Equivalent based questions (Valency Factors)

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Dhamnekar Winod
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Equivalent based questions (Valency Factors)

Post by Dhamnekar Winod »

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?
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Re: Equivalent based questions (Valency Factors)

Post by ChenBeier »

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
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Re: Equivalent based questions (Valency Factors)

Post by Dhamnekar Winod »

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.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Equivalent based questions (Valency Factors)

Post by ChenBeier »

Show your calculations.
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Re: Equivalent based questions (Valency Factors)

Post by Dhamnekar Winod »

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?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Equivalent based questions (Valency Factors)

Post by ChenBeier »

Correct answers
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Re: Equivalent based questions (Valency Factors)

Post by Dhamnekar Winod »

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?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Equivalent based questions (Valency Factors)

Post by ChenBeier »

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
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