[cruise1521]

Question is:
calculate the number of moles of the H2O2 in the 10.00mL sample in your titration from volume of KMnO4 used and the stoichiometry of the reaction.

My data is that is takes 18.05mL KMnO4 to titrate 10.00mL H2O2. KMnO4 has 0.20M.


this is what I worked out. Would (liters of solution) be only the volume used to titrate or would it also include the volume of H2O2? I did my calculations below only using the KMnO4 volume it took to titrate. Would this also be the stoichiometry because I am going from KMnO4 to H2O2.

M=moles solute/liters solution
(M)(liters solution)=moles solute

 
(0.20M KMnO4)(0.01805L KMnO4) = 3.61xE-3 = 3.6xE-3 moles KMnO4

(5molH2O2 / 2molMnO4)(3.6xE-3moles KMnO4) = 9.0xE-3moles H2O2

[sdekivit]

during this titration, a redox reaction occurs. The first thing you always need to do is write down the reaction equation.

(btw: is the KMnO4 acidified or not?)

[Yggdrasil]

The calculations you did are correct.  Since you are calculating the moles of KMnO4 added to the solution, you only need to take into account the volume of the KMnO4 solution used and not the volume of the H2O2 solution.  Stoichiometry refers to the ratio of KMnO4 to H2O2 in the balanced chemical reaction, which you use correctly in your calculations to convert from moles of KMnO4 to H2O2.