Limiting Limiting Reagents Reagents

and and Percent YieldPercent Yield

What Is a Limiting Reagent?What Is a Limiting Reagent?

• Many cooks follow a recipe when Many cooks follow a recipe when making a new dish.making a new dish.

• When a cook prepares to cook When a cook prepares to cook he/she needs to know that he/she needs to know that sufficient amounts of all the sufficient amounts of all the ingredients are available.ingredients are available.

• Let’s look at a recipe for the Let’s look at a recipe for the formation of a double formation of a double cheeseburger:cheeseburger:

1 hamburger bun1 hamburger bun

1 tomato slice1 tomato slice

1 lettuce leaf1 lettuce leaf

2 slices of cheese2 slices of cheese

2 burger patties2 burger patties

• If you want to make 5 double If you want to make 5 double cheese burgers:cheese burgers: How many hamburger buns How many hamburger buns

do you need?do you need? How many hamburger How many hamburger

patties do you need?patties do you need? How many slices of cheese How many slices of cheese

do you need?do you need? How many slices of tomato How many slices of tomato

do you need?do you need?

• How many double cheeseburgers How many double cheeseburgers can you make if you start with:can you make if you start with: 1 bun, 2 patties, 2 slices of 1 bun, 2 patties, 2 slices of

cheese, 1 tomato slicecheese, 1 tomato slice 2 buns, 4 patties, 4 slices of 2 buns, 4 patties, 4 slices of

cheese, 2 tomato slicescheese, 2 tomato slices 1 mole of buns, 2 moles 1 mole of buns, 2 moles

of patties, 2 moles of of patties, 2 moles of cheese, 1 mole of cheese, 1 mole of tomato slicestomato slices

10 buns, 20 patties, 2 slices 10 buns, 20 patties, 2 slices of cheese, 10 tomato slicesof cheese, 10 tomato slices

• We can’t make anymore than 1 We can’t make anymore than 1 double cheeseburger with our double cheeseburger with our ingredients.ingredients.– The slices of cheese The slices of cheese limits the numberlimits the number

of cheeseburgers we can make.of cheeseburgers we can make.• If one of our ingredients gets used up If one of our ingredients gets used up

during our preparation it is called the during our preparation it is called the limiting reactant (LR)limiting reactant (LR)

• The LR limits the amount of product The LR limits the amount of product we can form; in this case double we can form; in this case double cheeseburgers.cheeseburgers.

• It is equally impossible for a chemist It is equally impossible for a chemist to make a certain amount of a to make a certain amount of a desired compound if there isn’t desired compound if there isn’t enough of one of the reactants.enough of one of the reactants.

• As we’ve been learning, a As we’ve been learning, a balanced chemical rxn is a balanced chemical rxn is a chemist’s recipe.chemist’s recipe.– Which allows the chemist to Which allows the chemist to

predict the amount of product predict the amount of product formed from the amounts of formed from the amounts of ingredients availableingredients available

• Let’s look at the reaction equation Let’s look at the reaction equation for the formation of ammonia:for the formation of ammonia:

NN22(g) + 3H(g) + 3H22(g) (g) 2NH 2NH33(g)(g)• When 1 mole of NWhen 1 mole of N22 reacts with 3 reacts with 3

moles of Hmoles of H22, 2 moles of NH, 2 moles of NH33 are are produced.produced.

• How much NHHow much NH33 could be made if 2 could be made if 2 moles of Nmoles of N22 were reacted with 3 were reacted with 3 moles of Hmoles of H22? ?

•The amount of HThe amount of H22 limits the amount limits the amount of NHof NH33 that can be made. that can be made.– From the amount of NFrom the amount of N22 available we available we

can make can make 4 moles of NH4 moles of NH33

– From the amount of HFrom the amount of H22 available we available we can only make can only make 2 moles of NH2 moles of NH33..

•HH22 is our limiting reactant here. is our limiting reactant here.– It runs out before the NIt runs out before the N2 2 is used up.is used up.

•Therefore, at the end of the reaction Therefore, at the end of the reaction there should be Nthere should be N2 2 left over.left over.– When there is reactant left over it is When there is reactant left over it is

said to be in said to be in excessexcess..

NN22(g) + 3H(g) + 3H22(g) (g) 2NH 2NH33(g)(g)

• How much NHow much N22 will be left over after will be left over after the reaction?the reaction?–In our rxn it takes In our rxn it takes 1 mol1 mol of N of N22 to react to react all of all of 3 mols3 mols of H of H22, so there must be , so there must be 1 mol1 mol of N of N22 that remains unreacted. that remains unreacted.

• We can use our new stoich We can use our new stoich calculation skills to determine 3 calculation skills to determine 3 possible types of LR type calculations.possible types of LR type calculations.

