Determining The Excess Reactant In The Reaction 2 NBr3 + 3 NaOH

Introduction

In chemical reactions, reactants are not always present in perfect stoichiometric ratios. This means that one reactant might be present in a greater amount than required for the reaction to proceed completely, while another reactant might be limiting, thus controlling the amount of product formed. Identifying the excess reactant is crucial in understanding reaction yields and optimizing chemical processes. This article will delve into the process of determining the excess reactant in the given chemical equation:

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

Given that there are 40 moles of $NBr_3$ and 48 moles of $NaOH$, we will explore how to identify which reactant is in excess and how this affects the reaction's outcome. This is a fundamental concept in stoichiometry, which is the quantitative study of reactants and products in chemical reactions. Understanding stoichiometry allows chemists and students alike to predict the amounts of products formed and reactants consumed in a chemical reaction, ensuring efficient and safe chemical processes.

Stoichiometry Fundamentals

Before we dive into the specifics of the given reaction, let's briefly review some key concepts in stoichiometry. Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. At the heart of stoichiometry is the balanced chemical equation, which provides the mole ratios of reactants and products. These mole ratios are essential for determining the limiting and excess reactants.

The limiting reactant is the reactant that is completely consumed in a chemical reaction. It limits the amount of product that can be formed. Once the limiting reactant is used up, the reaction stops, regardless of the amount of other reactants present. Conversely, the excess reactant is the reactant that is present in a greater amount than required for the reaction to react completely with the limiting reactant. Some of the excess reactant will be left over after the reaction is complete.

To determine the limiting and excess reactants, we typically compare the mole ratio of the reactants available to the mole ratio required by the balanced chemical equation. This comparison helps us identify which reactant would be used up first and which would be left over. In this article, we will apply these principles to the reaction between $NBr_3$ and $NaOH$ to identify the excess reactant.

Understanding the Chemical Equation

The given chemical equation is:

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

This balanced equation tells us that two moles of nitrogen tribromide ($NBr_3$) react with three moles of sodium hydroxide ($NaOH$) to produce one mole of nitrogen gas ($N_2$), three moles of sodium bromide ($NaBr$), and three moles of hypobromous acid ($HOBr$). The coefficients in front of each chemical formula represent the stoichiometric coefficients, which are the mole ratios of the reactants and products. These ratios are the key to determining the limiting and excess reactants.

Mole Ratios

From the balanced equation, we can establish the following mole ratios:

• 2 moles of $NBr_3$ react with 3 moles of $NaOH$.
• 2 moles of $NBr_3$ produce 1 mole of $N_2$.
• 2 moles of $NBr_3$ produce 3 moles of $NaBr$.
• 2 moles of $NBr_3$ produce 3 moles of $HOBr$.
• 3 moles of $NaOH$ produce 1 mole of $N_2$.
• 3 moles of $NaOH$ produce 3 moles of $NaBr$.
• 3 moles of $NaOH$ produce 3 moles of $HOBr$.

These ratios allow us to compare the amounts of reactants needed for the reaction to proceed to completion. For example, if we have 2 moles of $NBr_3$, we would need 3 moles of $NaOH$ to react completely. If we have more than 3 moles of $NaOH$, then $NaOH$ is in excess, and $NBr_3$ is the limiting reactant. Conversely, if we have less than 3 moles of $NaOH$, then $NaOH$ is the limiting reactant, and $NBr_3$ is in excess. Understanding these stoichiometric relationships is crucial for solving problems related to limiting and excess reactants.

Calculating the Reactant Requirements

Given that we have 40 moles of $NBr_3$ and 48 moles of $NaOH$, we need to determine which reactant is the limiting reactant and which is the excess reactant. To do this, we can use the mole ratio from the balanced chemical equation to calculate the amount of one reactant required to react completely with the given amount of the other reactant.

Determining NaOH Required for $NBr_3$

First, let's calculate the amount of $NaOH$ required to react completely with 40 moles of $NBr_3$. From the balanced equation, we know that 2 moles of $NBr_3$ react with 3 moles of $NaOH$. We can set up a proportion to find the amount of $NaOH$ needed:

$\frac{3 \text{ moles } NaOH}{2 \text{ moles } NBr_3} = \frac{x \text{ moles } NaOH}{40 \text{ moles } NBr_3}$

Solving for x, we get:

$x = \frac{3 \text{ moles } NaOH}{2 \text{ moles } NBr_3} \times 40 \text{ moles } NBr_3 = 60 \text{ moles } NaOH$

This calculation shows that 60 moles of $NaOH$ are required to react completely with 40 moles of $NBr_3$. However, we only have 48 moles of $NaOH$ available. This indicates that $NaOH$ is the limiting reactant, and $NBr_3$ is the excess reactant.

Determining $NBr_3$ Required for $NaOH$

To further confirm our conclusion, let's calculate the amount of $NBr_3$ required to react completely with 48 moles of $NaOH$. Again, we use the mole ratio from the balanced equation:

$\frac{2 \text{ moles } NBr_3}{3 \text{ moles } NaOH} = \frac{y \text{ moles } NBr_3}{48 \text{ moles } NaOH}$

Solving for y, we get:

$y = \frac{2 \text{ moles } NBr_3}{3 \text{ moles } NaOH} \times 48 \text{ moles } NaOH = 32 \text{ moles } NBr_3$

This calculation shows that 32 moles of $NBr_3$ are required to react completely with 48 moles of $NaOH$. Since we have 40 moles of $NBr_3$ available, which is more than the required 32 moles, $NBr_3$ is indeed in excess, and $NaOH$ is the limiting reactant. This dual calculation approach provides a robust confirmation of the limiting and excess reactants in the given reaction scenario.

Identifying the Excess Reactant

Based on the calculations in the previous section, we have determined that 40 moles of $NBr_3$ require 60 moles of $NaOH$ for complete reaction. Since we only have 48 moles of $NaOH$ available, $NaOH$ is the limiting reactant. Conversely, 48 moles of $NaOH$ require 32 moles of $NBr_3$ for complete reaction. We have 40 moles of $NBr_3$ available, which is more than the required 32 moles. Therefore, $NBr_3$ is the excess reactant.

Excess Amount of $NBr_3$

To find the amount of $NBr_3$ that will be left over after the reaction is complete, we subtract the amount of $NBr_3$ that reacts (32 moles) from the initial amount (40 moles):

$40 \text{ moles } NBr_3 - 32 \text{ moles } NBr_3 = 8 \text{ moles } NBr_3$

This calculation indicates that 8 moles of $NBr_3$ will remain in excess after the reaction has gone to completion. This excess amount is crucial for understanding the reaction’s efficiency and the composition of the final reaction mixture.

Conclusion

In the given chemical reaction:

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

with 40 moles of $NBr_3$ and 48 moles of $NaOH$, we have determined that the excess reactant is $NBr_3$. The calculations showed that $NaOH$ is the limiting reactant, and $NBr_3$ is present in excess. Specifically, 8 moles of $NBr_3$ will remain unreacted after all the $NaOH$ has been consumed.

Understanding the concept of limiting and excess reactants is fundamental in stoichiometry and chemical reactions. It allows us to predict the amount of product that can be formed and the amount of reactant that will be left over. This knowledge is essential in various applications, including industrial chemistry, pharmaceutical research, and environmental science. By correctly identifying the limiting and excess reactants, we can optimize chemical processes, minimize waste, and ensure efficient use of resources.

In summary, by using the balanced chemical equation and the given amounts of reactants, we can accurately determine the excess reactant, which in this case is $NBr_3$. This process highlights the importance of stoichiometry in predicting and understanding chemical reactions.