2nbr3 + 3naoh = n2 +3nabr + 3hobr

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

n2
nbr3
naoh
hobr

The theoretical proportion is given by the balanced chemical equation:

2 mol NBr / 3 mol  Na OH

Then x mol NaOH / 40 mol NBr3 = 3mol NaOH/2 mol NBr3

Solve for x, x = 40 * 3/2 = 60 mol NaOH.

Given that there are 48 mol NaOH (less than 60) this is the limitant reactant and the other is the excess reactant.

NBr3..