A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 was dissolved if the titration required 21.3 mL of the KMnO4 solution

see explanation below

Explanation:

First to all, this is a redox reaction, and the reaction taking place is the following:

2KMnO4 + 3H2SO4 + 5H2O2 > 2MnSO4 + K2SO4 + 8H2O + 5O2

According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:

MhVh = MpVp

h would be the hydrogen peroxide, and p the permanganate.

But like it was stated before, the mole ratio is 5:2 so:

5MhVh = 2MpVp

Replacing moles:

5nh = 2MpVp

Now, we just have to replace the given data:

nh = 2MpVp/5

nh = 2 * 1.68 * 0.0213 / 5

nh = 0.0143 moles

Now to get the mass, we just need the molecular mass of the peroxide:

MM = 2*1 + 2*16 = 34 g/mol

Finally the mass:

m = 0.0143 * 34

m = 0.4862 g