1. Consider the chemical equation. If there are 40 mol of NBr3 and 48 mol of NaOH, what is the limiting reactant?
2NBr3 + 3NaOH --> N2 + 3NaBr + 3HOBI
ON2
ONBr3
O NaOH
O HOBI

NaOH

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2NBr3 + 3NaOH —> N2 + 3NaBr + 3HOBr

The limiting reactant can be obtained as follow:

From the balanced equation above,

2 moles of NBr3 reacted with 3 moles of NaOH.

Therefore, 40 moles of NBr3 will react with = (40 x 3)/2 = 60 moles of NaOH.

From the above calculations,

A higher amount of NaOH i.e 60 moles than what was given from the question i.e 48 moles is required to react completely with 40 moles of NBr3. Therefore, NaOH is the limiting reactant and NBr3 is the excess reactant.