Nitric oxide reacts with `Br_(2)` and gives nitrosyl bromide as per reaction gives below:
`2NO(g)+Br_(2)(g)hArr 2NOBr(g)`
When 0.087 mol of NO and 0.0437 mol of `Br_(2)` are mixed in a closed container at constant pressure, 0.0518 mol of NOBr is obtained at equilibrium. calculate equilibrium amount of NO and `Br_(2)`.

Explanation

`2NO(g)+Br_(2)(g) hArr 2NOBr(g) ,` It is given that 0.0518 mol of Nobr is obtained at equilibrium. As per the above equation, 2 mol of NOBr is formed from 2 mol of NO and 1 mol of `Br_(2)`.
Therefore, the formation of 0.0518 mol of NOBr requires 0.0518 mol of NO and `(1)/(2)xx0.0518=0.0259 ` mol of `Br_(2)`.
Hence, at equilibrium the amount of NO=0.087-0.0518=0.0352 mol and that of `Br_(2)=0.0437-0.0259=0.0178mol`.