If there are 40 mol of nbr3 and 48 mol of naoh, what is the excess reactant

              Excess Reagent =  NBr₃

The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,

                       2 NBr₃ + 3 NaOH   →   N₂ + 3 NaBr + 3 HBrO

Calculating the Limiting Reagent,

According to Balance equation,

               2 moles NBr₃ reacts with  =  3 moles of NaOH

So,

           40 moles of NBr₃ will react with  =  X moles of NaOH

Solving for X,

                       X  =  (40 mol × 3 mol) ÷ 2 mol

                       X  =  60 mol of NaOH

It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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