Nitric Oxide Formation Calculation From Ammonia Reaction

by Rajiv Sharma 57 views

Hey guys! Today, we're diving into a fascinating chemistry problem involving the formation of nitric oxide (NO) from the reaction of ammonia (NH3) with oxygen (O2). This is a classic stoichiometry question that helps us understand how to calculate the amount of product formed in a chemical reaction, especially when we consider the percent yield. Let's break it down step by step!

Understanding the Chemical Reaction

Before we jump into the calculations, let's first understand the chemical reaction we're dealing with. The balanced chemical equation is:

4 NH3 + 5 O2 → 4 NO + 6 H2O

This equation tells us that four molecules of ammonia (NH3) react with five molecules of oxygen (O2) to produce four molecules of nitric oxide (NO) and six molecules of water (H2O). The coefficients in front of each chemical formula are crucial because they represent the molar ratios of the reactants and products. Molar ratios are the key to solving stoichiometry problems, as they allow us to convert between the amounts of different substances in the reaction. In this specific reaction, the molar ratio between ammonia (NH3) and nitric oxide (NO) is 4:4, which simplifies to 1:1. This means that for every one mole of ammonia that reacts, one mole of nitric oxide is produced.

This reaction is an example of an oxidation reaction, where ammonia is being oxidized to nitric oxide. The oxygen acts as the oxidizing agent, accepting electrons from the ammonia. Nitric oxide is a crucial molecule in various biological processes and industrial applications. It plays a role in cell signaling, immune response, and vasodilation in the body. Industrially, nitric oxide is an important intermediate in the production of nitric acid, which is used in fertilizers and explosives. Understanding this reaction is not just an academic exercise; it has real-world implications in various fields.

The balanced equation is the foundation for all our calculations. It ensures that we are working with the correct proportions of reactants and products. Without a balanced equation, our stoichiometric calculations would be inaccurate. So, always double-check that your equation is balanced before proceeding with any calculations. This involves ensuring that the number of atoms of each element is the same on both sides of the equation. For example, in this case, we have 4 nitrogen atoms, 12 hydrogen atoms, and 10 oxygen atoms on both sides of the equation.

Step 1: Calculate the Molar Mass of Ammonia (NH3) and Nitric Oxide (NO)

To convert grams of ammonia to moles, we need to calculate its molar mass. The molar mass is the mass of one mole of a substance, which is numerically equal to its atomic or molecular weight in grams per mole (g/mol). To calculate the molar mass of ammonia (NH3), we add the atomic masses of one nitrogen atom (N) and three hydrogen atoms (H):

  • Atomic mass of N ≈ 14.01 g/mol
  • Atomic mass of H ≈ 1.01 g/mol

Molar mass of NH3 = 14.01 g/mol + 3(1.01 g/mol) = 14.01 g/mol + 3.03 g/mol = 17.04 g/mol

Next, we need to calculate the molar mass of nitric oxide (NO). We add the atomic masses of one nitrogen atom (N) and one oxygen atom (O):

  • Atomic mass of N ≈ 14.01 g/mol
  • Atomic mass of O ≈ 16.00 g/mol

Molar mass of NO = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol

Molar mass is a fundamental concept in chemistry because it allows us to convert between mass and moles. Moles are the chemist's way of counting atoms and molecules, as they provide a consistent way to relate the macroscopic world (grams) to the microscopic world (atoms and molecules). Without knowing the molar masses, we cannot accurately determine the number of moles present in a given mass of a substance. The molar mass calculation involves summing the atomic masses of all the atoms in the chemical formula, which are readily available in the periodic table. These values are essential for stoichiometric calculations, as they link mass, moles, and the balanced chemical equation.

Understanding molar mass is also crucial for preparing solutions of specific concentrations. In the lab, we often need to prepare solutions with a certain molarity, which is the number of moles of solute per liter of solution. To do this accurately, we need to convert the desired molarity to grams using the molar mass. The concept of molar mass is thus a cornerstone of quantitative chemistry, underpinning many experimental procedures and calculations. It is not just a number; it is a vital tool for chemists in their daily work.

Step 2: Convert Grams of Ammonia to Moles

Now that we have the molar mass of ammonia, we can convert the given mass of ammonia (34 g) to moles. To do this, we use the following formula:

Moles = Mass / Molar mass

Moles of NH3 = 34 g / 17.04 g/mol ≈ 1.995 moles

So, we have approximately 1.995 moles of ammonia.

