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The volume of oxygen liberated from 0.68...

The volume of oxygen liberated from `0.68g` of `H_(2)O_(2)` is :

A

`112 ml`

B

`224 ml`

C

`56 ml`

D

`336 ml`

Text Solution

Verified by Experts

The correct Answer is:
B
We know that
`2H_(2)O_(2) rarr 2H_(2)O+O_(2)`
`2xx34g` `" "22400ml`
`:' 2 xx 34 g = 68 g ` of `H_(2)O_(2) ` liberates
`22400 ml` `O_(2)` at `STP`
`:. 0.68 g ` of `H_(2)O_(2)` liberates
`=(0.68 xx22400)/(68)=224 ml`
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