For the microscale titration between KMnO4 and H2O2 the balanced net ionic equat

Question

For the microscale titration between KMnO4 and H2O2 the balanced net ionic equation is 2 MnO4-(aq) + 5 H2O2(aq) + 6 H+(aq) --> 2 Mn2+(aq) + 5 O2(g) + 8 H2O(l)

-the concentration of KMnO4 is 4.19 M/M

-the initial mass of KMnO4 solution in pipet is 4.6716g

-the final mass of KMnO4 solutino in pipet is 3.8704 g

-mass of solution of KMnO4 solution titrated is 0.8012 g

-initial mass of H2O2 solution in pipet is 5.1094 g

-final mass of H2O2 solution in pipet is 4.5904 g

-mass of H2O2 solution added is 0.5190 g

1.) determine the mass of KMnO4 reacted.

2.) determine te moles of KMnO4 reacted

3.)determine the moles of H2O2 reacted

4.) determine the mass of H2O2 reacted

please show steps..

Explanation / Answer

KMnO4 used = 4.19 % of 0.8012 g = 0.03357 g

So, moles of KMnO4 = 0.03357/[39+55+(4x16)] = 2.125 x 10^-4

Moles of H2O2 = 5/2 * 0.0002125 = 5.312*10^-4

Mass of H2O2 reacted = [2x1 + 2x16] x 5.312*10^-4 g = 0.01806 g

Conc. of H2O2 M/M = 100 * 0.01806/0.5190 = 3.48%

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