Nitric oxide reacts with $Br_{2}$ and gives nitrosyl bromide as per the reaction given below:
$2 NO_{(g)} + Br_{2(g)} \longleftrightarrow 2 NOBr_{(g)}$
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 the equilibrium amount of $NO$ and $Br_{2}$.

The given reaction is:
$2 NO_{(g)} + Br_{2(g)} \longleftrightarrow 2 NOBr_{(g)}$
From the stoichiometry,$2 \ mol$ of $NOBr$ are formed from $2 \ mol$ of $NO$ and $1 \ mol$ of $Br_{2}$.
Therefore,$0.0518 \ mol$ of $NOBr$ are formed from $0.0518 \ mol$ of $NO$ and $\frac{0.0518}{2} = 0.0259 \ mol$ of $Br_{2}$.
Initial amounts are: $[NO]_{initial} = 0.087 \ mol$ and $[Br_{2}]_{initial} = 0.0437 \ mol$.
The amount of $NO$ present at equilibrium is:
$[NO]_{eq} = 0.087 - 0.0518 = 0.0352 \ mol$.
The amount of $Br_{2}$ present at equilibrium is:
$[Br_{2}]_{eq} = 0.0437 - 0.0259 = 0.0178 \ mol$.