Answer:
6.11 mL of H₂C₂O₄ are needed
Step-by-step 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