Molarity of KMnO4 solution: 0.1000M. Balanced Redox Rxn: 5H 2 O 2 +6H + +2MnO 4

Question

Molarity of KMnO4 solution: 0.1000M. Balanced Redox Rxn: 5H2O2+6H++2MnO4->2Mn2++5O2+8H2O

1st. we added 10mL of H2O2, 50mL of H2O, & 15mL of H2SO4 to a flask.

2nd. we titrated KMnO4 to the flask, until it turned pink, using the volumes in the data table below.

Given this information, complete the table below.

DATA Trial 1 Trial 2 Trial 3 Mass of sample solution 9.822g 9.903g 9.78g Volume of KMno4 used 37.73ml 37.3ml 37.01ml moles of KMnO4 used 3.773*10^-3 mol 3.73*10^-3 mol 3.701*10^-3 mol Moles of H2O2 in sample Grams of H2O2 in sample Percent by mass of H2O2

Explanation / Answer

The reaction is very important as it gives us the stoichiometric ratio of the reactants. The reaction is:

2MnO4- + 5H2O2 + 6H+ ------------> 2Mn2+ + 5O2 + 8H2O

(Remember: The H+ ions in this reaction are coming from the sulphuric acid.)

This reaction tells us that for 5 moles of H2O2, 2 moles of MnO4- react.

Using this stoichiometric ratio to our advantage, we can calculate the moles of H2O2.

Trial 1: 5 moles of H2O2 / 2 moles of MnO4- X 3.773 X 10-3 moles of KMnO4 = 9.433 X 10-3 moles of H2O2.

Trial 2: 5 moles of H2O2 / 2 moles of MnO4- X 3.73 X 10-3 moles of KMnO4 = 9.325 X 10-3 moles of H2O2.

Trial 3: 5 moles of H2O2 / 2 moles of MnO4- X 3.701 X 10-3 moles of KMnO4 = 9.253 X 10-3 moles of H2O2.

Now that we have the number of moles of H2O2 that reacted with KMnO4, we can multiply the moles by the molar mass of H2O2 to calculate the mass of H2O2 in grams. Molar mass of H2O2 = 34.015 g/mol.

Trial 1: 34.015 g/mol X 9.433 X 10-3 moles of H2O2 = 0.321 g H2O2.

Trial 2: 34.015 g/mol X 9.325 X 10-3 moles of H2O2 = 0.317 g H2O2.

Trial 3: 34.015 g/mol X 9.253 X 10-3 moles of H2O2 = 0.315 g H2O2.

Mass percent of H2O2 = mass of H2O2 in grams / mass of solution X 100

Trial 1: Mass percent =  0.321 g H2O2 / 9.822 g X 100 = 3.268 %

Trial 2: Mass percent =  0.317 g H2O2 / 9.903 g X 100 = 3.201 %

Trial 3: Mass percent =  0.315 g H2O2 / 9.822 g X 100 = 3.207 %

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