Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below:
`2NO_((g))+Br_(2(g))hArr2NOBr_((g))`.
When `0.087 "mole"` of `NO` and `0.0437 "mole"` of `Br_(2)` are mixed in a closed container at constant temperature, `0.0518 "mole"` of `NOBr` is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

Explanation

`0.0518 mol` of `NOBr` is formed from `0.0518 mol` of `NO` and `0.0518//2=0.0259 mol of Br_(2)`.
`:.` At equilibrium,
amount of `NO=0.087-0.0518=0.0352 mol`
amount of `Br_(2)=0.0437-0.0259=0.0178 mol`