consider the chemical equation.\n2nbr₃ + 3naoh → n₂ + 3nabr + 3hobr\nif there are 40 mol of nbr₃ and 48 mol…

Explanation:

Step1: Determine mole - ratio from equation

The mole - ratio of $NBr_3$ to $NaOH$ from the equation $2NBr_3 + 3NaOH\rightarrow N_2+3NaBr + 3HOBr$ is $\frac{n_{NBr_3}}{n_{NaOH}}=\frac{2}{3}$.

Step2: Calculate moles of $NBr_3$ needed for given $NaOH$

If we have $n_{NaOH} = 48$ mol, the moles of $NBr_3$ required for complete reaction of $NaOH$ is $n_{NBr_3\ required}=\frac{2}{3}\times n_{NaOH}$. Substituting $n_{NaOH}=48$ mol, we get $n_{NBr_3\ required}=\frac{2}{3}\times48 = 32$ mol.

Step3: Compare available and required moles of $NBr_3$

We have $n_{NBr_3\ available}=40$ mol. Since $n_{NBr_3\ available}(40\ mol)>n_{NBr_3\ required}(32\ mol)$, $NBr_3$ is in excess.

Answer:

B. $NBr_3$