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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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Given: `0.087 "mol of" NO "and"0.0437 "mol of" Br_(2)`
Substitution & Calculation:
`0.0518` mol of `NOBr` is formed from `0.0518 "mol of" NO "and" 0.0518//2=0.0259 "mol of" Br_(2)`.
`therefore "At equilibrium Amoun of" NO=0.087-0.0518=0.0352 "mol"`
Amount of `Br_(2)=0.0437-0.0259=0.0178 "mol"`

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