2(8H+ + MnO4- + 5e- --> Mn2+ + 4H2O)
5(H2O2 --> O2 + 2H+ + 2e-)
--------------------------------------…
16H+ + 2MnO4- + 5H2O2 --> 2Mn2+ + 5O2 + 8H2O + 10H+
5(H2O2 --> O2 + 2H+ + 2e-)
--------------------------------------…
16H+ + 2MnO4- + 5H2O2 --> 2Mn2+ + 5O2 + 8H2O + 10H+
The balanced chemical equation for this reaction is:
2 KMnO4 + 3 H2O2 + 2 H2SO4 --> 2 MnSO4 + 3 O2 + K2SO4 + 4 H2O
This equation tells you that it takes 2 moles of KMnO4 to react with 3 moles of H2O2 in the presence of 2 moles of H2SO4.
Now, you can set up a simple proportion to find out the number of moles of KMnO4 required:
2 moles of KMnO4 / 3 moles of H2O2 = x moles of KMnO4 / 5 moles of H2O2
Cross-multiplying, you get:
2 moles of KMnO4 * 5 moles of H2O2 = 3 moles of H2O2 * x moles of KMnO4
10 moles of KMnO4 = 3 moles of H2O2 * x moles of KMnO4
Dividing both sides by 3 moles of H2O2:
10 moles of KMnO4 / 3 moles of H2O2 = x moles of KMnO4
So, the number of moles of KMnO4 required to react with 5 moles of H2O2 in acidic medium is approximately 3.33 moles.