Question

Calculate the volume of 0.0200 M KMnO₄ needed to titrate 2.00mL of 0.882 M H₂O₂.

Answer 1

Balanced equation for the reaction is:
2 MnO4-(aq) + 5 H2O2(aq) + 6 H+(aq) --> 2 Mn2+(aq) + 5 O2(g) + 8 H2O(l)

Number of moles of H2O2 = M(H2O2)* V(H2O2)
                                                       = 0.882 M * 2*10^-3 L
                                                       = 1.764*10^-3 mol

5 mol of H2O2 reacts with 2 mol of MnO4-

So,
number of moles of KMNO4 required = 2*1.764*10^-3 / 5 = 7.056*10^-4 mol

use:
number of moles of KMNO4 = M(KMNO4 )* V(KMNO4 )
7.056*10^-4 = 0.02 * V
V=0.0353 L
   = 35.3 mL
Answer: 35.3 mL