Lesson 16: Theoretical Yield Actual Yield Percent Yield · O are produced, what is the percent...
Transcript of Lesson 16: Theoretical Yield Actual Yield Percent Yield · O are produced, what is the percent...
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Lesson 16:
Theoretical Yield
Actual Yield
Percent Yield
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Do Now (5pts) April 3, 2019• Copy down info from CJ board.
• Take out calculators to be checked.
• Answer questions in Box 1 of Lesson 16 note packet.
• Progress reports will be handed out next class. Quiz correction due date will be announced then.
• Who needs to make a make-up quiz? Please come see me at the podium.
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2 1 4 1
2 flags + 1 pole + 4 wheels + 1 car body 1 car
If you had 62 flags, 37 poles, 102 wheels, and 30 car bodies, how many complete
toy cars could you make? What reagent is limiting?
62 flags / 2 per car = 3137 poles / 1 per car = 37102 wheels / 4 per car = 25.530 car bodies / 1 per car = 30
Has the smallest equivalent, the wheels are the limiting reagentYou can only make 25 complete cars
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SiO2 (s) + 4HF (g) SiF4 (g) + 2H2O (l)If you started with 0.910 mol of silicon dioxide and 3.51 mol of HF, what is the limiting reactant?What is the theoretical yield of water (in moles)?
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HW review
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Review: Limiting and Excess Reagents
One reactant almost always limits the amount of product produced in a reaction.
Once one of the reactant is used up, ____________________________________________.
The substance that is used up first is the ________________________________.
Identifying the limiting reagent: calculate the ___________________________________.
The limiting reagent has the lesser equivalent (you will run out of that reactant
first).
then no more product can be formedlimiting reagent
molar equivalents
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1SiO2 + 4HF → 2 SiF4 + 2 H2O
4.5 mol 6.0 mol
What is the limiting reagent?
Amount that you
start with
Co-efficient of the
reactant
4.5
1vs.
6.0
4
4.5 1.5These
numbers are called “ molar equivalents”
This number is SMALLER, therefore it is the limiting
reagent.
HF
HF
SiO2
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Amount that you
start with
Co-efficient of the
reactant
Finding the “molar equivalent” of each reactant:
This has to be in compounds, molecules, or moles.
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N2H4 (l) + 2 H2O2 (l) → N2 (g) + 4 H2O (g)
0.750 mol 5.2 mol What is the limiting reagent?
Amount that you
start with
Co-efficient of the
reactant
0.750
1vs.
5.2
2
0.750 2.6These
numbers are called “ molar equivalents”
This number is SMALLER, therefore it is the limiting
reagent.
H2O2
N2H4
N2H4
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Let’s look at #3 on p.3
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If 500 mol CO and 750 mol H2 are present, which is the limiting reagent?
500 mol CO1
750 mol H2
2
500 375 H2 is the limiting reagent
Methanol, CH3OH is the simplest of alcohols. It is synthesized by the
reaction of hydrogen and carbon monoxide.
CO (g) + 2H2 (g) CH3OH
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Methanol, CH3OH is the simplest of alcohols. It is synthesized by the
reaction of hydrogen and carbon monoxide.
CO (g) + 2H2 (g) CH3OH
How many moles of the excess reagent remain unchanged?
750 mol H2
mol H2
mol COmol CO×
1=
2
1375 will be used up
Amount we started with – amount consumed in reaction
500 mol CO – 375 mol CO = 125 mol CO
Start with the limiting reagent:
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Methanol, CH3OH is the simplest of alcohols. It is synthesized by the
reaction of hydrogen and carbon monoxide.
CO (g) + 2H2 (g) CH3OH
How many moles of CH3OH are formed?
750 mol H2
mol H2
mol CH3OH×
1=
2
Start with the limiting reagent:
1 mol CH3OH375 will be produced
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Let’s look at the second problem on p.3
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The black oxide of iron, Fe3O4, occurs in nature as the mineral magnetite. The substance can also be used in the laboratory by the reaction between red-hot iron and steam according to the following reaction:
3Fe(s) + 4H2O (g) Fe3O4(s) + 4H2 (g)
When 36 g of H2O are mixed with 67.0 g Fe, which is the limiting reagent?
36 g H2O
18.016 g H2O
1 mol H2O
67.0 g Fe
55.85 g Fe
1 mol Fe
×
×
1
1
=
=
2 mol H2O
1.19964 mol Fe
2 mol H2O
4
1.19964 mol Fe
3
=
=
0.5
0.39988
LR
FIRST – find how many moles of each reactant you have.THEN – calculate the mole equivalents.
molar mass
molar mass
mole equivalent
mole equivalent
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What mass in grams of black iron oxide is produced?
3Fe(s) + 4H2O (g) Fe3O4(s) + 4H2 (g)LR
36 g
This is a grams grams conversion.
