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2NBr3 + 3NaOH = N2 +3NaBr + 3HOBr


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


N2

NBr3

NaOH

HOBr

Verified Answer :

Edufirst

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..

 

Answer:

NBr3

Explanation:

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