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
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.
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
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)?
HW review
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
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
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.
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
Let’s look at #3 on p.3
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
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:
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
Let’s look at the second problem on p.3
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
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:
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
Lesson 22:
Theoretical Yield
Actual Yield
Percent Yield
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
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
The ______________________________________________obtained from a
reaction is called the actual yield (or experimental yield) of that
product.
measured amount of a product
ACTUAL YIELD
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
Let’s look at the #1 on p.5
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:
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
Let’s look at the #2 on p.5
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.
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
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
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!
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 =