# Limiting Reactants and Percent Yield

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Limiting Reactants and Percent Yield
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 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. 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

2. The one that makes the least product is the limiting reagent.
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?

Cu is Limiting Reactant
If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2Cu + S ® Cu2S Cu is Limiting Reactant 1 mol Cu 1 mol Cu2S g Cu2S 10.6 g Cu 63.55g Cu 2 mol Cu 1 mol Cu2S = 13.3 g Cu2S = 13.3 g Cu2S 1 mol S 1 mol Cu2S g Cu2S 3.83 g S 32.06g S 1 mol S 1 mol Cu2S = 19.0 g Cu2S

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

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

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!

7. Still another example If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced? How much excess reagent will remain? Al CuSO4 -> Al2(SO4)3 + Cu Balanced? 2 Al CuSO4 -> Al2(SO4)3 + 3Cu

2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu
1st do Al Now CuSO4 3mol Cu 2mol Al 1mol Al 63.55g Cu 10.3g Al 26.98g Al 1 mol Cu = 36.4g Cu 1mol CuSO4 51.7g CuSO4 3mol Cu 63.55g Cu 3 mol CuSO4 159.56g CuSO4 1 mol Cu =20.6g Cu

So, limiting reactant? Al: 36.4g Cu CuSO4: 20.6g Cu Limiting reactant: CuSO4 Excess: Al

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

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%

measured in lab calculated on paper

B. Example 1 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu What is the actual yield? What is the theoretical yield? What is the percent yield?

2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu Actual yield? 6.78g Why? “cause it told me already!!!! Theoretical yield? Stoichiometry!!!! 3.92g Al 1mol Al 3 mol Cu 63.55g Cu 26.98g Al 1 mol Cu 2mol Al = 13.85g Cu

So, Theoretical yield is 13.85g Cu
What is percent yield? Actual yield Theoretical yield 6.87g Cu 13.85g Cu X 100% X 100% = 48.95%

Example 2 When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g
Theoretical Yield: 1 mol K2CO3 g 45.8 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

K2CO3 + 2HCl  2KCl + H2O + CO2 49.4 g 45.8 g actual: 46.3 g Theoretical Yield = 49.4 g KCl 46.3 g 49.4 g % Yield =  100 = 93.7%

Yet another example If 36.8g of C6H6 react with excess of Cl2 And the actual yield of C6H5Cl is 38.8g, what is the percent yield? 1st find theoretical yield of C6H5Cl 1 mol C6H6 1mol C6H5Cl 112.5g C6H5Cl 36.8g C6H6 1mol C6H6 78.06g C6H6 1mol C6H5Cl = 53.03g C6H5Cl

Percent yield 38.8g 53.03g X 100% = 73.17%

One more time! If 1.85g of Al react with excess CuSO4 to produce 3.70g of Cu, what is the percent yield? Al + CuSO4 -> Al2(SO4) Cu Balance 2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu Now find theoretical yield

2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu
1mol Al 63.55g Cu 3mol Cu 1.85g Al 26.98g Al 2mol Al 1mol Cu = 6.54g Cu Now percent yield 3.70g 6.45g X 100% = %