Moles Of Nitrogen Needed For Ammonia Formation Stoichiometry Calculation

Jul 13, 2025 by ADMIN 73 views

Stoichiometry of Ammonia Formation Moles of Nitrogen Needed

In the realm of chemistry, understanding the quantitative relationships between reactants and products in chemical reactions is paramount. This is where the concept of stoichiometry comes into play. Stoichiometry allows us to predict the amount of reactants needed and products formed in a given reaction. In this comprehensive article, we'll delve into a specific chemical reaction: the formation of ammonia ( $NH_3$ ) from nitrogen ( $N_2$ ) and hydrogen ( $H_2$ ). Our primary goal is to determine the number of moles of nitrogen required to completely convert a given amount of hydrogen. This is a quintessential stoichiometry problem that demonstrates the practical application of mole ratios derived from balanced chemical equations.

The cornerstone of stoichiometric calculations is the balanced chemical equation. This equation provides the precise molar ratios between reactants and products. For the synthesis of ammonia, the balanced equation is:

N_2 + 3H_2 \rightarrow 2NH_3

This equation tells us that one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia gas. This 1:3:2 ratio is the key to solving our problem. We will use this ratio as a conversion factor to calculate the moles of nitrogen needed for the complete reaction of 6.34 moles of hydrogen. We will also discuss the importance of understanding balanced equations and mole ratios for solving various stoichiometry problems.

Understanding the mole concept is essential for this calculation. A mole is simply a unit of measurement for the amount of a substance. It represents Avogadro's number ( $6.022 \times 10^{23}$ ) of entities, such as atoms, molecules, or ions. In stoichiometry, the mole allows us to relate the masses of reactants and products to the number of particles involved in the reaction. By working with moles, we can accurately predict the quantities of substances needed or produced in a chemical reaction. This concept is not only fundamental in chemistry but also has practical applications in various fields, such as industrial chemistry and environmental science.

In the following sections, we will systematically solve the problem, providing a step-by-step explanation of the calculations involved. We will also discuss common mistakes to avoid and offer additional practice problems to reinforce your understanding of stoichiometry. Whether you are a student learning chemistry or someone interested in the practical applications of chemical reactions, this article will provide you with the knowledge and skills to tackle similar stoichiometry problems with confidence.

Stoichiometric Calculation: Moles of Nitrogen Required

To accurately determine the moles of nitrogen ( $N_2$ ) needed to completely convert 6.34 moles of hydrogen ( $H_2$ ) into ammonia ( $NH_3$ ), we must utilize the stoichiometric coefficients from the balanced chemical equation:

N_2 + 3H_2 \rightarrow 2NH_3

From this equation, we observe that one mole of nitrogen reacts with three moles of hydrogen. This 1:3 mole ratio between $N_2$ and $H_2$ is the key to our calculation. The mole ratio serves as a conversion factor that allows us to translate from moles of one substance to moles of another within the context of a balanced chemical reaction. To calculate the moles of nitrogen required, we will multiply the given moles of hydrogen by the mole ratio of $N_2$ to $H_2$ .

Let's break down the calculation step by step:

Identify the given quantity: We are given 6.34 moles of $H_2$ .
Identify the desired quantity: We want to find the moles of $N_2$ .
Apply the mole ratio: From the balanced equation, the mole ratio of $N_2$ to $H_2$ is 1:3. This means that for every 3 moles of $H_2$ that react, 1 mole of $N_2$ is required.
Set up the conversion: To convert moles of $H_2$ to moles of $N_2$ , we multiply the given moles of $H_2$ by the mole ratio:
$\text{Moles of } N_2 = \text{Moles of } H_2 \times \frac{\text{Moles of } N_2}{\text{Moles of } H_2}$
Plug in the values:
$\text{Moles of } N_2 = 6.34 \text{ mol } H_2 \times \frac{1 \text{ mol } N_2}{3 \text{ mol } H_2}$
Calculate the result:
$\text{Moles of } N_2 = \frac{6.34}{3} \text{ mol } N_2$ $\text{Moles of } N_2 = 2.11333... \text{ mol } N_2$
Round to the appropriate significant figures: Since the given quantity (6.34 mol) has three significant figures, we round our answer to three significant figures as well:
$\text{Moles of } N_2 \approx 2.11 \text{ mol } N_2$

Therefore, 2.11 moles of nitrogen are needed to completely convert 6.34 moles of hydrogen into ammonia. This calculation demonstrates the power of stoichiometry in predicting the quantitative relationships between reactants in a chemical reaction. By understanding and applying mole ratios derived from balanced chemical equations, we can accurately determine the amounts of substances required for complete reactions, which is crucial in various chemical processes.

