Browse by Stream
QnA
Home
QnA
get app
Go To Your Study Dashboard
  1. Home
  2. Engineering
  3. Try this! - Some basic concepts in chemistry - JEE Main-3
# Engineering

In order to oxidise a mixture of one mole of each of FeC_{2}O_{4},Fe_{2}(C_{2}O_{4})_{3},FeSO_{4} and Fe_{2}(SO_{4})_{3} in acidic medium, the number of moles of KMnO_{4} required is :

  • Option 1)

    1.5

  • Option 2)

    3

  • Option 3)

    1

  • Option 4)

    2

Answers (1)
S solutionqc

\\FeC_{2}O_{4}+KMnO_{4}\rightarrow Fe^{+3}+CO_{2}+Mn^{2+}\\V.F=3\; \; \;\; \; \; V.F=5

\Rightarrow +3 is the n=factor of FeC_{2}O_{4} and not Fe

\Rightarrow In this compound, carbon is in +3 oxidation state and Fe is in +2 oxidation state.

\RightarrowIn acidic medium , Fe^{2+} is converted to Fe^{+3} and C_{2}O_{4}^{2-} is  converted to CO_{2}. So oxidation number of carbon changes from +3\; to\; +4. As, there are two carbon atoms, the net change in O. Number is +2 and adding to this, the increase in oxidation state of iron, the final increase in O. Number is +3.

Hence n-factor of FeC_{2}O_{4} is +3.

1\times 3= mole of KMnO_{4}\times5\Rightarrow mole=\frac{3}{5}

\\Fe_{2}(C_{2}O_{4})_{3}+KMnO_{4}\rightarrow Fe^{3+}+CO_{2}+Mn^{2+}\\ \; \; V.F=6\; \; \; \; \; \; \; \; \; \; \; V.F=5

1\times 6=mole \times5\; of KMnO_{4}\; \; \; \; \Rightarrow mole\: of\; KMnO_{4}=\frac{6}{5}

\\\rightarrow FeSO_{4}+KMnO_{4}\rightarrow Fe^{+3}+SO_{4}^{2-}+Mn^{2+}\\V.F=1\; \! \; \; \; \; \; V.F=5

1\times 1 = mole of KMnO_{4}\times 5

mole of KMnO_{4}=\frac{1}{5}

\rightarrow Fe_{2}(SO_{4})_{3} doesn't oxidise

Total moles of KMnO_{4}=\frac{3}{5}+\frac{6}{5}+\frac{1}{5}=2


Option 1)

1.5

Option 2)

3

Option 3)

1

Option 4)

2

Similar Questions

Latest Asked Questions

Most Viewed Questions

Preparation Products

JEE Main Rank Booster 2022

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes(Classes will start from Last week of September), Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main 2022

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 4999/-
Buy Now
Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 2/-
Buy Now
Knockout JEE Main 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes(Classes will start from Last week of September), Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 24999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
Start Now
 
Exams
Articles
Questions