High School

A solution of hydrogen peroxide, \(\text{H}_2\text{O}_2\), is titrated with a solution of potassium permanganate, \(\text{KMnO}_4\). The reaction is:

\[ 5 \text{H}_2\text{O}_2(aq) + 2 \text{KMnO}_4(aq) + 3 \text{H}_2\text{SO}_4(aq) \rightarrow K_2\text{SO}_4(aq) + 2 \text{MnSO}_4(aq) + 5 \text{O}_2(g) + 8 \text{H}_2\text{O}(l) \]

It requires 51.7 mL of 0.145 M \(\text{KMnO}_4\) to titrate 40.0 g of the solution of hydrogen peroxide. What is the mass percentage of \(\text{H}_2\text{O}_2\) in the solution?

Answer :

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Final answer:

The mass percentage of H2O2 in the solution is approximately 99.37%.

Explanation:

To calculate the mass percentage of H2O2 in the solution, we need to first calculate the moles of KMnO4 used in the titration.

Given:

  1. Volume of KMnO4 solution used = 51.7 mL
  2. Concentration of KMnO4 solution = 0.145 M

We can use the formula:

moles = concentration x volume (in liters)

Converting the volume to liters:

51.7 mL = 51.7/1000 L = 0.0517 L

Calculating the moles of KMnO4:

moles of KMnO4 = 0.145 M x 0.0517 L = 0.0075 moles

Next, we need to calculate the moles of H2O2 in the solution.

Given:

  1. Mass of H2O2 solution = 40.0 g

We can use the molar mass of H2O2 to convert the mass to moles:

Molar mass of H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol

Calculating the moles of H2O2:

moles of H2O2 = 40.0 g / 34.02 g/mol = 1.176 moles

Finally, we can calculate the mass percentage of H2O2 in the solution:

mass percentage = (moles of H2O2 / total moles) x 100

total moles = moles of H2O2 + moles of KMnO4

total moles = 1.176 moles + 0.0075 moles = 1.1835 moles

mass percentage = (1.176 moles / 1.1835 moles) x 100 = 99.37%

Learn more about calculating mass percentage of a compound in a solution here:

https://brainly.com/question/34219130

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