7.9 Nitric oxide reacts with Br2 and gives nitrosyl bromide as per the reaction given below:

2NO (g) + Br2 (g) 2NOBr (g)

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amount of NO and Br2.

The given reaction is: 
                              2NOg+Brg2NOBrg2mol      1 mol        2 mol

Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr is formed from 0.0518 mol of NO.
Again, 2 mol of NOBr are formed from 1 mol of Br.    
Therefore, 0.0518 mol of NOBr is formed from 0.05182 mol of Br, or 0.0259 mol of NO.
The amount of NO and Br present initially is as follows:
[NO] = 0.087 mol [Br2] = 0.0437 mol
Therefore, the amount of NO present at equilibrium is:
[NO] = 0.087 – 0.0518 = 0.0352 mol
And, the amount of Br present at equilibrium is:
[Br2] = 0.0437 – 0.0259 = 0.0178 mol