# Redox Titration Question

## Homework Statement

How many molecules of hydrogen peroxide were in the whole bottle?
(See Data)

## Homework Equations

Data.
Balanced equation.
6H+ + 2Mno4- + 5 h2o2 --> 2Mn2+ + 5O2 + 8 H2O
I used 26.85 mL of MnO4- the first trial, and 29.80 mL in the second trial.
10.000 g of H2O2 used in the solution being titrated.
3.951 g of KMnO4 solid used to make the standard solution.
The molarity of the KMnO4 solution is .1000 M

## The Attempt at a Solution

I first tried to find the mass percent of hydrogen peroxide I used in the solution.
26.85 mL MnO4 - conversion to liters ---> .0269
X / .0269 L = .1000 M
X= .00269 mol MnO4-
Convert to grams hydrogen peroxide.
.00269 * (5/2 molar ratio) * 34.02 molar mass of H2O2 = .23 g H2O2
.23 / 10.000 g h2o2 used = .023
.023 * 100 = 2.3% by mass H2O2

I did the same for the second trial with 29.80, and ended up with 2.6% mass H2O2.
The average of the mass percents was around 2.5 % H2O2

Where do I go from here? Did I calculate the mass percentage of hydrogen peroxide correctly?

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Borek
Mentor
10.000 g of H2O2 used in the solution being titrated.
Please elaborate. There was either 10 g of hydrogen peroxide, or 0.23 g of hydrogen peroxide.

Do you mean 10 g of SOLUTION of hydrogen peroxide?

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Please elaborate. There was either 10 g of hydrogen peroxide, or 0.23 g of hydrogen peroxide.

Do you mean 10 g of SOLUTION of hydrogen peroxide?

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I meant we used 10 g of hydrogen peroxide as the analyte.
The .023 g was calculated using stoichiometry from the mol of MnO4-.