0.5 mol of `H_(2) and 0.5` mole of `I_(2)` react in 10 litre flask at `448^(@)C`. The equilibrium brium constant `K_(C)` is 50 for
`H_(2)(g)I_(2)(g) hArr 2HI(g)`
Calcualte mole of `I_(2)` at equilibrium.

Explanation

`H_(2_((g)))+I_(2_((g))) hArr 2HI_((g))`
`H_(2_((g)))+I_(2_((g))) hArr 2HI_((g))`
Initial `0.5 " " 0.5 " 0`
At equilibrium `0.5-x 0.5-x " " 2x`
`K_(C)=([HI]^(2))/([H_(2)][I_(2)])`
`50=((2x)^(2))/((0.5-x)(0.5-x))`
`(2x)/(0.5-x)=7` (approxex)
`x=(9)/(3.5)=0.3888`
`:.` No of moels of `I_(2)=0.5-0.3888`
`=0.111` moles