For Reaction 7-1, how many milliliters of 0.165 0 M KMnO4 are needed to react with 108.0 mL of 0.165 0 M oxalic acid

43.2 mL

Explanation:

The reaction is:

5H₂C₂O₄ + 2MnO₄⁻ + 6H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O      

We have:

C(KMnO₄) = 0.1650 M

V(KMnO₄) =?

C(H₂C₂O₄) =  0.165 M

V(H₂C₂O₄) = 108.0 mL

From equation (1) we have that 5 moles of H₂C₂O₄ react with 2 moles of MnO₄⁻, so the number of moles of KMnO₄ is:

Now, we can find the volume of KMnO₄ as follows:

Therefore, are needed 43.2 mL of KMnO₄.

I hope it helps you!      

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