Page 1 of 1
Equivalent based questions (Valency Factors)
Posted: Sat Sep 11, 2021 1:15 am
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?
Re: Equivalent based questions (Valency Factors)
Posted: Sat Sep 11, 2021 2:03 am
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
Re: Equivalent based questions (Valency Factors)
Posted: Sat Sep 11, 2021 7:42 am
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.
Re: Equivalent based questions (Valency Factors)
Posted: Sun Sep 12, 2021 3:33 am
by ChenBeier
Show your calculations.
Re: Equivalent based questions (Valency Factors)
Posted: Mon Sep 13, 2021 4:05 am
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?
Re: Equivalent based questions (Valency Factors)
Posted: Mon Sep 13, 2021 9:11 am
by ChenBeier
Correct answers
Re: Equivalent based questions (Valency Factors)
Posted: Mon Sep 13, 2021 10:23 am
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?
Re: Equivalent based questions (Valency Factors)
Posted: Mon Sep 13, 2021 11:40 am
by ChenBeier
No your answer is correct
Ratio is 1 K
2Cr
2O
7 to 6 Na
2C
2O
4
We have 50 ml 0.3 M Na
2C
2O
4 what is 15 mmol
So we need 1/6 of 15 mmol = 2,5 mmol K
2Cr
2O
7
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