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 as follows:

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

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

If It requires 51.7 mL of 0.145 M KMnO, to titrate 40.0 g of the solution of hydrogen peroxide, then the mass percentage of H2O2 in the solution is found to be 0.32%.

First, we need to determine the moles of KMnO4 used. Moles = volume (L) x concentration (mol/L). We convert the given volume to liters: 51.7 mL = 0.0517 L. Then, we calculate the moles of KMnO4: Moles KMnO4 = 0.0517 L x 0.145 mol/L = 0.0075 mol.

Next, we use the stoichiometry of the reaction to find the moles of H2O2: Moles H2O2 = (0.0075 mol KMnO4)(1 mol H2O2/2 mol KMnO4) = 0.00375 mol.

Finally, we calculate the mass of H2O2: Mass H2O2 = (0.00375 mol H2O2)(34.0147 g/mol)

= 0.128 g.

The mass percentage of H2O2 in the solution is given by: (Mass H2O2 / Mass of solution) x 100%. Using the given mass of the solution (40.0 g), the mass percentage of H2O2 is: (0.128 g / 40.0 g) x 100% = 0.32%.

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