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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)
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...
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-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 +
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 +
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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