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..

 

It conforms what ever surface its in.