A student titrates a 10.0 mL sample of H2O2. The titration required a 22.25 mL of 0.195 M KMnO4. Calculate the concentration (molarity) of the H2O2. Show all work.

See explanation below

Explanation:

In this case, we first need to write the general equation of the peroxide and the manganate which is:

H2O2 + KMnO4 > MnO2 + KOH + O2 + H2O

Now, we need to balance this. We can balance through REDOX, oxidation number or ion electron mode. Either method you use, the balanced reaction is the following:

3H2O2 + 2KMnO4 > 2MnO2 + 2KOH + 3O2 + 2H2O

Now that we have this, we can say that:

3 moles H2O2 = 2 moles KMnO4

moles can be expressed in terms of concentration and volume

n = M*V

Replacing this, above we have:

3 M* V H2O2 = 2 M* V KMnO4

We have volume and concentration of permanganate, and volume of peroxyde, so we can solve for the concentration:

M H2O2 = 2 * M * V KMnO4 / 3 V H2O2

Replacing the values we have:

M H2O2 = 2 * 0.195 * 22.25 / 3 * 10

M H2O2 = 0.283 M

i wish i could but this one is weird give me abc and i can solve it.