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General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 6.8 Percent Yield and Limiting Reactants Chapter 6 Chemical Reactions and Quantities.

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Presentation on theme: "General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 6.8 Percent Yield and Limiting Reactants Chapter 6 Chemical Reactions and Quantities."— Presentation transcript:

1 General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 6.8 Percent Yield and Limiting Reactants Chapter 6 Chemical Reactions and Quantities © 2013 Pearson Education, Inc. Lectures

2 © 2013 Pearson Education, Inc. Chapter 6, Section 8 2 Theoretical, Actual, and Percent Yield  Theoretical yield is the maximum amount of product, which is calculated using the balanced equation.  Actual yield is the amount of product obtained when the reaction takes place.  Percent yield is the ratio of actual yield to theoretical yield.

3 © 2013 Pearson Education, Inc. Chapter 6, Section 8 3 Guide to Calculations for Percent Yield

4 © 2013 Pearson Education, Inc. Chapter 6, Section 8 4 Calculating Percent Yield On a space shuttle, LiOH is used to absorb exhaled CO 2 from breathing air to form LiHCO 3. What is the percent yield of the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO 3 ?

5 © 2013 Pearson Education, Inc. Chapter 6, Section 8 5 Calculating Percent Yield On a space shuttle, LiOH is used to absorb exhaled CO 2 from breathing air to form LiHCO 3. What is the percent yield of the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO 3 ? Step 1 State the given and needed quantities. Analyze the Problem.

6 © 2013 Pearson Education, Inc. Chapter 6, Section 8 6 Calculating Percent Yield What is the percent yield of the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO 3 ? Step 2 Write a plan to calculate the theoretical yield and the percent yield. Calculation of theoretical yield: grams molar moles mole-mole moles molar grams LiOH mass LiOH factor LiHCO 3 mass LiHCO 3 Calculation of percent yield:

7 © 2013 Pearson Education, Inc. Chapter 6, Section 8 7 Calculating Percent Yield What is the percent yield of the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO 3 ? Step 3 Write the molar mass for the reactant and the mole–mole factor from the balanced equation.

8 © 2013 Pearson Education, Inc. Chapter 6, Section 8 8 Calculating Percent Yield What is the percent yield of the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO 3 ? Step 3 Write the molar mass for the reactant and the mole–mole factor from the balanced equation.

9 © 2013 Pearson Education, Inc. Chapter 6, Section 8 9 Calculating Percent Yield What is the percent yield of the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO 3 ? Step 4 Solve for the percent yield ratio by dividing the actual yield (given) by the theoretical yield and multiplying the result by 100%. Calculation of theoretical yield:

10 © 2013 Pearson Education, Inc. Chapter 6, Section 8 10 Calculating Percent Yield What is the percent yield of the reaction if 50.0 g of LiOH gives 72.8 g of LiHCO 3 ? Step 4 Solve for the percent yield ratio by dividing the actual yield (given) by the theoretical yield and multiplying the result by 100%. Calculation of percent yield:

11 © 2013 Pearson Education, Inc. Chapter 6, Section 8 11 Limiting Reactant A limiting reactant in a chemical reaction is the substance that is used up first and limits the amount of product that can form. The other reactant, called the excess reactant, is left over.

12 © 2013 Pearson Education, Inc. Chapter 6, Section 8 12 Example: Limiting Reactant How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.

13 © 2013 Pearson Education, Inc. Chapter 6, Section 8 13 Example: Limiting Reactant How many peanut butter sandwiches could be made from 8 slices of bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item.

14 © 2013 Pearson Education, Inc. Chapter 6, Section 8 14 Calculating Moles of Product from Limiting Reactant Calculate the moles of product (CH 3 OH) that can form given 3.00 moles of CO reacts with 5.00 moles of H 2. What is the limiting reactant? a. What mole–mole equalities will be needed in the. calculation? b. What are the mole–mole factors from these. equalities? c. What is the number of moles of CH 3 OH from each. reactant? d. What is the limiting reactant for the reaction?

15 © 2013 Pearson Education, Inc. Chapter 6, Section 8 15 Calculating Moles of Product from Limiting Reactant Calculate the moles of product (CH 3 OH) that can form given 3.00 moles of CO reacts with 5.00 moles of H 2. What is the limiting reactant? a. What mole–mole equalities will be needed in the calculation? 1 mole of CO = 1 mole of CH 3 OH 2 moles of H 2 = 1 mole of CH 3 OH

16 © 2013 Pearson Education, Inc. Chapter 6, Section 8 16 Calculating Moles of Product from Limiting Reactant Calculate the moles of product (CH 3 OH) that can form given 3.00 moles of CO reacts with 5.00 moles of H 2. What is the limiting reactant? b. What are the mole–mole factors from these equalities?

