Limiting Reagents and Percent Yield

1 / 24

# Limiting Reagents and Percent Yield - PowerPoint PPT Presentation

Limiting Reagents and Percent Yield. What Is a Limiting Reagent?. Many cooks follow a _________ when making a new dish. When a cook prepares to cook he/she needs to know that sufficient amounts of all the ______________ are available.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Limiting Reagents and Percent Yield' - zamora

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Limiting Reagents and Percent Yield

What Is a Limiting Reagent?
• Many cooks follow a _________ when making a new dish.
• When a cook prepares to cook he/she needs to know that sufficient amounts of all the ______________ are available.
• Let’s look at a recipe for the formation of a double cheeseburger:

1 hamburger bun

1 tomato slice

1 lettuce leaf

2 slices of cheese

2 burger patties

How many hamburger buns do you need?

• If you want to make 5 double cheese burgers:
• How many hamburger patties do you need?
• How many slices of cheese do you need?
• How many slices of tomato do you need?

1 bun, 2 patties, 2 slices of cheese, 1 tomato slice

• 2 buns, 4 patties, 4 slices of cheese, 2 tomato slices
• How many double cheeseburgers can you make if you start with:
• 1 mole of buns, 2 moles of patties, 2 moles of cheese, 1 mole of tomato slices
• 10 buns, 20 patties, 2 slices of cheese, 10 tomato slices
• The ______________ limits the number of cheeseburgers we can make.
• If one of our ingredients gets used up during our preparation it is called the ______________________________
• The ____ limits the amount of product we can form; in this case double cheeseburgers.
• It is equally impossible for a chemist to make a certain amount of a desired compound if there isn’t enough of one of the reactants.
As we’ve been learning, a ___________ chemical rxn is a chemist’s recipe.
• Which allows the chemist to predict the amount of product formed from the amounts of ingredients available
• Let’s look at the reaction equation for the formation of ammonia:
• When _ mole of N2 reacts with _ moles of H2, _ moles of NH3 are produced.
• How much NH3 could be made if 2 moles of N2 were reacted with 3 moles of H2?
The amount of H2 limits the amount of NH3 that can be made.
• From the amount of N2 available we can make ______________________
• From the amount of H2 available we can only make ____________________.
• ___ is our limiting reactant here.
• It runs out before the ___is used up.
• Therefore, at the end of the reaction there should be ___left over.
• When there is reactant left over it is said to be in ______________.
How much N2 will be left over after the reaction?
• In our rxn it takes _____ of N2 to react all of _______ of H2, so there must be _______ of N2 that remains unreacted.
• We can use our new stoich calculation skills to determine 3 possible types of LR type calculations.
• Determine which of the reactants will run out first (limiting reactant)
• Determine amount of product
• Determine how much excess reactant is wasted
Limiting Reactant Problems:

Given the following reaction:

2Cu + S  Cu2S

• What is the limiting reactant when 82.0 g of Cu reacts with 25.0 g S?
• What is the maximum amount of Cu2S that can be formed?
• How much of the other reactant is wasted?
Our 1st goal is to calculate how much _ would react if all of the __ was reacted.
• From that we can determine the ___________________________.
• Then we can use the Limiting Reactant to calculate the amount of product formed and the amount of excess reactant left over.
• 82g Cu
• mol Cu
• mol S
• g S

2Cu + S  Cu2S

1mol S

1molCu

32.1g S

82.0gCu

• So if all of our 82.0g of Copper were reacted completely it would require only ______________ of Sulfur.
• Since we initially had 25g of S, we are going to run out of the Cu, the limiting reactant) & end up with _____________

2molCu

1mol S

63.5gCu

=___________

________

• _______ being our Limiting Reactant is then used to determine how much product is produced.
• The amount of Copper we initially start with limits the amount of product we can make.