Converting grams to moles is a crucial step in stoichiometry because chemical reactions occur on a molar basis, not on a mass basis. The balanced chemical equation represents the molar ratios of reactants and products, so we need to know the number of moles of each reactant to determine the theoretical yield of the products. The formula we used, Moles = Mass / Molar mass, is a simple yet powerful tool that allows us to bridge the gap between the macroscopic world (grams) and the microscopic world (moles).

In this case, we started with 34 grams of ammonia and converted it to approximately 1.995 moles. This value now represents the amount of ammonia in terms of moles, which is directly related to the number of molecules. This conversion is essential for using the molar ratios from the balanced chemical equation. The more accurately we determine the number of moles, the more accurate our subsequent calculations will be. Errors in this step can propagate through the rest of the calculation, so it's important to be meticulous and use the correct molar mass value.

Furthermore, the concept of moles is not just limited to stoichiometry. It is also fundamental in other areas of chemistry, such as gas laws, solution chemistry, and chemical kinetics. The mole concept provides a universal way to quantify the amount of a substance, regardless of its physical state or chemical nature. It is a cornerstone of quantitative chemistry, allowing chemists to make accurate measurements and predictions about chemical reactions and processes. Understanding this conversion is essential for anyone studying chemistry or related fields.

Step 3: Use the Molar Ratio to Find Moles of Nitric Oxide

From the balanced equation, we know that the molar ratio of NH3 to NO is 4:4, which simplifies to 1:1. This means that for every 1 mole of NH3 that reacts, 1 mole of NO is produced. Therefore:

Moles of NO = Moles of NH3 = 1.995 moles

So, theoretically, 1.995 moles of nitric oxide should be produced.

The molar ratio is the heart of stoichiometry, providing the direct link between the amounts of reactants and products in a chemical reaction. In this step, we used the molar ratio to determine the theoretical yield of nitric oxide. Because the ratio between ammonia and nitric oxide is 1:1, the number of moles of nitric oxide produced is equal to the number of moles of ammonia reacted. This simplifies the calculation, but it's important to remember that molar ratios can be different for other reactions, and careful attention must be paid to the coefficients in the balanced equation.

The molar ratio is derived directly from the balanced chemical equation, which represents the conservation of mass in a chemical reaction. The coefficients in the balanced equation tell us the relative numbers of moles of each substance involved in the reaction. These coefficients are crucial for calculating the molar ratios and determining the stoichiometric amounts of reactants and products. Incorrectly using the molar ratio is a common mistake in stoichiometry problems, so it's important to double-check the balanced equation and ensure the correct coefficients are used.

In this case, the 1:1 molar ratio made the calculation straightforward, but in reactions with more complex ratios, it's essential to set up the calculation carefully. The molar ratio allows us to predict the amount of product formed from a given amount of reactant, assuming the reaction goes to completion. However, in real-world scenarios, reactions may not proceed to 100% completion, and we need to consider the percent yield, which we will address in the next step. Understanding the molar ratio is not just about solving stoichiometry problems; it's about grasping the fundamental principles of chemical reactions and their quantitative relationships.

Step 4: Convert Moles of Nitric Oxide to Grams

Now, we convert the moles of nitric oxide back to grams using its molar mass:

Mass = Moles × Molar mass

Mass of NO = 1.995 moles × 30.01 g/mol ≈ 59.87 g

This is the theoretical yield of nitric oxide, assuming the reaction goes to completion.

Converting moles back to grams allows us to express the amount of product in a more practical unit. In the lab, we typically measure the mass of substances, so converting moles to grams gives us a value that we can directly compare to experimental results. The formula Mass = Moles × Molar mass is the inverse of the formula we used to convert grams to moles, and it completes the cycle of converting between mass and moles. This conversion is essential for determining the yield of a reaction and for comparing the theoretical yield to the actual yield.

In this step, we found that the theoretical yield of nitric oxide is approximately 59.87 grams. This is the maximum amount of nitric oxide that could be produced from 34 grams of ammonia, assuming that all of the ammonia reacts and there are no losses in the process. However, in reality, reactions rarely go to 100% completion, and there are often side reactions or losses during product isolation. This is why we need to consider the percent yield, which takes into account the efficiency of the reaction.