1.1. Determine which of the reactants Determine which of the reactants will run out first (limiting reactant)will run out first (limiting reactant)

2.2. Determine amount of product Determine amount of product 3.3. Determine how much excess Determine how much excess

reactant is wastedreactant is wasted

Limiting Reactant Problems:Limiting Reactant Problems:

Given the following reaction:Given the following reaction:2Cu + S 2Cu + S Cu Cu22SS

• What is the limiting reactant when What is the limiting reactant when 82.0 g of Cu reacts with 25.0 g S?82.0 g of Cu reacts with 25.0 g S?

• What is the maximum amount of What is the maximum amount of CuCu22S that can be formed? S that can be formed?

• How much of the other reactant is How much of the other reactant is wasted?wasted?

• Our 1Our 1stst goal is to calculate how goal is to calculate how much S would react if all of the Cu much S would react if all of the Cu was reacted.was reacted.

• From that we can determine the From that we can determine the limiting reactant (LR).limiting reactant (LR).

• Then we can use the Limiting Then we can use the Limiting Reactant to calculate the amount Reactant to calculate the amount of product formed and the amount of product formed and the amount of excess reactant left over.of excess reactant left over.

82g Cu82g Cumol Cumol Cu mol Smol S g Sg S

2Cu + S Cu2S2Cu + S Cu2S

82.0gCu82.0gCu1molCu1molCu

63.5gCu63.5gCu

1mol S1mol S

2molCu2molCu

32.1g S32.1g S

1mol S1mol S

=20.7 g S=20.7 g S•So if all of our 82.0g of Copper were So if all of our 82.0g of Copper were

reacted completely it would require reacted completely it would require only only 20.7 grams20.7 grams of Sulfur. of Sulfur.–Since we initially had 25g of S, we are Since we initially had 25g of S, we are going to run out of the Cu, the limiting going to run out of the Cu, the limiting reactant) & end up with reactant) & end up with 4.3 grams of S4.3 grams of S

• Copper being our Limiting Copper being our Limiting Reactant is then used to Reactant is then used to determine how much product is determine how much product is produced.produced.– The amount of Copper we initially The amount of Copper we initially

start with limits the amount of start with limits the amount of product we can make.product we can make.

82.0gCu82.0gCu1molCu1molCu

63.5gCu63.5gCu

1molCu1molCu22SS

2molCu2molCu22SS________________159gCu159gCu22SS

1molCu1molCu22SS

= 103 g Cu= 103 g Cu22SS

• So the reaction between 82.0g of So the reaction between 82.0g of Cu and 25.0g of S can only Cu and 25.0g of S can only produce 103g of Cuproduce 103g of Cu22S. S. – The Cu runs out before the S and The Cu runs out before the S and

we will end up wasting 4.7 g of the we will end up wasting 4.7 g of the S.S.

Ex 2: Hydrogen gas can be Ex 2: Hydrogen gas can be produced in the lab by the rxn of produced in the lab by the rxn of

Magnesium metal with HCl Magnesium metal with HCl according to the following rxn according to the following rxn

equation: Mg + 2HCl equation: Mg + 2HCl MgCl MgCl22 + H + H22−What is the LR when 6.0 g HCl What is the LR when 6.0 g HCl reacts with 5.0 g Mg? What is the reacts with 5.0 g Mg? What is the maximum amount of Hmaximum amount of H22 that can that can be formed? And how much of the be formed? And how much of the other reactant is wasted?other reactant is wasted?

5.0g Mg5.0g Mg 1molMg1molMg

24.3gMg24.3gMg

2molHCl2molHCl

1molMg1molMg

36.5gHCl36.5gHCl

1molHCl1molHCl

= 15.0g HCl= 15.0g HCl

5.0g Mg 5.0g Mg mol Mg mol Mg 2mol HCl 2mol HCl g HClg HCl

• So if 5.0g of Mg were used up it So if 5.0g of Mg were used up it would take would take 15.0g HCl15.0g HCl, but we only , but we only had 6.0g of HCl to begin with.had 6.0g of HCl to begin with.−Therefore, the 6.0g of HCl will run Therefore, the 6.0g of HCl will run

out before the 5.0g of Mg, so out before the 5.0g of Mg, so HCl is HCl is our Limiting Reactantour Limiting Reactant..

6.0g HCl6.0g HCl 1molHCl1molHCl

36.5gHCl36.5gHCl

1molH1molH22

2molHCl2molHCl

2.0gH2.0gH22

1molH1molH22

= 0.164 g H2 produced= 0.164 g H2 produced

6.0g HCl6.0g HCl2mol HCl 2mol HCl 1mol H2 1mol H2 g H2g H2

6.0g HCl6.0g HCl 1molHCl1molHCl

36.5gHCl36.5gHCl

1molMg1molMg

2molHCl2molHCl

24.3gMg24.3gMg

1molMg1molMg

= 1.997 g Mg= 1.997 g Mg

6.0g HCl6.0g HCl2mol HCl 2mol HCl 1mol Mg 1mol Mg g Mgg Mg

- 5.0 g Mg- 5.0 g Mg= 3.01g Mg extra= 3.01g Mg extra

Calculating Percent YieldCalculating Percent Yield• In theory, when a teacher gives an In theory, when a teacher gives an

exam to the class, every student exam to the class, every student should get a grade of 100%.should get a grade of 100%.