67.0 g Fe1 ×
g Fe
mol Fe
mol Fe
mol Fe3O4
mol Fe3O4
g Fe3O4
______ g Fe3O4
× × =
molar mass molar massMolar ratio of reactant to product in
balanced reaction
55.851 1
3 1231.55
92.6
92.592
67.0 g
ER
Start with the limiting reagent:
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What mass in grams of excess reagent remains when the reaction is completed?
3Fe(s) + 4H2O (g) Fe3O4(s) + 4H2 (g)
LR
36 g
Start with the limiting reagent:
67 g Fe1 g Fe
mol Fe
mol Fe
mol H2O× × ×
mol H2O
g H2O=55.85
134
118.016 29 g H2O
consumed during the reaction
Amount we started with – amount consumed in reaction
36g water to start – 29 g used up = 7 g water left over
67.0 g
ER
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Lesson 22:
Theoretical Yield
Actual Yield
Percent Yield
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Theoretical yield is the _________________________________________________________
when the limiting reagent is completely consumed.
Theoretical yield _____________________________________________________________ and is
calculated using stoichiometry ________________________________________________.
amount of the limiting reagent
from the limiting reagent.
depends solely on the
beginning
THEORETICAL YIELD
maximum amount of product that can be formed
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Just as the __________________allows us to calculate limiting and excess reagents and
masses, we can also ____________________ the mass of products in any given reaction.
The theoretical yield represents _____________________________________________________
THEORETICAL YIELD
predict
mole ratio
the maximum quantity of product possible if the reaction proceeds to completion.
This is what you’ve been calculating all along
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The ______________________________________________obtained from a
reaction is called the actual yield (or experimental yield) of that
product.
measured amount of a product
ACTUAL YIELD
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PERCENT YIELD
Comparing the theoretical and actual yield helps chemists determine the
reaction’s _________________________. efficiency
actual yieldtheoretical yield × 100Percentage yield =
• The percent yield represents the ____________of the actual yield to the theoretical yield.
ratio
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Let’s look at the #1 on p.5
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CH4 + 2O2 CO2 + 2H2O
How many grams of H2O are expected when 8.00 grams of CH4 reacts
with an excess of O2?
8.00 g CH4
g CH4
mol CH4
mol CH4
mol H2Omol H2O
g H2O
g H2O
Theoretical Yield
× × × =1 16.042
1
1
2
1
18.016
18.0
Start with the limiting reagent:
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If only 17.0 grams of H2O are produced, what is the percent yield of this
reaction?
CH4 + 2O2 CO2 + 2H2O
g H2O Theoretical Yield18.0
g H2O ACTUAL Yield17.0
actual yieldtheoretical yield
× 100Percentage yield =
17.0 / 18.0 = 0.9444× 100
94.4 % yield
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Let’s look at the #2 on p.5
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C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)52.6 g 60.2 gStarting amounts:
Which is the limiting reagent?
WHAT DO I DO?First, convert from grams to moles
using molar mass.
Then, calculate mole equivalents.
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C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)52.6 g 60.2 gStarting amounts:
Which is the limiting reagent?
52.6 g C6H6
78.108 g C6H6
1 mol C6H6 0.673 mol C6H6
60.2 g Cl2
70.9 g Cl2
1 mol Cl20.849 mol Cl2
×
×
1
1
=
=
0.673 /1 = 0.673
0.849 /1 = 0.849
LR ER
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C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)52.6 g 60.2 gStarting amounts:
LR ERDetermine the amount of excess reagent reacted and amount remaining.
Start with the limiting reagent:
52.6 g C6H6
78.108 g C6H6
1 mol C6H6
1 mol C6H6
1 mol Cl2
1 mol Cl2
70.9 g Cl2× × ×
g chlorineconsumed in reaction
1
47.760.2 g to start – 47.7 g consumed = 12.5 g of Cl2 in excess
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Calculate the theoretical yield of product chlorobenzene.
C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)52.6 g 60.2 gStarting amounts:
LR ER
52.6 g C6H6
78.108 g C6H6
1 mol C6H6
1 mol C6H6
1 mol C6H5Cl× × ×
1 1 mol C6H5Cl
112.55 g C6H5Cl
= 80.98 g C6H5Cl
Start with the limiting reagent:
Theoretically made if the reaction was 100% efficientBUT REACTIONS ARE NOT ALWAYS 100% EFFICIENT!
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C6H6 (l) + Cl2(g) C6H5Cl (l) + HCl (g)
If 73.9 g of chlorobenzene (C6H5Cl) was actually produced, what was the percent yield?
Actual yield = 73.9 g C6H5Cl
Theoretical yield = 80.98 g C6H5Cl
73.9 / 80.98 = 0.91257× 100
91.3 % yield
actual yieldtheoretical yield
× 100Percentage yield =