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Report for Laboratory Exercise #4
Synthesis of Tetramethylammonium Triiodide and Tetramethylammonium
Pentaiodide
_
Using Microsoft Word, students are to insert responses in all yellow highlighted areas. It is
recommended that the report be completed without changing font size, column width, row
width, margins and highlights. The completed report must be uploaded to the 101 Brightspace
site as a .pdf file by the due date posted on Brightspace. All answers must be the student’s own
work without assistance from others. Only reports which are completed using the template will
be marked.
Name:
Ton Tran Lab Section: B02 Quad
4 Date: November 2, 2021
Abstract
By reacting I
2
with NMe
4
I in solution with control of the reaction stoichiometry, Me
4
NI
3
and
Me
4
NI
5
were synthesised. The % yield for Me
4
NI
3
was 58.8%, and the % yield for Me
4
NI
5 was
64.2%. Data/Results
Table 1. Experimental
data and calculated values for the preparation of Me
4
NI
3
and Me
4
NI
5
Synthesis
Me
4
NI
3
Me
4
NI
5
NMe
4
I (g)
0.141g
0.320g
I
2
(g)
1.78g
0.808
actual yield (g)
0.527g
1.129g
theoretical yield (g)
0.896g
1.759g
% yield
58.8%
64.2%
2
Algebraic Equations (see page 12 of the lab manual)
a)
Balanced chemical equations for formation of each of the products: a.
Me
4
NI + I
2
Me
3
NI
3
b.
Me
4
NI + 2 I
2
Me
3
NI
5
b)
Theoretical yield for each of the products =
a.
LR = Limiting Reagent
b.
C = Coefficient
c.
P = Product
i.
LR * (LR C / P C)
c)
% yield of each of the products =
a.
E = Experimental b.
M = Mass
i.
E M Me
3
NI
3
/ (LR * (LR C / P C))
ii.
E M Me
3
NI
5
/ (LR * (LR C / P C))
Discussion Respond to the following:
Discuss the meaning of the % yield, do not just give a formula and comment on the % yield observed for each of the products, include the actual values (max 4 lines).
The % yield is produced yielded in experiment vs the max product yield. Me
4
NI + I
2
Me
3
NI
3 could have a total yield if done 100% correct, 0.896g. Experimental Me
4
NI + I
2
Me
3
NI
3 had a total yield of 0.527g. Only 58.8% of the max 0.896g was yielded. Me
4
NI + I
2
Me
3
NI
3 can have a
total yield if done 100% correct, 1.759g. Only 64.2% of the max 1.759g was yielded
Provide the limiting reagent for each of the preparations. Briefly state how the limiting reagent was determined for each of the preparations
. (
max. 3 lines).
Find minimum number of moles of both reactants for reaction to work. Smaller mol = Limiting reagent.
Me
4
NI + I
2
Me
3
NI
3
Limiting reagent was I
2
. Me
4
NI 2.49*10
-3
g/mol > I
2 1.97*10
-3
g/mol
Me
4
NI + 2 I
2
Me
3
NI
5
, Limiting reagent was Me
4
NI. Me
4
NI 2.48*10
-3
g/mol < 2 I
2 2.56*10
-3
g/mol
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prod03-cnow-owl.cengagenow.com
Fe(s) + 2HCI(aq)→→→FeCl₂(s) + H₂(g)
AH° = -7.4 kJ and AS° = 107.9 J/K
D2L
D2L
D2L
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terface/acellus_engine.htm?ClassID=1467683063
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Compound
1
2
Et
onal Academy of Science. All Rights Reserved.
Mass Fe
(grams)
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20.98
[20:1
Enter the answer that belongs in the green box.
Iron Ratio Cmpd 2
Mass O
(grams)
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Remaining Time: 42 minutes, 23 seconds.
✓ Question Completion Status:
What is the skeletal formula, for the aldehyde or ketone formed when the following alcohol is oxidized [O]. If no reaction, write none.
OA
08.
OH
OC.
OD.
0
C-H
[0]
C-OH
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
Betalin
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Table 2: Molarity of H2O2 and KI and Reaction Rate
Trial
H2O2 Concentration, M
KI Concentration, M
Reaction Rate(Reciprocal Slope)
1
0.29 M
0.40 M
14.08
2
0.29 M
0.20M
25
3
0.023 M
0.40 M
20
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Order with respect to H2O2:
Order with respect to KI:
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