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*00259 "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" `.