Nitric onxide reacts with bromine to from nitrosyl bromide as follows
`2NO(g)+Br_2(g)iff2NOBr(g)`
When 0.087 mol of NO and 0.0437 mole of `Br_2` are mixed in a closed containar at constant temperature,0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and `Br_2`.

Explanation

2 moles of NO gives 2 mole of NOBr
`therefore` Number of moles NO reacted to from 0.0518 mol of NOBr=0.0518
Number of moles of `Br_2` reacted= `0.0518/2=0.0259`
`therefore` Number of moles of NO remaining at equilibrium=0.087-0.0518=0.0352 mol
`therefore` Number of moles `Br_2` at equilibrium =0.0437-0.0259=0.0178 mol