Chemistry Forum

a) What volume of 0.2 M K₂Cr₂O₇ is required to oxidize 50 ml of 0.3 M Na₂C₂O₄ in acidic medium?

b)How many moles of KMnO₄ will be required to react completely with 1 mole of K₂C₂O₄ in acidic medium?

c) What volume of 0.1 M KMnO₄ is required to oxidize 100 ml of 0.3M FeC₂O₄ (Ferrous Oxalate) in acidic medium?

d) How many moles of FeC₂O₄ is required to reduce 2 mol of KMnO₄ in acidic medium?

e)How many moles of K₂Cr₂O₇ are required to react completely with 2.5 moles of Cu₂S in acidic medium?

Solutions:
How to answer all these questions? What are the answers to these questions?

Develop the redox equation. Then you can see the ratio of the moles easily.

Hint acidic medium means, if oxygen will be transferred.

Oxidation add H₂O and get H+
Reduction add H+ and get H₂O

Oxalate will get CO₂

My attempt to answer all these questions:


a) 75 mL of 0.2M K₂Cr₂O₇ is required to oxidize 50 mL of 03M Na₂C₂O₄ in acidic medium.

b) 0.4 mole of KMnO₄ is required to react completely with 1 mole of K₂C₂O₄ in acidic medium.

c) 120 mL of 0.1M KMnO₄ is required to oxidize 100 mL of 0.3M FeC₂O₄ in acidic medium.

d) 5 moles of FeC₂O₄ are required to reduce 2 moles of KMnO₄ in acidic medium.

e) 5 moles of K₂Cr₂O₇ are required to react completely with 2.5 moles of Cu₂S in acidic medium.

Are these above answers correct ?
Let me know. Thanks.

Show your calculations.

Answer to a) Redox equation is K₂Cr₂O₇ + 6Na₂C₂O₄ + 14H^+ → 12CO₂ + 2Cr + 2K^+ + 12Na^+ + 7H₂O
From the aforesaid redox equation, we can say that to oxidize 6 moles of Na₂C₂O₄ (Sodium Oxalate), one mole of Potassium Dichromate (K₂Cr₂O₇) is required. So, for 0.3M Sodium Oxalate 0.05 M Potassium Dichromate is required. We have 0.2 M K₂Cr₂O₇ solution.


Molar Mass of Potassium Dichromate is 294.181 grams. To oxidize 50 ml of 0.3 M Na₂C₂O₄, we require \(\frac{14.70905}{58.8362}\times \frac{50 ml }{1000 ml}= 12.5 ml\) of 0.2 M Potassium Dichromate.

Answer to b) Redox equation is 2KMnO₄ + 5 K₂C₂O₄ + 16 H^+ → 10CO₂ + 8H₂O + 12 K^+ + 2Mn^2+

From the above redox equation, we can say that to oxidize 5 moles of K₂C₂O₄(Potassium Oxalate), 2 moles of Potassium permanganate (KMnO₄) are required. So for 1 mole of K₂C₂O₄ , \(\frac25\) mole of KMnO₄ is required.

Answer to c) Redox equation is 2KMnO₄ + 5 FeC₂O₄ + 16 H^+ →10 CO₂ + 8 H₂O + 2 K^+ + 2 Mn^2+ + 2 Fe^2+

From the aforesaid redox equation, we can say that to oxidize 5 moles of Ferrous Oxalate(FeC₂O₄), 2 moles of KMnO₄ (Potassium Permanganate) are required . So for 0.3 M of FeC₂O₄ , 0.12 M KMnO₄ is required. But we have 0.1M of KMnO₄. Molar mass of KMnO₄ is 158.032 grams . So we require to oxidize 100 ml of 0.3 M of FeC₂O₄, 120 ml of 0.1 M of KMnO₄.

Answer to d)5 moles of ferrous Oxalate (FeC₂O₄) are required to reduce 2 moles of Potassium Permanganate in acidic medium.

Answer to e) Redox equation is K₂Cr₂O₇ + 3Cu₂S + 14 H^+ → 6Cu^2+ + 3S + 7 H₂O + 2K^+ + 2 Cr
So we can say that \(\frac56\) moles of K₂Cr₂O₇ is required to react completely with 2.5 moles of Cu₂S in acidic medium


Are these answers correct?

Correct answers

I think my answer to a) is wrong. Your answer 75 ml of K₂Cr₂O₇ is correct.

But I have another answer as well
2) Using this formula v.f.1 × M₁ × volume1= v.f.2× M₂ × volume2
Plugging in the values available with us into this formula, we have 6×0.2M× V₁=2×0.3M × 50 ml

= 25 ml of Potassium Dichromate

What is wrong with this answer?

No your answer is correct

Ratio is 1 K₂Cr₂O₇ to 6 Na₂C₂O₄

We have 50 ml 0.3 M Na₂C₂O₄ what is 15 mmol


So we need 1/6 of 15 mmol = 2,5 mmol K₂Cr₂O₇

The molarity is 0,2 M so we need 2,5mmol/0,2 mmol/ml = 12,5 ml

Your question

we have 6×0.2M× V₁=2×0.3M × 50 ml

Where does the 2 comes from, it is the same like my equation above without 2