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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.

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`{:(,,2NO_((g)),+,Br_(2(g)),hArr,2NOBr_((g))),(,"Initial mole",0.0870,,0.0437,,-),(,"Equilibrium",(0.087-2a),,(0.0437-a),,2a):}`
Given: `2a=0.0518`
`:. a=0.0259`
At equilibrium, `NO=0.0870-0.0518`
`=0.0352 "mole"`
`:. Br_(2)=0.0437-0.0259=0.0178 "mole"`

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