How many milliliters of 0.021 moles of oxalic acid is needed to react with 80 mL of 0.11 moles of KMnO4

6.11 mL of H₂C₂O₄ are needed

Explanation:

We determine the reaction which is a redox one, in acidic medium

C₂O₄⁻² + MnO₄⁻ → CO₂ + Mn²⁺

C₂O₄⁻² → 2CO₂ + 2e⁻ Oxidation

Carbon changes the oxidation state, from + 3 to + 4

5e⁻ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O Reduction

We add 4 water to the product side, in order to balance the oxygen and, we have 8H + in the reactant side, in order to balance the H

Mn changes the oxidation state from + 7 to + 2

(C₂O₄⁻² → 2CO₂ + 2e⁻).5

5C₂O₄⁻² → 10CO₂ + 10e⁻

(5e⁻ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O).2

10e⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O

10e⁻ + 2MnO₄⁻ + 16H⁺ + 5C₂O₄⁻² → 2Mn²⁺ + 8H₂O + 10CO₂ + 10e⁻

The electrons are cancelled, so the balanced reaction is:

2KMnO₄ + 6HCl + 5H₂C₂O₄ → 2MnCl₂ + 8H₂O + 10CO₂ + 2KCl

Concentration of KMnO₄ = 0.11 mol / 0.080mL = 1.375M

Imagine that the reactants are in molar concentration (mol/L)

Ratio in stoichiometry is 2:5

2 moles of KMnO₄ react to 5 moles of oxalic acid

Then, 1.375 moles of KMnO₄ will react to (1.375 moles. 5) / 2 = 3.437 M

Concentration of H₂C₂O₄ = 3.437 M (mol/L)

3.437 mol/L = 0.021 moles / Volume (L)

0.021 moles / 3.437 mol/L = Volume (L) → 0.00611 L

0.00611 L. 1000 mL / 1L = 6.11 mL