17 © 2013 Pearson Education, Inc. Chapter 6, Section 8 17 Calculating Moles of Product from Limiting Reactant Calculate the moles of product (CH 3 OH) that can form given 3.00 moles of CO reacts with 5.00 moles of H 2. What is the limiting reactant? c. What is the number of moles of CH 3 OH from each reactant?

18 © 2013 Pearson Education, Inc. Chapter 6, Section 8 18 Calculating Moles of Product from Limiting Reactant Calculate the moles of product (CH 3 OH) that can form given 3.00 moles of CO reacts with 5.00 moles of H 2. What is the limiting reactant? d. What is the limiting reactant for the reaction? The limiting reactant is H 2, which produces the smaller number of moles of product, 2.50 moles of CH 3 OH.

19 © 2013 Pearson Education, Inc. Chapter 6, Section 8 19 Calculating Moles of Product from Limiting Reactant Calculate the moles of product (CH 3 OH) that can form given 3.00 moles of CO reacts with 5.00 moles of H 2. What is the limiting reactant?

20 © 2013 Pearson Education, Inc. Chapter 6, Section 8 20 Calculating Mass of Product from Limiting Reactant

21 © 2013 Pearson Education, Inc. Chapter 6, Section 8 21 Calculating Mass of Product from Limiting Reactant How many grams of CO can be produced from a mixture of 70.0 g of SiO 2 and 50.0 g of C? Step 1 State the given and needed quantities. Analyze the Problem.

22 © 2013 Pearson Education, Inc. Chapter 6, Section 8 22 Calculating Mass of Product from Limiting Reactant How many grams of CO can be produced from a mixture of 70.0 g of SiO 2 and 50.0 g of C? Step 2 Use coefficients to write mole–mole factors; write molar mass factors, if needed.

23 © 2013 Pearson Education, Inc. Chapter 6, Section 8 23 Calculating Mass of Product from Limiting Reactant How many grams of CO can be produced from a mixture of 70.0 g of SiO 2 and 50.0 g of C? Step 2 Use coefficients to write mole–mole factors; write molar mass factors, if needed.

24 © 2013 Pearson Education, Inc. Chapter 6, Section 8 24 Calculating Mass of Product from Limiting Reactant How many grams of CO can be produced from a mixture of 70.0 g of SiO 2 and 50.0 g of C? Step 3 Calculate the number of moles of product from each reactant and determine the limiting reactant.

25 © 2013 Pearson Education, Inc. Chapter 6, Section 8 25 Calculating Mass of Product from Limiting Reactant How many grams of CO can be produced from a mixture of 70.0 g of SiO 2 and 50.0 g of C? Step 4 Use the molar mass to convert the smaller number of moles of product to grams. 1 mole of CO = 28.0 g of CO

26 © 2013 Pearson Education, Inc. Chapter 6, Section 8 26 Learning Check If 48.2 grams of Ca are mixed with 31.0 grams of N 2, how many grams of Ca 3 N 2 can be produced?

27 © 2013 Pearson Education, Inc. Chapter 6, Section 8 27 Solution If 48.2 grams of Ca are mixed with 31.0 grams of N 2, how many grams of Ca 3 N 2 can be produced? Step 1 State the given and needed quantities. Analyze the Problem.

28 © 2013 Pearson Education, Inc. Chapter 6, Section 8 28 Solution If 48.2 grams of Ca are mixed with 31.0 grams of N 2, how many grams of Ca 3 N 2 can be produced? Step 2 Use coefficients to write mole–mole factors; write molar mass factors, if needed. 1 mole of Ca = 40.1 g of Ca 1 mole of N 2 = 28.0 g of N 2

29 © 2013 Pearson Education, Inc. Chapter 6, Section 8 29 Solution If 48.2 grams of Ca are mixed with 31.0 grams of N 2, how many grams of Ca 3 N 2 can be produced? Step 2 Use coefficients to write mole–mole factors; write molar mass factors, if needed.

30 © 2013 Pearson Education, Inc. Chapter 6, Section 8 30 Solution If 48.2 grams of Ca are mixed with 31.0 grams of N 2, how many grams of Ca 3 N 2 can be produced? Step 3 Calculate the number of moles of product from each reactant and determine the limiting reactant. The limiting reactant is Ca, which produces mole Ca 3 N 2.

31 © 2013 Pearson Education, Inc. Chapter 6, Section 8 31 Solution If 48.2 grams of Ca are mixed with 31.0 grams of N 2, how many grams of Ca 3 N 2 can be produced? Step 4 Use the molar mass to convert the smaller number of moles of product to grams.


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