Question

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

Answer 1

43.2 mL

Step-by-step 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:

`n_{KMnO_(4)} = (2)/(5)*n_{H_(2)C_(2)O_(4)} = (2)/(5)*(0.165 (mol)/(L)*0.108 L) = 7.13 \cdot 10^(-3) moles`

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

`V = (n)/(C) = (7.13 \cdot 10^(-3) moles)/(0.165 M) = 0.0432 L = 43.2 mL`

Therefore, are needed 43.2 mL of KMnO₄.

I hope it helps you!