_molCu2S

1molCu

___gCu2S

____gCu

_molCu2S

1molCu2S

____gCu

= ___________

So the reaction between 82.0g of Cu and 25.0g of S can only produce ____g of Cu2S.
• The Cu runs out before the S and we will end up wasting ____ g of the S.

Ex 2: Hydrogen gas can be produced in the lab by the rxn of Magnesium metal with HCl according to the following rxn equation: Mg + 2HCl  MgCl2 + H2

• What is the LR when 6.0 g HCl reacts with 5.0 g Mg? What is the maximum amount of H2 that can be formed? And how much of the other reactant is wasted?

5.0g Mg 

• mol Mg 
• 2mol HCl
• g HCl

___gHCl

_molHCl

1molMg

___g Mg

1molHCl

_molMg

___gMg

= _________

• So if 5.0g of Mg were used up it would take _________, but we only had 6.0g of HCl to begin with.
• Therefore, the 6.0g of HCl will run out before the 5.0g of Mg, so ___________ _________________.

6.0g HCl

• 2mol HCl 
• 1mol H2 
• g H2

___gH2

_molH2

1molHCl

___g HCl

1molH2

_molHCl

____gHCl

= _____ g H2 produced

• 6.0g HCl
• 2mol HCl 
• 1mol Mg
• g Mg

____gMg

_molMg

1molHCl

___g HCl

1molMg

_molHCl

____gHCl

= _____ g Mg

- 5.0 g Mg

= 3.01g Mg extra

Calculating Percent Yield
• In theory, when a teacher gives an exam to the class, every student should get a grade of 100%.
• Your exam grade, expressed as a percent, is a quantity that shows how well you did on the exam compared with how well you could have done if you had answered all questions correctly
This calc is similar to the percent yield calc that you do in the lab when the product from a chemical rxn is less than you expected based on the ___________________________.
• You might have assumed that if we use stoich to calculate that our rxn will produce 5.2 g of product, that we will actually recover 5.2 g of product in the lab.
• This assumption is as _____________ as assuming that all students will score 100% on an exam.
When an equation is used to calculate the amount of product that is possible during a rxn, a value representing the ______________________ is obtained.
• The _______________ is the maximum amount of product that could be formed from given amounts of reactants.
• In contrast, the amount of product that forms when the rxn is carried out in the lab is called the ____________.
• The ______ yield is often less than the _____________ yield.
The percent yield is the ratio of the actual yield to the theoretical yield as a percent
• It measures the measures the ____________ of the reaction

______ yield

Percent yield=

x 100

__________ yield

• What causes a percent yield to be less than 100%?
Rxns don’t always go to ___________; when this occurs, less than the expected amnt of product is formed.
• Impure reactants and competing side rxns may cause unwanted products to form.
• Actual yield can also be lower than the theoretical yield due to a loss of ____________ during filtration or transferring between containers.
• If a wet precipitate is recovered it might weigh _____________ due to incomplete drying, etc.

Calcium carbonate is synthesized by heating,as shown in the following equation: CaO + CO2 CaCO3

• What is the theoretical yield of CaCO3 if 24.8 g of CaO is heated with 43.0 g of CO2?
• What is the percent yield if 33.1 g of CaCO3 is produced?

Determine which reactant

is the limiting and then decide

what the theoretical yield is.

24.8gCaO

• molCaO
• mol CaCO3
• gCaCO3
• 24.8gCaO
• molCaO
• mol CO2
• gCO2

24.8 g

CaO

1molCaO

1mol CO2

__ g CO2

__g CaO

1mol CaO

1molCO2

= ____gCO2

1mol CaO

___g CaCO3

24.8 g

CaO

1molCaCO3

__g CaO

1mol CaO

1molCaCO3

= ____ g CaCO3

_____________

• ____ is our LR, so the reaction should theoretically produce 44.3 g of CaCO3 (How efficient were we?)
• Our percent yield is:

33.1 g CaCO3

Percent yield=

x 100

44.3 g CaCO3

Percent yield = ______