If there are 40 mol of nbr3 and 48 mol of naoh, what is the excess reactant?

First, write the chemical reaction and balance it:

 NBr3 +  NaOH ---> N2 +  NaBr +  HOBr

 2 NBr3 + 3 NaOH ---> N2 + 3 NaBr + 3 HOBr

Then the theoretical proportion is 2 NBr3 : 3 NaOH

The reactants are in the proportion 40 NBr : 48 NaOH

From the theoretical proportion you can make [2 NBr3 / 3 NaOH] * 48 NaOH = 32 NaOH.

Given that you have more than 32 moles of NaOH, this is the excess reactant.

Sorry i dont understand the question can you rephase it?