Question

The density of a 0.0122 M KMnO4 is 1.037 g/mL. Suppose 26.35 g of 0.0122 M KMnO4 are required to titrate 1.072 g of a household H2O2 solution.

a)Calculate the mL of MnO4^- added to reach the endpoint.
b)Calculate the moles of MnO4^- added to reach the endpoint.
c)Calculate the number of moles of H2O2 in sample.
d)Calculate the number of grams of H2O2 in the sample.
e)Calculate the %m/m H2O2 in the household H2O2 solution.


I have no idea where to start for this. Any help is appreciated! Thank you.

Answer 1

And I have no intention of working this problem BUT I certainly can tell you where to start.
You ALWAYS start by writing and balancing the equation.
H2O2 + KMnO4 ==>

a. You have density and grams KMnO4 solution, convert g to mL.
b. You now have mL and molarity KMnO4, mols = M x L = ?
c. You have grams of H2O2. mols = grams/molar mass
d. You have mols from c. grams = mols x molar mass
e. $ = (grams H2O2/mass sample)*100 = ?

There, I've done the whole thing for you. Check my thinking. Sometimes I get ahead of myself and don't read a problem right. Post your work if you get stuck.