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If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?

User Mservidio
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2 Answers

4 votes

Answer:

NaOH

Step-by-step explanation:

User Codutie
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First, write the chemical reaction and balance it:

NBr3 + NaOH ---> N2 + NaBr + HOBr

2 NBr3 + 3 NaOH ---> N2 + 3 NaBr + 3 HOBr

Then the theoretical proportion is 2 NBr3 : 3 NaOH

The reactants are in the proportion 40 NBr : 48 NaOH

From the theoretical proportion you can make [2 NBr3 / 3 NaOH] * 48 NaOH = 32 NaOH.

Given that you have more than 32 moles of NaOH, this is the excess reactant.






User Mwijnands
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