Chem107_Sect508_Lect7_final
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Date
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Instructor: Prof. Michael B. Hall
Office Hrs: T 5:10 – 6:00 Outside ILCB 226 for General Questions
Th 5:10 – 6:00 In ILCB 207 for General Questions
M 5:00 – 6:00 By Zoom for Individual Conferences (1
st
one next M) Chemistry Connection Point: Mon - Fri 8:00 am - 5:00 pm, ILSQ 3rd Floor Lobby
REMINDERS & NEW ANNOUNCEMENTS:
OWL homework (HW) due Sunday, 9/17, at 10pm. Good practice for Quiz. Next Quiz - Monday 9/18 on Unit 2 – Chemical Reactions (last week and this week)
This quiz covers more material, so you should increase your study time.
ARE YOU HAVING LAB THIS WEEK?
HAVE YOU DONE THE LABQUIZ? DO YOU KNOW
THE DRESS CODE? DO YOU HAVE GOOGLES? DO YOU HAVE A NOTEBOOK?
Chemistry 107 - Section 508
Fall 2023 – Lecture #7
Reaction Stoichiometry
Mass to Mass Conversions
1.
Converting from grams of one chemical species to grams of another chemical species in a reaction.
2A + C ®
3B + D
Reaction Stoichiometry
Chp. 4.2
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Dehydration of Sucrose
Demonstration
When sucrose, C
12
H
22
O
11
, is dehydrated pure carbon forms. Write a balanced reaction for the dehydration of sucrose and calculate the mass of carbon formed if 80 g of sucrose is dehydrated. A. 2.80 g
B. 6.66 g C. 33.68 g
D. 0.23 g
E. 2.57 g
https://www.youtube.com/watch?v=GuHViMhYbRc&feature=youtu.be
C
12
H
22
O
11
(s)
®
12 C
(s) + 11
H
2
O
(g)
Combustion in Automobiles
Chp. 3, Chp. 4
incomplete combustion in air (80% N
2
, 20% O
2
)
C
8
H
18
(g) + N
2,
O
2
(g) ®
NO
x
(g) + CO(g) + C
x
H
y
O
z
(g) + CO
2
(g) + H
2
O(g)
toxins and pollutants
octane
air
?
Stoichiometry: Gram to Gram Conversions
Chp. 4.2
When 10.5 g of acetaldehyde (MW 44.05 g/mol) reacts with excess oxygen in a palladium catalyzed oxidation, how many grams of water (MW 18.01 g/mol) are formed?
2C
2
H
4
O
(g) + 5O
2
(g) ®
4CO
2
(g) + 4H
2
O(g)
A. 4.29 g
B. 8.59 g
C. 17.2 g D. 10.5 g
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Reaction Stoichiometry
Limiting Reagents
1. Identifying limiting reagents in chemical reactions.
2. Converting from grams of one chemical species to grams of another chemical species in a reaction where there is a limiting reagent.
3.
Determining how much of an excess reagent remains.
Limiting Reagent Concept
Chp. 4.3
Limiting Reagent – reactant that is the first to be completely consumed in a reaction.
Excess Reagent – any reactant that remains once the limiting
reagent has been completely consumed.
2H
2
(g) + O
2
(g) ®
2H
2
O(l)
Limiting Reagents
Chp. 4.3
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Mg
(
s
)
+ 2HCl
(
aq
)
→ H
2
(
g
)
+ MgCl
2
(
aq
)
1.5 g of magnesium is reacted with 30 mL (Flask 1), 62.5 mL (Flask 2) and 90 mL (Flask 3) of 2.0 M HCl.
Limiting Reagent Concept
Demo
https://youtu.be/wbgoNGAzoKo
Mg
(
s
)
+ 2HCl
(
aq
)
→ H
2
(
g
)
+ MgCl
2
(
aq
)
1.5 g of magnesium is reacted with 30 mL (Flask 1), 62.5 mL (Flask 2) and 90 mL (Flask 3) of 2.0 M HCl.
0.062 mol Mg
0.062 mol HCl
0.062 mol Mg
0.124 mol HCl
0.062 mol Mg
0.186 mol HCl
1
2
3
What is the limiting reagent in
Reaction 3?
