Download - Notes #14, Ch. 9
![Page 1: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/1.jpg)
Notes #14, Ch. 9
Stoichiometry and Theoretical Yield
![Page 2: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/2.jpg)
What is Stoichiometry? It is a way to calculate how much of a
chemical is consumed or produced in a chemical reaction using Dimensional Analysis.
![Page 3: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/3.jpg)
Propanol (rubbing alcohol) C3H7OH,
burns in air. Write and balance the reaction:
2C3H7OH + 9O2 6CO2 + 8H2O
What are the Equivalents?2 mols C3H7OH 9 mols O2 6 mols CO2 8 mols
H2O
This reaction will be used in the next few examples.
![Page 4: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/4.jpg)
Example 1:
How many moles of water are produced from the combustion of 20.0 g propanol?
The arrow: 20.0 g C3H7OH ? mols H2O
Chemistry Tip of the Year: “When in doubt, go to moles!” Why? So you can use the mole ratios in
the reaction as equivalents.
![Page 5: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/5.jpg)
Solution:
So, 1.33 mols of H2O is produced
OH mols 1.33OHHC mol 2
OH mols 8
OHHC 60.0g
OHHC mol 1OHHC 20.0g 2
73
2
73
7373
![Page 6: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/6.jpg)
Example 2:
How many grams of oxygen is needed to react with 12.0 grams of propanol?
12.0 g C3H7OH ? g O2
![Page 7: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/7.jpg)
Solution:
So, 28.8 grams of oxygen is needed for the 12.0 grams of propanol to burn
22
2
73
2
73
7373 O g8.28
O mol 1
O 32.0g
OHHC mol 2
O mols 9
OHHC 60.0g
OHHC mol 1OHHC 12.0g
2O g
![Page 8: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/8.jpg)
Example 3:
If 25.3 g of carbon dioxide is produced, how many grams of propanol was burned?
25.3g CO2 ? g C3H7OH
![Page 9: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/9.jpg)
Solution:
So 11.5 grams of propanol was burned to produce the 25.3 grams of CO2.
OHHC g5.11OHHC mol 1
OHHC 60.0g
CO mol 6
OHHC mol 2
CO 44.0g
CO mol 1CO 25.3g 73
73
73
2
73
2
22
![Page 10: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/10.jpg)
The Theoretical Yield Concept
What if the reacts aren’t present in the exact ratio needed?
![Page 11: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/11.jpg)
What theoretical yield is all about:Ordinarily, reactants are not present in the exact
ratio needed to complete a reaction. One reactant is used up, or “limits the reaction.”
The other reactant has some left over, unused.
![Page 12: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/12.jpg)
Steps for calculating theoretical yield:
1. Calculate the yield as if the first reactant was limiting how much product is made.
2. Calculate the yield as if the second reactant was limiting the amount of product.
3. The Lesser of these two yields is the theoretical yield (T.Y.), or how much product you could ideally make.
4. The reactant in the D.A. which leads to the T.Y. is used up and is your “limiting reactant.”
![Page 13: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/13.jpg)
Hot Dog Example:
8 buns + 10 wieners ? hot dogs?
![Page 14: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/14.jpg)
Step 1. 8 buns ? hot dogs
hotdogs 8bun 1
hotdog 1Buns 8
![Page 15: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/15.jpg)
Step 2. 10 wieners ? hot dogs
hotdogs 01 wiener1hotdog 1
wieners10
![Page 16: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/16.jpg)
Step 3: TY & LR
Possible Yield from step 1 was 8 hot dogsPossible Yield from step 2 was 10 hot dogs
You can make lesser yield, 8 hot dogs. This is the Theoretical YieldThe buns are the limiting reactant.There are extra wienies.
![Page 17: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/17.jpg)
Skateboard Factory Example:
You inherit the remnants of a skateboard store which went under.
You have:167 wheels93 wheel “trucks”71 boards
How many skateboards can you make?
![Page 18: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/18.jpg)
So, 41 skateboards is the theoretical yield.The limiting reactant is the wheels.
Skatebds 46 trucks2
Skatebd 1 trucks93
Skatebds 41 wheels4Skatebd 1
wheels167
Skatebds 71board 1
Skatebd 1 boards 71
Skateboard 1 wheels4 board 1 trucks2
![Page 19: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/19.jpg)
Chemistry Example:
2Ag + I2 2AgIIf you start with 1.00 g of each reactant,
how much silver iodide could you make?
![Page 20: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/20.jpg)
2Ag + I2 2AgIStep 1. 1.00 g Ag ? g AgI
AgIgAgImolAgIg
AgmolAgImol
AggAgmol
gAg 18.2 1 235
2 2
108 1
00.1
Possible Yield: 2.18 g AgI
![Page 21: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/21.jpg)
2Ag + I2 2AgIStep 2. 1.00g I2 ? g AgI
AgIgAgImolAgIg
ImolAgImol
IgImol
gI 85.1 1 235
1 2
542 1
00.122
22
Possible Yield: 1.85 g AgI
![Page 22: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/22.jpg)
Step 3: So, the theoretical Yield is 1.85 g AgI, lesser of the two.Limiting Reactant is iodine, it’s all used
up. There’s left over silver.The actual yield is always a little less, due
to unwanted side reactions and loss.
![Page 23: Notes #14, Ch. 9](https://reader036.vdocument.in/reader036/viewer/2022062321/56813460550346895d9b4709/html5/thumbnails/23.jpg)
Percent Yield Example:
If you do the reaction in lab, mixing 1.0 grams of each reactant, and you get 1.50 g AgI in your filter paper, what is the actual percent yield?
%1.8110085.1
50.1%
gAgI
gAgIyield
100% lTheoretica
Actualyield