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How many moles of O(2) will be liberated...

How many moles of `O_(2)` will be liberated by one mole of `CrO_(5)` is the following reaction:
`CrO_(5) + H_(2) SO_(4) rarr Cr_(2) (SO_(4))_(3) + H_(2)O + O_(2)`

A

`5//2`

B

`5//4`

C

`9//2`

D

`7//2`

Text Solution

Verified by Experts

The correct Answer is:
D
`2CrO_(5)+3H_(2)SO_(4) rarr Cr_(2)(SO_(4))_(3)+3H_(2)O+7//2O_(2)`
`1 mole CrO_(5)` liberates `rarr 7//2 mol e` of `O_(2)`
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How many moles of O_(2) will be liberated by one mole of CrO_(5) is the following reaction: CrO_(5) + H_(2) SO_(4) rarr Cr_(2) (SO_(4))_(5) + H_(2)O + O_(2)

In a balanced redox reaction net gain of electron (s) is equal to net loss of electrons (s). n_("factor") is a reaction specific parameter and for intermolecular redox reaction n-factor of oxidising reducing agent is the no. of moles of electron gained /lost by one mole of compound. Consider the following reaction: CrO_(5)+H_(2)SO_(4) to Cr_(2)(SO_(4))_(3)+H_(2)O+O_(2) One mole of CrO_(5) will liberate how many moles of O_(2) ?

The equivalent weight of H_(2)SO_(4) in the following reaction is Na_(2)Cr_(2)O_(7)+3SO_(2)+H_(2)SO_(4)rarr3Na_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+H_(2)O

What is the value of n in the following equation : Cr(OH)_(4)^(-)+OH^(-) rarr CrO_(4)^(2-)+H_(2)O+n e ?

Na_(2)SO_(3)+H_(2)O_(2) rarr Na_(2)SO_(4) +H_(2)O , in reaction

Which of the following statements is/are correct about the reaction. 4CrO_(5)+6H_(2)SO_(4)to2Cr_(2)(SO_(4))_(3)+6H_(2)O+7O_(2)

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