Consider a different titration for this exercise. Potassium permanganate (KMnO4)

Question

Consider a different titration for this exercise. Potassium permanganate (KMnO4) is the titrant and hydrogen peroxide (H2O2) the analyte according to the following balanced chemical equation. 2 MnO4 + 5 H2O2 + 6 H + 2 Mn2+ + 5 O2 + 8 H2O

(a) What is the stoichiometry of MnO4 to H2O2?

8:3

5:2

2:1

1:1

1:2

2:5

3:8

(b) Complete the following table for this titration. Titration of hydrogen peroxide with potassium permanganate.

concentration of MnO4 0.556 M

volume H2O2 solution 17.72 mL

mass H2O2 solution 17.80 g

volume of MnO4 solution 15.55 mL

mmol of MnO4 ___?__mmol

mmol of H2O2 __?__mmol

mass of H2O2 __?__g

mass % of H2O2 in the original sample __?__%

molarity of H2O2 in the original sample __?__M

Explanation / Answer

1)

2MnO4 + 5 H2O2 + 6 H + 2 Mn2+ + 5 O2 + 8 H2O

so the ans. is 2:5

2)

M1V1/n2 = M2V2/n1
0.556*15.55/5 = M2*17.72/2
M2 = 0.2 M
Molarity of H2O2 is 0.2M

moles of H2O2 = Molarity*volume in lt
               =0.2*0.01772 = 3.54*10^-3 moles
milli moles of H2O2 = 3.54*10^-6 milli moles.
....................................................
moles of MnO4^- = Molarity*volume in lt = 0.556*0.01555
                 = 8.64*10^-3 moles

milii moles = 8.64*10^-6 milli moles
.....................................................
mass of H2O2 = no.of moles * molecler wt.
             = 0.2* 34 = 6.4 gm.
.......................................................
mass % =( mass of H2O2/mass of H2O2 sample )*100
       = (6.4/17.8)*100 = 35.95%

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