Calculate the percent yield for the reaction shown below by constructing a BCA table, determining the maximum grams of product that can be produced, and calculating the percent yield. Complete Parts 1-3 before submitting your answer. MnO₂ + 4 HCI → MnCl₂ + Cl₂ + 2 H₂O 1 Before (mol) Change (mol) After (mol) 2 NEXT > The reaction was started with 1.00 mol of MnO₂ (86.94 g/mol) and 95.0 g of HCI (MW 36.458 g/mol). Set up the table below that represents 100% yield with the given reaction conditions. Ignore the water side product. MnO₂ 3 + 4 HCI MnCl₂ + CI₂
Calculate the percent yield for the reaction shown below by constructing a BCA table, determining the maximum grams of product that can be produced, and calculating the percent yield. Complete Parts 1-3 before submitting your answer. MnO₂ + 4 HCI → MnCl₂ + Cl₂ + 2 H₂O 1 Before (mol) Change (mol) After (mol) 2 NEXT > The reaction was started with 1.00 mol of MnO₂ (86.94 g/mol) and 95.0 g of HCI (MW 36.458 g/mol). Set up the table below that represents 100% yield with the given reaction conditions. Ignore the water side product. MnO₂ 3 + 4 HCI MnCl₂ + CI₂
Chemistry by OpenStax (2015-05-04)
1st Edition
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Chapter4: Stoichiometry Of Chemical Reactions
Section: Chapter Questions
Problem 65E: Freon-12, CCl2F2, is prepared from CCl4 by reaction with HF. The other product of this reaction is...
Related questions
Question
![-0.651
-36.46
Before
(mol)
Change
(mol)
After (mol)
0
Calculate the percent yield for the reaction shown below by
constructing a BCA table, determining the maximum grams of product
that can be produced, and calculating the percent yield. Complete Parts
1-3 before submitting your answer.
0.349
3
NEXT >
The reaction was started with 1.00 mol of MnO₂ (86.94 g/mol) and 95.0 g of HCI (MW
36.458 g/mol). Set up the table below that represents 100% yield with the given
reaction conditions. Ignore the water side product.
1.00
Question 69 of 70
-0.349
MnO₂ + 4 HCI-
2
MnO₂
+
95.0
MnCl₂ + Cl₂ + 2 H₂O
2
4.00
2
4 HCI
-95.0
-4.00
MnCl₂
+
2.61
1.30
CI₂
-2.61
-1.30
Submit
RESET
0.651
36.46
+](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99091e55-84ac-4709-a57b-335d8cd227fa%2F88974b76-0199-4310-b76e-1ec44bcfa123%2F1xy4xea_processed.png&w=3840&q=75)
Transcribed Image Text:-0.651
-36.46
Before
(mol)
Change
(mol)
After (mol)
0
Calculate the percent yield for the reaction shown below by
constructing a BCA table, determining the maximum grams of product
that can be produced, and calculating the percent yield. Complete Parts
1-3 before submitting your answer.
0.349
3
NEXT >
The reaction was started with 1.00 mol of MnO₂ (86.94 g/mol) and 95.0 g of HCI (MW
36.458 g/mol). Set up the table below that represents 100% yield with the given
reaction conditions. Ignore the water side product.
1.00
Question 69 of 70
-0.349
MnO₂ + 4 HCI-
2
MnO₂
+
95.0
MnCl₂ + Cl₂ + 2 H₂O
2
4.00
2
4 HCI
-95.0
-4.00
MnCl₂
+
2.61
1.30
CI₂
-2.61
-1.30
Submit
RESET
0.651
36.46
+
![0
113
1
Calculate the percent yield for the reaction shown below by
constructing a BCA table, determining the maximum grams of product
that can be produced, and calculating the percent yield. Complete Parts
1-3 before submitting your answer.
MnO₂ + 4 HCI → MnCl₂ + Cl₂ + 2 H₂O
NEXT >
< PREV
Based on your table from the previous step, determine the maximum number of
grams of Cl₂ (MW 70.90 g/mol) that can be prepared.
massci₂
Question 69 of 70
185
1
92.3
=
2
46.2
g
3
23.8
26.6
Submit
RESET
27.2
+](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99091e55-84ac-4709-a57b-335d8cd227fa%2F88974b76-0199-4310-b76e-1ec44bcfa123%2Fwwyem6u_processed.png&w=3840&q=75)
Transcribed Image Text:0
113
1
Calculate the percent yield for the reaction shown below by
constructing a BCA table, determining the maximum grams of product
that can be produced, and calculating the percent yield. Complete Parts
1-3 before submitting your answer.
MnO₂ + 4 HCI → MnCl₂ + Cl₂ + 2 H₂O
NEXT >
< PREV
Based on your table from the previous step, determine the maximum number of
grams of Cl₂ (MW 70.90 g/mol) that can be prepared.
massci₂
Question 69 of 70
185
1
92.3
=
2
46.2
g
3
23.8
26.6
Submit
RESET
27.2
+
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