAN YOU / HAVE YOU? • Identifythe limiting reactantand calculate the mass of a product, given the reaction equation and reactant data. • Include: theoretical yield, experimental yield • Additional KEY Terms • Actual yield

More Related