Consider the chemical equation. 2NBr3 + 3NaOH Right arrow. N2 + 3NaBr + 3HOBr If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?

NBr₃ is the excess reactant.

Explanation:

We can solve this problem by calculating how many moles of NaOH would react completely with 40 moles of NBr₃, using the stoichiometric coefficients of the reaction:

As the required number of NaOH moles is higher than the available number (60 required vs 48 available), NaOH is the limiting reactant.

As such, the excess reactant is NBr₃.

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.

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