A. Mg
B. HCl
C. Neither reactant is limiting
Limiting Reagent Concept
Demo
2A + C ®
3B + D
Limiting Reagent Reaction Stoichiometry
Chp. 4.3
Grams A
Grams C
Moles A
Moles C
Moles Limiting Reactant
Moles B
Grams B
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2NH
3
(g) + CO
2
(g) ®
(NH
2
)
2
CO(aq) +H
2
O(l)
How many grams of urea, (NH
2
)
2
CO, are formed when 637.2 g of NH
3
are reacted with 1142 g of CO
2
? How much of the excess reactant remains?
Limiting Reagents
Chp. 4.3
1. Determine the limiting reagent by comparing moles NH
3
to moles CO
2
.
2. Use the moles (or grams to moles) of the limiting reagent to determine the moles then grams of urea formed.
3. Use the moles (or grams to moles) of the limiting reagent to determine the moles then grams of excess reagent reacted.
4. Subtract the grams of excess reagent reacted from the initial grams of excess reagent to determine the remaining excess reactant.
2NH
3
(g) + CO
2
(g) ®
(NH
2
)
2
CO(aq) +H
2
O(l)
What is the limiting reagent when 637.2 g of NH
3
are allowed to react with 1142 g of CO
2
to form urea as a product? How many grams of urea, (NH
2
)
2
CO, are formed?
Limiting Reagents
Chp. 4.3
Determine the limiting reagent by comparing moles NH
3
to moles CO
2
.
Use the moles of the limiting reagent to determine the moles then grams of urea formed.
2NH
3
(g) + CO
2
(g) ®
(NH
2
)
2
CO(aq) +H
2
O(l)
How many grams of the excess reagent remains?
Limiting Reagents
Chp. 4.3
Use the moles (or grams to moles) of the limiting reagent to determine the moles then grams of excess reagent reacted.
Subtract the grams of excess reagent reacted from the initial grams of excess reagent to determine the remaining excess reactant.
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Stoichiometry: Limiting Reagents
Chp. 4.3
Imagine that 20.0 g of octane (MW 114.23 g/mol) is reacted with 50.0 g oxygen (MW 31.98 g/mol), what is the limiting reagent?
2C
8
H
18
(g) + 25O
2
(g) ®
16CO
2
(g) + 18H
2
O(g)
A. octane
B. oxygen
Is this a rich or lean fuel environment?
A. lean B. rich
Stoichiometry: Limiting Reagents
Chp. 4.3
Imagine that 20.0 g of octane (MW 114.23 g/mol) is reacted with 50.0 g oxygen (MW 31.98 g/mol), how many grams of carbon dioxide (MW 44.01 g/mol) are formed?
2C
8
H
18
(g) + 25O
2
(g) ®
16CO
2
(g) + 18H
2
O(g)
A. 61.64 g
B. 7.71 g
C. 68.8 g D. 44.0 g
Stoichiometry: Limiting Reagents
Chp. 4.3
Imagine that 20.0 g of octane (MW 114.23 g/mol) is reacted with 50.0 g oxygen (MW 31.98 g/mol), how much of the excess reagent is lef
t over
?
2C
8
H
18
(g) + 25O
2
(g) ®
16CO
2
(g) + 18H
2
O(g)
A. 5.74 g
B. 35.8 g
C. 30.0 g D. 14.3 g
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Reaction Stoichiometry
Percent Yield
1.
Defining and determining actual and thoeretical yields for chemical reactions.
2.
Defining and calculating the percent yield for a chemical reaction.
Percent Yield = x 100
Actual Yield
Theoretical Yield
Percent Yield
Chp. 4.4
What is the theoretical yield (maximum possible yield) when you pop 100 kernels of popcorn?
What is the actual yield in the example above?
What percentage of the kernels popped?
Percent Yield
Chp. 4.4
What is the theoretical yield (maximum possible yield) when you pop 100 kernels of popcorn?
A. 18 kernals
B. 82 kernals
C. 100 kernals
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Calculate the difference between AH and AU when 1.0 mol Sn (gray) of density 5.75 g cm3
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