Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, , can be titrated against a solution of potassium permanganate, . The following equation represents the reaction:

2KMnO4+H202-> 3H2SO4->O2+2MnSO4+ K2SO4+4H2O

A certain amount of hydrogen peroxide was dissolved in 100.mL of water and then titrated with 1.68M KMnO4. How much H2O2 was dissolved if the titration required 12.3mL of the KMnO4 solution?

To determine the amount of hydrogen peroxide (H2O2) dissolved in water, we can use the stoichiometry of the reaction between hydrogen peroxide and potassium permanganate.

From the balanced equation:

2 KMnO4 + H2O2 -> 3 H2SO4 + O2 + 2 MnSO4 + K2SO4 + 4 H2O

We can see that 2 moles of KMnO4 react with 1 mole of H2O2.

First, let's determine the number of moles of potassium permanganate (KMnO4) used in the titration. We know the concentration of the KMnO4 solution is 1.68 M, and the volume used is 12.3 mL.

moles of KMnO4 = concentration * volume
= 1.68 M * 0.0123 L
= 0.0206 moles

Since the stoichiometry is 2:1 between KMnO4 and H2O2, the number of moles of H2O2 used in the reaction is half the number of moles of KMnO4 used:

moles of H2O2 = 0.0206 moles / 2
= 0.0103 moles

Now we need to convert moles of H2O2 to grams. The molar mass of H2O2 is 34.02 g/mol.

mass of H2O2 = moles * molar mass
= 0.0103 moles * 34.02 g/mol
= 0.3503 grams

So, approximately 0.3503 grams of H2O2 were dissolved in 100 mL of water.

To determine the amount of H2O2 dissolved in the solution, we can use the balanced chemical equation provided:

2KMnO4 + H2O2 -> 3H2SO4 -> O2 + 2MnSO4 + K2SO4 + 4H2O

Given that the volume of the KMnO4 solution used in the titration is 12.3 mL and its molarity is 1.68 M, we can calculate the number of moles of KMnO4 used:

Moles of KMnO4 = Volume of KMnO4 solution (in L) x Molarity of KMnO4 solution (in mol/L)
= 12.3 mL / 1000 mL/L x 1.68 mol/L
= 0.02064 mol

According to the balanced chemical equation, the stoichiometric ratio between KMnO4 and H2O2 is 2:1. This means that for every 2 moles of KMnO4, there is 1 mole of H2O2.

Moles of H2O2 = Moles of KMnO4 / 2
= 0.02064 mol / 2
= 0.01032 mol

Finally, to determine the amount of H2O2 in grams, we need to multiply the number of moles by its molar mass. The molar mass of H2O2 is:

2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol

Mass of H2O2 = Moles of H2O2 x Molar mass of H2O2
= 0.01032 mol x 34.02 g/mol
= 0.351 g

Therefore, approximately 0.351 grams of H2O2 were dissolved in the solution.

moles KMnO4 = M x L = ??

moles H2O2 = 1/2 x moles KMnO4.
g H2O2 = moles H2O2 x molar mass H2O2.