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.

Answer: NBr3 ...