Question

Consider the chemical equation. 2NBr3 + 3NaOH mc008-1.jpg N2 + 3NaBr + 3HOBr If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?

Answer 1

Answer:- $NBr_3$ is excess reactant.

Solution:- The given balanced equation is:

$2NBr_3+3NaOH\rightarrow N_2+3NaBr+3HOBr$

From this equation, there is 2:3 mol ratio between $NBr_3$ and NaOH.

Let's calculate the moles of $NBr_3$  required to react with given moles of NaOH.

$48molNaOH(\frac{2molNBr_3}{3molNaOH})$

= $32molNBr_3$

from above calculations, 32 moles of $NBr_3$ are required where as it's 40 moles are available. So, $NBr_3$ is the excess reactant.

Excess moles of $NBr_3$ = 40 - 32 = 8