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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

2 Answers

2 votes

Answer:

NBr3

Step-by-step explanation:

User Justadampaul
by
7.1k points
3 votes
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..


User MobIT
by
7.1k points