The equilibrium constant for the reaction:
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` at `725K`
is `6.0xx10^(-2)`. At equilibrium `[H_(2)]=0.25 "mol" L^(-1)` and `[NO_(3)]=0.06 "mol" L^(-1)`
Calculate the equilirbium concentration of `N_(2)`.

Explanation

Given
`N_(2_((g)))+3H_(2_((g))) hArr 2NH_(3_((g)))`
`K_(c)=6xx10^(-2)`
`[H_(2)]=0.25` moles/lit
`[NH_(3)]=0.06` moles/lit
`K_(C)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
`6xx10^(-2)=((0.06)^(2))/((N_(2))(0.025)^(3))`
`[N_(2)]=((0.06)^(2))/(6xx10^(-2)xx(0.25)^(3))`
`=(0.0036)/(0.015625xx6xx10^(-2))`
`=(0.0036)/(0.09375xx10^(-2)`
`=3.84` mole/lit