Nitric oxide reacts with `Br_(2)` and gives nitrosul bromide as per reaction given 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 temperature `0.0518` mol of `NOBr` is obtained at equilibrium. Calculate equilibrium amount of `NO` and `Br_(2)`.

Explanation

Suppose x moles of NO react till equilibrium is attained.
` {:(,, 2NO(g)+, Br_2(g), hArr , 2NOBr(g)),(" Initial no. of moles ",,0.087 ,, 0.0437 ,-),(" Moles at equilibrium",, 0.087 -x,,0.0437-(x)/(2) ,x):}`
Given that ` " " x= 0.0518 `
No. of moles of NO(g) at equilibrium =0.087 - x
` " "= 0.087 - 0.0518`
` " " = 0.0352`
and number of moles of `Br_2 (g)` at equilibrium
` " " = 0.0437 -(x)/(2)`
` " " = 0.0437 - ( 0.0518)/(2)`
` " " = 0.0178`