Understanding the conversion between moles and grams is also crucial for preparing solutions of specific concentrations. When making a solution, we need to weigh out a certain mass of solute and dissolve it in a specific volume of solvent. To do this accurately, we need to convert the desired molarity to grams using the molar mass. The ability to convert between moles and grams is thus a fundamental skill in chemistry, essential for both theoretical calculations and practical laboratory work. The theoretical yield is a benchmark against which we measure the success of an experiment, and it provides a goal to strive for when optimizing reaction conditions.

Step 5: Calculate the Actual Yield Considering the Percent Yield

However, the problem states that the reaction has a 98% yield. This means that we only obtain 98% of the theoretical yield. To calculate the actual yield, we use the following formula:

Actual Yield = (Percent Yield / 100) × Theoretical Yield

Actual Yield of NO = (98 / 100) × 59.87 g = 0.98 × 59.87 g ≈ 58.67 g

So, the actual amount of nitric oxide formed is approximately 58.67 grams.

Percent yield is a critical concept in chemistry that reflects the efficiency of a chemical reaction. In the real world, reactions rarely proceed to 100% completion due to various factors such as side reactions, incomplete reactions, and losses during product isolation and purification. The percent yield provides a measure of how much product is actually obtained compared to the maximum possible amount (the theoretical yield). It is calculated using the formula: Percent Yield = (Actual Yield / Theoretical Yield) × 100.

In this problem, the reaction has a 98% yield, which means that only 98% of the theoretical amount of nitric oxide is produced. This is a relatively high yield, indicating that the reaction is quite efficient under the given conditions. To calculate the actual yield, we multiplied the theoretical yield (59.87 g) by the percent yield (98%) and divided by 100. This gave us an actual yield of approximately 58.67 grams of nitric oxide.

The percent yield is an important metric for evaluating the success of a chemical reaction and for optimizing reaction conditions. Chemists often strive to maximize the percent yield by carefully controlling factors such as temperature, pressure, reaction time, and catalyst use. A low percent yield may indicate that the reaction is not proceeding efficiently or that there are significant losses during product isolation. Understanding and calculating the percent yield is essential for both laboratory work and industrial chemical processes. It helps in assessing the economic viability of a reaction and in making decisions about process optimization. The percent yield is not just a number; it's a reflection of the real-world challenges and complexities of chemical reactions.

Final Answer

Therefore, the amount of nitric oxide (NO) formed from 34g of ammonia (NH3) with a 98% yield is approximately 58.67 grams. None of the provided options (A) 12.5 g of NO is correct.

In this article, we have comprehensively solved the problem of calculating the amount of nitric oxide formed from the reaction of ammonia with oxygen, considering a 98% yield. We started by understanding the balanced chemical equation and then meticulously worked through the steps:

  1. Calculating the molar masses of ammonia and nitric oxide.
  2. Converting grams of ammonia to moles.
  3. Using the molar ratio to find the moles of nitric oxide.
  4. Converting moles of nitric oxide to grams (theoretical yield).
  5. Calculating the actual yield considering the percent yield.

By following these steps, we arrived at the final answer of approximately 58.67 grams of nitric oxide. This detailed solution not only provides the correct answer but also reinforces the fundamental concepts of stoichiometry, molar mass, molar ratios, and percent yield. Understanding these concepts is crucial for success in chemistry and related fields. Stoichiometry problems like this one are not just about plugging numbers into formulas; they are about understanding the quantitative relationships between reactants and products in chemical reactions. Each step in the solution builds upon the previous one, and a clear understanding of each concept is necessary for solving more complex problems. The ability to confidently tackle stoichiometry problems is a valuable skill for anyone pursuing a career in science or engineering.

Moreover, the problem we solved has real-world applications. Nitric oxide is an important molecule in various biological processes and industrial applications. The reaction between ammonia and oxygen is used in the production of nitric acid, which is a key ingredient in fertilizers and explosives. Therefore, understanding the stoichiometry of this reaction is not just an academic exercise; it has practical implications in various industries. By mastering stoichiometry, we gain the ability to predict and control the outcomes of chemical reactions, which is essential for both research and industrial processes. The principles we have discussed in this article can be applied to a wide range of chemical reactions, making stoichiometry a cornerstone of chemistry education.

"repair-input-keyword": "What amount of nitric oxide (NO) is formed from 34g of ammonia (NH3) in reaction with oxygen (O2) with a 98% yield? The reaction equation is: 4 NH3 + 5 O2 → 4 NO + 6 H2O.",