Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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This was what was done before, and it was incorrect. Tried the same method for the next set of numbers just to check and it was also incorrect. 

Step 1
Given :
Volume of Na2C204= 46 ml
Molarity of Na2C204 = 0.04726 M
Volume of potassium permanganate = 39.26 ml
Step 2
Calculations:
Reaction given as :
1
+ 5 H2C204 + 6H+
--- > Mn²+ + 10CO2 + 8H20
Moles of NazC204
= Molarity x volume
Moles of Na2C204 = 0.04726 M x 46 x10-³L
Moles of Naz C204 = 2. 1739 x 10-3 moles
5 x Moles of Na2C204
= 2 x Moles of KMNO4
5х 2.1739 х 10-3 тoles
2 x Molarity x 39. 26 x 10-³L
Molarity
0. 1384 M
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Transcribed Image Text:Step 1 Given : Volume of Na2C204= 46 ml Molarity of Na2C204 = 0.04726 M Volume of potassium permanganate = 39.26 ml Step 2 Calculations: Reaction given as : 1 + 5 H2C204 + 6H+ --- > Mn²+ + 10CO2 + 8H20 Moles of NazC204 = Molarity x volume Moles of Na2C204 = 0.04726 M x 46 x10-³L Moles of Naz C204 = 2. 1739 x 10-3 moles 5 x Moles of Na2C204 = 2 x Moles of KMNO4 5х 2.1739 х 10-3 тoles 2 x Molarity x 39. 26 x 10-³L Molarity 0. 1384 M
Titration of 53.00 mL of 0.04746 M Na2 C204 required 39.33 mL of a potassium permanganate solution.
2Mn O4 + 5H2C2O4 + 6H† –→ 2MN²+ + 10CO2 (g) + 8H2O
Calculate the molar concentration of the KMNO4 solution.
Concentration
M
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Transcribed Image Text:Titration of 53.00 mL of 0.04746 M Na2 C204 required 39.33 mL of a potassium permanganate solution. 2Mn O4 + 5H2C2O4 + 6H† –→ 2MN²+ + 10CO2 (g) + 8H2O Calculate the molar concentration of the KMNO4 solution. Concentration M
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