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SCORPION-xisa [38]
2 years ago
6

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

Chemistry
1 answer:
alexira [117]2 years ago
8 0

Answer:

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

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