Stoichiometry So far In all our reactions we assume both reactants get used up We assume both reactants provided are pure We rarely see these two Limiting reagents Percent Purity Percent Yield ID: 380729
Download Presentation The PPT/PDF document "Limiting Reagent" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Limiting Reagent
Stoichiometry
!Slide2
So far
In all our reactions, we assume both reactants get used up
We assume both reactants provided are pure
We rarely see these two.
Limiting reagents
Percent Purity
Percent Yield
So in a chemical reaction, we typically only use up one reactant and the other one is usually left over (excess)Slide3
Baking example
You have 1 bag of flour
1 bag of sugar
Same size
You will use up your bag of flour before you can use up your bag of sugar.
So your bag of flour will be your limiting reagent
Your bag of sugar is your excessSlide4
Why?
We have excess amounts so that the one that is limiting will get used up
If it is too expensive, we want to use it up
If it is harmful to the environment, we want to react it to a less harmful chemical
It can also be unavoidable as we usually do have chemical that is in high quantities
Air is all around us and if the reaction involves air, it is in excessSlide5
Key Terms
Excess Reagent = reagent that is in excess, will not get used up in the reaction
Limiting Reagent = reagent that will get used up fully in the reaction
To find, we need to convert to MOLES of product.
We cannot tell just by given the mass of our reactants.
The amount of products we form will be called the theoretical yield.Slide6
Simulation!Slide7
Note!
We refer to moles, not mass for limiting reagent.
Just because something has a smaller mass doesn’t mean it is limiting!
A limiting reagent question will always show enough information to get to moles for both reactants!Slide8
Steps to solve
Convert mass of both reagents to moles of reagents
Convert both moles of reagent to the moles of ONE PRODUCT!
Whichever one is less will be the limiting reagent
Whichever one is in excess will have some remaining, we can calculate what is leftover if we find the moles of excess used up and subtract from starting!Slide9
Example
2Al + 3I
2
→ AlI
3
Determine the limiting reagent and theoretical yield of the product if
A) 1.20 mol Al and 2.40 mol iodine
B) 1.20g of Al and 2.40g of iodine
C) How many grams of Al are left over in part b?Slide10
Example
15.00g of aluminum sulfide and 10.00g of water react until the limiting reagent is used up
Al
2
S
3
+ 6H
2
O → 2Al(OH)
3
+ 3H
2
S
A) Which is the limiting reagent?
B) What is the mass of H
2
S which can be formed from the limiting reagent?
C) How much excess reagent remains after the reaction is complete?Slide11
Percentage Yield and Percentage PuritySlide12
Assumptions
It is assumed in a chemical reaction that our products are always produced 100% pure. That means whatever we started with will all turn into our pure product
So we make many assumptions
What we start with will all turn into product
Percentage yield
What we make is 100% pure
Percentage puritySlide13
Not always the case
We will get percentages of both
This is due to
Impure products
Reactants not measured exactly
Recovery of products is difficult
Human error in transfer of stepsSlide14
Percentage yield
The percent of what we actually get compared to what we expect
The higher the better!
Percentage yield = actual yield / theoretical yield x 100%
Actual yield = the actual amount of product formed in the experiment
Theoretical yield = the amount of product expected from calculation
NOTE – No way to get below 0% or above 100%Slide15
Percentage Purity
The measure of how much desired product is present in a mass of impure product
Percentage purity = mass of desired product / mass of impure product x 100%
The higher the better as it means our product is pure!
Very important in chemistry and in not so legal things (breaking bad)
No way we can go below 0% or over 100%
100% purity is very tough to achieve!Slide16
Steps to solving
Do your calculation as if we are doing any
stoichiometry
calculation
Assume everything is 100% percentage
Apply the percentage yield and purity calculation at the very end.Slide17
Example
When 45.8g of K
2
CO
3
are reacted completed with excess
HCl
, 46.3g of
KCl
are produced. Water and carbon dioxide are also formed. Calculate the theoretical yield and the percent yield of
KCl
.
K
2
CO
3
+ 2HCl → 2KCl + H
2
O + CO
2
Find our limiting first!Slide18
Example
Consider the reaction below
3Mg(OH)
2
+ 2H
3
PO
4
→ Mg
3
(PO
4
)
2
+ 6H
2
O
Calculate the mass of Mg
3
(PO
4
)
2
that will be formed from the reaction of 15.0g Mg(OH)
2
that has a percentage purity of 85% with excess H
3
PO
4
Assume that the Mg(OH)
2
is 100% pure first!Slide19
Example
What mass of K
2
CO
3
is produced when 1.50g of KO
2
is reacted with an excess of CO
2
according to the reaction
4KO
2
+ 2CO
2
→ 2K
2
CO
3
+ 3O
2
If our reaction has a percent yield of 76.0%?Slide20
Magic Triangle
You can make magic triangles for both calculation
%y = a/t
%p = d/
iSlide21
Homework
Page 133 #26-32 (odd)
Page 137 #33-38 (odd)
Limiting reagent worksheet
Stoichiometry
Review worksheets