Home
Class 11
CHEMISTRY
Four moles of acidified KMnO(4) reacts w...

Four moles of acidified `KMnO_(4)` reacts with excess of `H_(2)O_(2)` to produce 2X moles of `O_(2)` Then X is.....

Text Solution

Verified by Experts

The correct Answer is:
5
`2KMnO_(4)+3H_(2)SO_(4)+4H_(2)O_(2) to K_(2)SO_(4)+2MnSO_(4)+8H_(2)O+5O_(2)`
`2-5`
`4-?`
Promotional Banner

Topper's Solved these Questions

  • HYDROGEN AND ITS COMPOUNDS

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE-I) (LEVEL-I (MAIN) Straight Objective Type Questions )|10 Videos

  • HYDROGEN AND ITS COMPOUNDS

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE-I) (LEVEL-II (ADVANCED) Straight Objective Type Questions )|29 Videos

  • HYDROGEN AND ITS COMPOUNDS

    AAKASH SERIES|Exercise LECTURE SHEET EXERCISE -I (LEVEL-II (Matrix Matching Type Questions))|2 Videos

  • HYDROCARBONS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE SHEET ( LEVEL-II (PRACTICE SHEET (ADVANCED) (MATRIX MATCHING TYPE QUESTIONS)))|5 Videos

  • IONIC EQUILIBRIUM

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL -II PRACTICE SHEET (ADVANCED) (Integer Type Questions))|8 Videos

Similar Questions

Explore conceptually related problems

Four moles of KMnO_(4) reacts with excess of hypo solution to produce - moles of Na_(2)SO_(4)

The number of moles of H_(2)O_(2) needed to reduce 2 mole of KMnO_(4) in acidic medium is.

H_(2)O_(2) reacts repidly in

2mol FeSO_(4) in acid medium are oxidised by x mole of KMnO_(4) , where as 2 moleof FeC_(2)O_(4) in acid medium are oxidized by y mole of KMnO_(4) . The ratioof x and y is

Number of moles of K_(2)Cr_(2)O_(7) reduced by one mole of Sn^(2+) ions is

If x' litres of O_(2) is released at STP from one litre of H_(2)O_(2) solution due to decomposition of H_(2)O_(2) then we lable such solution as 'x volume H_(2)O_(2) 2H_(2)O_(2) to 2H_(2)O+O_(2) 30% (w/v) H_(2)O_(2) is called perhydrol. How much volume of 11.2 volumes H_(2)O_(2) is required to completely oxidise one mole of lead sulphite into lead sulphite?

If x' litres of O_(2) is released at STP from one litre of H_(2)O_(2) solution due to decomposition of H_(2)O_(2) then we lable such solution as 'x volume H_(2)O_(2) 2H_(2)O_(2) to 2H_(2)O+O_(2) 30% (w/v) H_(2)O_(2) is called perhydrol. How much volume of 224 volumes H_(2)O_(2) can completely reduce 10 ml of 0.1M KMnO, Into

Number of moles of KMnO_(4) required to oxidize one mole of Fe(C_(2)O_(4)) in acidic medium is