• Your exam grade, expressed as a Your exam grade, expressed as a perc-ent, is a quantity that shows perc-ent, is a quantity that shows how well you did on the exam how well you did on the exam compared with how well you could compared with how well you could have done if you had answered all have done if you had answered all questions correctlyquestions correctly

• This calc is similar to the percent This calc is similar to the percent yield calc that you do in the lab yield calc that you do in the lab when the product from a chemical when the product from a chemical rxn is less than you expected rxn is less than you expected based on the balanced eqn.based on the balanced eqn.

• You might have assumed that if You might have assumed that if we use stoich to calculate that our we use stoich to calculate that our rxn will produce 5.2 g of product, rxn will produce 5.2 g of product, that we will actually recover 5.2 g that we will actually recover 5.2 g of product in the lab.of product in the lab.

• This assumption is as faulty as This assumption is as faulty as assuming that all students will assuming that all students will score 100% on an exam.score 100% on an exam.

• When an equation is used to When an equation is used to calculate the amount of product calculate the amount of product that is possible during a rxn, a value that is possible during a rxn, a value representing the representing the theoretical yieldtheoretical yield is is obtained.obtained.

• The theoretical yield is the The theoretical yield is the maximum amount of product that maximum amount of product that could be formed from given could be formed from given amounts of reactants.amounts of reactants.

• In contrast, the amount of product In contrast, the amount of product that forms when the rxn is carried that forms when the rxn is carried out in the lab is called the out in the lab is called the actual actual yieldyield..

• The actual yield is often less than The actual yield is often less than the theoretical yield.the theoretical yield.

• The percent yield is the ratio of the The percent yield is the ratio of the actual yield to the theoretical yield actual yield to the theoretical yield as a percentas a percent– It measures the measures the It measures the measures the

efficiencyefficiency of the reaction of the reaction

Percent yield=Percent yield=actual yieldactual yield

theoretical yieldtheoretical yieldx 100x 100

• What causes a percent yield to be What causes a percent yield to be less than 100%?less than 100%?

• Rxns don’t always go to Rxns don’t always go to completion; when this occurs, less completion; when this occurs, less than the expected amnt of than the expected amnt of product is formed.product is formed.– Impure reactants and competing Impure reactants and competing

side rxns may cause unwanted side rxns may cause unwanted products to form.products to form.

– Actual yield can also be lower Actual yield can also be lower than the theoretical yield due to a than the theoretical yield due to a loss of product during filtration or loss of product during filtration or transferring between containers.transferring between containers.

– If a wet precipitate is recovered it If a wet precipitate is recovered it might weigh heavy due to might weigh heavy due to incomplete drying, etc. incomplete drying, etc.

Calcium carbonate is synthesized by Calcium carbonate is synthesized by heating,as shown in the following heating,as shown in the following equation: CaO + COequation: CaO + CO22 CaCO CaCO33

• What is the theoretical yield of What is the theoretical yield of CaCOCaCO33 if 24.8 g of CaO is heated if 24.8 g of CaO is heated with 43.0 g of COwith 43.0 g of CO22??

• What is the percent yield if 33.1 g What is the percent yield if 33.1 g of CaCOof CaCO33 is produced? is produced?

Determine which reactant Determine which reactant is the limiting and then decide is the limiting and then decide

what the theoretical yield is.what the theoretical yield is.

24.8 24.8 gg

CaOCaO

1molCa1molCaOO

56g CaO56g CaO

1mol CO1mol CO22

1mol CaO1mol CaO44 g CO44 g CO22

1molCO1molCO22

= 19.5gCO2= 19.5gCO2

24.8gCaO24.8gCaOmolCaOmolCaOmol COmol CO22gCOgCO22

24.8 g24.8 gCaOCaO

1mol 1mol CaOCaO

56g CaO56g CaO

1molCaC1molCaCOO331mol 1mol

CaOCaO

100g CaCO100g CaCO33

1molCaCO1molCaCO33

= 44.3 g CaCO3= 44.3 g CaCO3

24.8gCaO24.8gCaOmolCaOmolCaOmol CaCOmol CaCO33gCaCOgCaCO33

LRLR

• CaO is our LR, so the reaction CaO is our LR, so the reaction should theoretically produce 44.3 should theoretically produce 44.3 g of CaCOg of CaCO3 3 (How efficient were (How efficient were we?)we?)

• Our percent yield is:Our percent yield is:Percent yield=Percent yield= 33.1 g CaCO333.1 g CaCO3

44.3 g CaCO344.3 g CaCO3

__________________________ x 100x 100

Percent yield = 74.7%Percent yield = 74.7%