The reaction, `CO(g)+3H_(2)(g) hArr CH_(4)(g)+H_(2)O(g)` is at equilibrium at `1300 K` in a `1 L` flask. It also contains `0.30 mol` of `CO, 0.10 mol` of `H_(2)` and `0.02` mol of `H_(2)O` and an unknown amount of `CH_(4)` in the flask. Determine the concentration of `CH_(4)` in the mixture. The equilibrium constant `K_(c )` for the reaction at the given temperature us `3.90`.

Explanation

`K_c=([CH_4][H_2O])/([CO][H_2]^2)`
`3.90=([CH_4]xx0.02)/((0.30)(0.10)^3)`
or `[CH_4]=(3.90xx0.30xx(0.10)^3)/0.02`
`=5.85xx10^(-2)` M