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Chapter 9 Limiting Reactants and Percent Yield A. Limiting Reactant 1. If you are given one dozen loaves of bread, a gallon of mustard, and three pieces.

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Presentation on theme: "Chapter 9 Limiting Reactants and Percent Yield A. Limiting Reactant 1. If you are given one dozen loaves of bread, a gallon of mustard, and three pieces."— Presentation transcript:

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2 Chapter 9 Limiting Reactants and Percent Yield

3 A. Limiting Reactant 1. If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of ham, how many ham sandwiches can you make? 2. The limiting reactant is the reactant you run out of first. 3. The excess reactant is the one you have left over.

4 4. The limiting reactant determines how much product you can make 5. What is the limiting reactant of your sandwich supplies? ham 6. What is the excess reactant of your sandwich supplies? gallon of mustard

5 B. How do you find out? 1. Do two stoichiometry problems. 2. The one that makes the least product is the limiting reagent. 3. For example Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

6 If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reactant

7 C. Clarify 1. Limiting Reactant –used up in a reaction –determines the amount of product 2. Excess Reactant –added to ensure that the other reactant is completely used up –cheaper & easier to recycle

8 3. Example a. Available Ingredients –4 slices of bread –1 jar of peanut butter –1/2 jar of jelly b. Limiting Reactant –bread c. Excess Reactants –peanut butter and jelly

9 E. Calculating Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: –limiting reactant 4. Do Stoichiometry!

10 7. Still another example If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? How much excess reagent will remain? Al + CuSO 4 -> Al 2 (SO 4 ) 3 + Cu Balanced? 2 Al + 3 CuSO 4 -> Al 2 (SO 4 ) 3 + 3Cu

11 1 st do Al Now CuSO g Al 1mol Al 3mol Cu 2mol Al 1 mol Cu 63.55g Cu = 36.4g Cu g CuSO 4 1mol CuSO 4 3 mol CuSO 4 3mol Cu 63.55g Cu 1 mol Cu =20.6g Cu 51.7g CuSO g Al

12 So, limiting reactant? Al: 36.4g Cu CuSO 4 : 20.6g Cu Limiting reactant: CuSO 4 Excess: Al

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14 A. Percent Yield 1. The amount of product made in a chemical reaction. 2. There are three types: a. Actual yield- what you get in the lab when the chemicals are mixed b. Theoretical yield- what the balanced equation tells what should be made

15 c. Percent yield c. Percent yield = Actual Theoretical 3. Details Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. X 100%

16 calculated on paper measured in lab

17 B. Example g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu What is the actual yield? What is the theoretical yield? What is the percent yield?

18 2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu Actual yield? 6.78g Why? “cause it told me already!!!! Theoretical yield? Stoichiometry!!!! 3.92g Al 1mol Al 26.98g Al 2mol Al 3 mol Cu 1 mol Cu 63.55g Cu = 13.85g Cu

19 So, Theoretical yield is 13.85g Cu What is percent yield? Actual yield Theoretical yield 6.87g Cu 13.85g Cu X100% X = 48.95%

20 Example 2 When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g ? g actual: 46.3 g

21 45.8 g K 2 CO 3 1 mol K 2 CO g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g? g actual: 46.3 g Theoretical Yield:

22 Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO g 49.4 g actual: 46.3 g

23 Yet another example If 36.8g of C 6 H 6 react with excess of Cl 2 And the actual yield of C 6 H 5 Cl is 38.8g, what is the percent yield? 1 st find theoretical yield of C 6 H 5 Cl 78.06g C 6 H 6 1 mol C 6 H 6 1mol C 6 H 5 Cl 36.8g C 6 H 6 1mol C 6 H 5 Cl 112.5g C 6 H 5 Cl = 53.03g C 6 H 5 Cl

24 Percent yield 38.8g 53.03g X100%= 73.17%

25 One more time! If 1.85g of Al react with excess CuSO 4 to produce 3.70g of Cu, what is the percent yield? Al + CuSO 4 -> Al 2 (SO 4 ) 3 + Cu Balance 2Al + 3CuSO 4 -> Al 2 (SO 4 ) 3 + 3Cu Now find theoretical yield

26 2Al + 3CuSO 4 -> Al 2 (SO 4 ) 3 + 3Cu 1.85g Al 26.98g Al 1mol Al 2mol Al 3mol Cu 63.55g Cu 1mol Cu = 6.54g Cu Now percent yield 3.70g 6.45g X 100%= 57.36%


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