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Chapter 9 Chemical Quantities in Reactions

On a space shuttle, the LiOH in the canisters removes CO 2 from the air. Chapter 9 Chemical Quantities in Reactions. 9.3 Limiting Reactants. Theoretical, Actual, and Percent Yield. Theoretical yield The maximum amount of product calculated using the balanced equation. Actual yield

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Chapter 9 Chemical Quantities in Reactions

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  1. On a space shuttle, the LiOH in the canisters removes CO2 from the air. Chapter 9 Chemical Quantities in Reactions 9.3 Limiting Reactants

  2. Theoretical, Actual, and Percent Yield Theoretical yield • The maximum amount of product calculated using the balanced equation. Actual yield • The amount of product obtained when the reaction takes place. Percent yield • The ratio of actual yield to theoretical yield. Percent yield = actual yield (g) x 100 theoretical yield (g)

  3. Calculating Percent Yield Suppose you prepare cookie dough to make 5 dozen cookies. The phone rings and you answer. While talking, a sheet of 12 cookies burns and you have to throw them out. The rest of the cookies are okay. What is the percent yield of edible cookies? Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield48 cookies x 100% = 80.% yield 60 cookies

  4. Calculations for Percent Yield

  5. Learning Check Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O2(g) 2CO(g) What is the percent yield if 40.0 g of CO is produced when 30.0 g of O2 is used? 1) 25.0% 2) 75.0% 3) 76.2%

  6. Solution STEP 1Write the given and needed quantities. Given 40.0 g of CO produced (actual) 30.0 g of O2 used Need percent yield of CO STEP 2 Write a plan to calculate the theoretical yield and the percent yield. g of O2 moles of moles of g of CO O2 CO (theoretical) Percent yield of CO = g of CO (actual) x 100% g of CO (theoretical)

  7. Solution (continued) STEP 3 Write the molar masses and the mole-mole factors from the balanced equation. 1 mol of O2 = 32.00 g of O2 1 mol O2 and 32.00 g O2 32.00 g O2 1 mol O2 1 mol of O2 = 2 mol of CO 1 mol O2 and 2 mol CO 2 mol CO1 mol O2 1 mol of CO= 28.01 g of CO 1 mol COand 28.01 g CO 28.01 g CO1 mol CO

  8. Solution (continued) STEP 4 Calculate the percent yield ratio by dividing the actual yield (given) by the theoretical yield x 100%. 30.0 g O2 x 1 mol O2 x 2 mol CO x 28.01 g CO 32.00 g O2 1 mol O2 1 mol CO = 52.5 g of CO (theoretical) Percent yield: 40.0 g CO (actual) x 100 = 76.2% yield (3) 52.5 g CO (theoretical)

  9. Learning Check When N2 and 5.00 g of H2 are mixed, the reaction produces 16.0 g of NH3. What is the percent yield for the reaction? N2(g) + 3H2(g) 2NH3(g) 1) 31.3 % 2) 56.7 % 3) 80.0 %

  10. Solution STEP 1Write the given and needed quantities. Given 16.0 g of NH3 produced (actual) 5.00 g of H2 Need percent yield of NH3 STEP 2 Write a plan to calculate the theoretical yield and the percent yield. g of H2 moles of H2 moles of NH3 g of NH3 (theoretical) Percent yield = g of NH3(actual) x 100% g of NH3(theoretical)

  11. Solution (continued) STEP 3 Write the molar masses and the mole-mole factors from the balanced equation. 1 mol of H2 = 2.016 g of H2 1 mol H2 and 2.016 g H2 2.016 g H2 1 mol H2 3 mol of H2 = 2 mol of NH3 3 mol H2 and 2 mol NH3 2 mol NH3 3 mol H2 1 mol of NH3 = 17.03 g of NH3 1 mol NH3 and 17.03 g NH3 17.03 g NH3 1 mol NH3

  12. Solution (continued) STEP 4 Calculate the percent yield ratio by dividing the actual yield (given) by the theoretical yield x 100%. 5.00 g H2 x 1 mol H2 x 2 mol NH3 x 17.03 g NH3 2.016 g H2 3 mol H2 1 mol NH3 = 28.2 g of NH3 (theoretical) Percent yield = 16.0 g NH3 x 100 = 56.7 % (2) 28.2 g NH3

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