Answer Evaluation and Stoichiometry Principles

In the previous section, we meticulously calculated that 2.11 moles of nitrogen ( $N_2$ ) are required to completely convert 6.34 moles of hydrogen ( $H_2$ ) into ammonia ( $NH_3$ ). Now, let's evaluate this result in the context of the given multiple-choice options and reinforce the underlying principles of stoichiometry.

The provided options are:

A. 1.02 mol
B. 2.11 mol
C. 12.68 mol
D. 19.02 mol

Our calculated answer, 2.11 mol, matches option B, which confirms the correctness of our calculation. This result is significant because it demonstrates the practical application of stoichiometry in predicting the quantitative relationships in chemical reactions. Let's delve deeper into why this answer is logical and how it aligns with the principles of stoichiometry.

Stoichiometry is fundamentally based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This law is reflected in balanced chemical equations, where the number of atoms of each element must be the same on both the reactant and product sides. The balanced equation for ammonia synthesis is:

N_2 + 3H_2 \rightarrow 2NH_3

This equation tells us that one molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia. At the macroscopic level, this translates to one mole of nitrogen reacting with three moles of hydrogen to produce two moles of ammonia. The mole ratio of $N_2$ to $H_2$ is 1:3, which is the crucial piece of information we used in our calculation.

If we consider the given amount of hydrogen, 6.34 moles, we can intuitively understand why approximately 2.11 moles of nitrogen are needed. Since the mole ratio is 1:3, the amount of nitrogen required should be one-third of the amount of hydrogen. A quick estimation confirms this: 6.34 divided by 3 is approximately 2.11. This mental check helps us validate our calculated result and ensures that it is within a reasonable range.

The other options provided are incorrect and highlight common mistakes students make when dealing with stoichiometry problems. For instance:

Option A (1.02 mol) is likely a result of incorrectly dividing the moles of hydrogen by a factor other than 3, or perhaps inverting the mole ratio.
Option C (12.68 mol) is likely obtained by multiplying the moles of hydrogen by a factor instead of dividing, possibly confusing the mole ratio.
Option D (19.02 mol) could be a result of more complex errors in applying the stoichiometric principles.

By correctly applying the mole ratio from the balanced equation, we avoided these errors and arrived at the correct answer. This exercise underscores the importance of carefully interpreting the balanced equation and using the mole ratios as conversion factors. Additionally, it's always a good practice to estimate the expected result to ensure that the calculated answer is reasonable.

Practical Applications and Further Practice on Ammonia Formation

Understanding the stoichiometry of ammonia formation is not just an academic exercise; it has significant practical applications in various industries and scientific fields. Ammonia ( $NH_3$ ) is a crucial compound in the production of fertilizers, which are essential for modern agriculture. The Haber-Bosch process, a widely used industrial process, synthesizes ammonia from nitrogen and hydrogen gases under high temperature and pressure conditions. The efficiency and yield of this process are heavily dependent on the stoichiometric relationships between the reactants.

In the Haber-Bosch process, the reaction is carefully controlled to ensure that the reactants are supplied in the correct mole ratio, maximizing the production of ammonia and minimizing waste. This highlights the importance of stoichiometry in optimizing chemical processes for industrial applications. Furthermore, understanding the stoichiometry of ammonia formation is vital in environmental science, as ammonia is a significant pollutant in water and air. By understanding the chemical reactions involving ammonia, scientists can develop strategies to mitigate its environmental impact.

To further solidify your understanding of stoichiometry and ammonia formation, let's consider some additional practice problems:

Problem 1: If 10.5 moles of nitrogen gas react completely with hydrogen gas, how many moles of ammonia gas will be produced?
Problem 2: How many grams of hydrogen gas are required to react completely with 56 grams of nitrogen gas to produce ammonia?
Problem 3: If 4 moles of ammonia are produced, how many moles of nitrogen and hydrogen were consumed?

Solving these problems requires applying the same stoichiometric principles we discussed earlier, including:

Identifying the balanced chemical equation: $N_2 + 3H_2 \rightarrow 2NH_3$
Determining the relevant mole ratios between reactants and products.
Using mole ratios as conversion factors to calculate the desired quantities.
Converting between moles and grams using molar masses when necessary.

By working through these practice problems, you will gain confidence in your ability to apply stoichiometry to various chemical reactions. Stoichiometry is a fundamental concept in chemistry, and mastering it will enable you to solve a wide range of quantitative problems in chemistry and related fields. Remember, the key to success in stoichiometry is a thorough understanding of balanced chemical equations, mole ratios, and the application of these concepts in problem-solving.

In conclusion, we have explored the stoichiometry of ammonia formation in detail, from calculating the moles of nitrogen required to react with a given amount of hydrogen to discussing the practical applications of this reaction. By understanding the principles of stoichiometry and practicing problem-solving, you can confidently tackle any stoichiometric challenge and appreciate the quantitative nature of chemical reactions.