Answer:
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:
[tex] n_{KMnO_{4}} = \frac{2}{5}*n_{H_{2}C_{2}O_{4}} = \frac{2}{5}*(0.165 \frac{mol}{L}*0.108 L) = 7.13 \cdot 10^{-3} moles [/tex]
Now, we can find the volume of KMnO₄ as follows:
[tex] V = \frac{n}{C} = \frac{7.13 \cdot 10^{-3} moles}{0.165 M} = 0.0432 L = 43.2 mL [/tex]
Therefore, are needed 43.2 mL of KMnO₄.
I hope it helps you!
