Nitrogen For 9.3 Moles Ammonia: Step-by-Step Calculation

Aug 5, 2025 by Sebastian Müller 57 views

$$N_2 + 3H_2 \rightarrow 2NH_3$: Calculating Nitrogen Requirements for Ammonia Synthesis$

Hey guys! Let's dive into a fascinating chemistry problem today. We're going to figure out how much nitrogen we need to produce a specific amount of ammonia. This is a classic stoichiometry question, and it’s super relevant in many industrial processes, especially in the production of fertilizers. So, grab your thinking caps, and let’s get started!

The Chemical Reaction: A Quick Overview

First, let's look at the balanced chemical equation:

$N_2 + 3H_2 \rightarrow 2NH_3$

This equation tells us a crucial story: one mole of nitrogen gas ( $N_2$ ) reacts with three moles of hydrogen gas ( $H_2$ ) to produce two moles of ammonia ( $NH_3$ ). This is the foundation for our calculations, guys. Understanding these molar ratios is the key to solving stoichiometry problems.

Now, let’s break this down even further. Think of it like a recipe. If you want to bake a cake, you need specific amounts of ingredients. In this case, nitrogen and hydrogen are our ingredients, and ammonia is the cake we're baking. The balanced equation is like the recipe card, telling us exactly how much of each ingredient we need. If we want to make a certain amount of cake (ammonia), we need to know how much flour (nitrogen) to use. So, the coefficients in front of each chemical formula in the balanced equation represent the number of moles of each substance involved in the reaction. This is super important because moles are how chemists count molecules – it's like using dozens to count eggs, but on a molecular scale!

Let's zoom in on the nitrogen and ammonia relationship. The equation shows a 1:2 ratio between $N_2$ and $NH_3$ . This means for every 1 mole of nitrogen we use, we get 2 moles of ammonia. This ratio is our golden ticket to solving the problem. If we know how many moles of ammonia we want, we can use this ratio to figure out how many moles of nitrogen we need. And once we have the moles of nitrogen, we can easily convert that to grams using the molar mass of nitrogen. So, remember this 1:2 ratio – it's going to be our guiding star in this calculation.

Understanding the balanced equation is not just about memorizing numbers; it's about understanding the fundamental chemistry happening. It tells us that the nitrogen and hydrogen atoms are rearranging themselves to form ammonia molecules. The coefficients ensure that we have the same number of each type of atom on both sides of the equation, which is a crucial principle in chemistry – the law of conservation of mass. This law basically says that matter can't be created or destroyed in a chemical reaction, it just changes form. So, the balanced equation is our way of keeping track of all the atoms and making sure we're not magically creating or destroying any!

Problem Statement: What Are We Trying to Find?

Okay, so here’s the question we're tackling: How many grams of nitrogen ( $N_2$ ) are required to create 9.3 moles of ammonia ( $NH_3$ )? We also have the molar mass of $N_2$ given as 28.02 g/mol. This piece of information is vital because it allows us to convert between moles and grams. Remember, moles are a chemist's way of counting particles, and grams are how we measure mass in the lab. So, the molar mass is the bridge between these two worlds.

Let’s rephrase the question to make sure we’re crystal clear on what we need to find. We're essentially asking, “If we want to make 9.3 moles of ammonia, what mass of nitrogen do we need to start with?” This helps us frame the problem in a way that's easier to tackle. We know our desired output (9.3 moles of ammonia), and we need to figure out the required input (grams of nitrogen). This is a typical stoichiometry problem, where we use the balanced chemical equation to relate the amounts of reactants and products.

Breaking down the problem statement is a crucial step in problem-solving. It’s like reading the instructions before you start building something from IKEA – it saves you a lot of time and frustration! By clearly identifying what we're looking for, we can map out a strategy to get there. In this case, we know we need to convert moles of ammonia to moles of nitrogen, and then convert moles of nitrogen to grams of nitrogen. This gives us a clear roadmap for our calculation.

It's also important to pay attention to the units given in the problem. We have moles of ammonia and the molar mass of nitrogen in grams per mole. Keeping track of the units is essential in chemistry because it helps us make sure we're doing the calculations correctly. If the units don't line up, we know we've made a mistake somewhere. So, always double-check your units, guys – it's a simple habit that can save you a lot of trouble!

Step-by-Step Solution: Cracking the Code

Now for the fun part – solving the problem! Here’s how we can break it down into manageable steps:

Step 1: Convert Moles of Ammonia to Moles of Nitrogen

Remember the 1:2 ratio we talked about? For every 2 moles of $NH_3$ , we need 1 mole of $N_2$ . We can use this as a conversion factor:

Moles of $N_2$ = Moles of $NH_3$ * (1 mole $N_2$ / 2 moles $NH_3$ )

Let’s plug in the values:

Moles of $N_2$ = 9.3 moles $NH_3$ * (1 mole $N_2$ / 2 moles $NH_3$ ) = 4.65 moles $N_2$

So, to make 9.3 moles of ammonia, we need 4.65 moles of nitrogen. See how we used the molar ratio from the balanced equation to convert from moles of ammonia to moles of nitrogen? This is a classic stoichiometry trick, and it's the heart of solving these types of problems.

Think of the conversion factor as a bridge that takes us from one unit to another. In this case, it's a bridge that connects the world of ammonia to the world of nitrogen. The ratio in the conversion factor comes directly from the balanced equation, which is why understanding the balanced equation is so crucial. If we didn't have the balanced equation, we wouldn't know the correct ratio, and our calculations would be way off!

It's also important to pay attention to the units in the conversion factor. We set it up so that the moles of ammonia cancel out, leaving us with moles of nitrogen. This is a good way to check that we've set up the conversion factor correctly. If the units don't cancel out, we know we need to flip the conversion factor around. So, always keep an eye on those units!

Step 2: Convert Moles of Nitrogen to Grams of Nitrogen

Now that we know the moles of $N_2$ , we can use the molar mass to find the grams. The molar mass of $N_2$ is 28.02 g/mol, which means 1 mole of $N_2$ weighs 28.02 grams. We can use this as another conversion factor:

Grams of $N_2$ = Moles of $N_2$ * (Molar mass of $N_2$ )

Let’s plug in the values:

Grams of $N_2$ = 4.65 moles $N_2$ * (28.02 g $N_2$ / 1 mole $N_2$ ) = 130.393 g $N_2$

Therefore, we need approximately 130.393 grams of nitrogen to create 9.3 moles of ammonia.

This step is all about converting from the chemist's world of moles to the lab's world of grams. Moles are a convenient way to count molecules, but in the lab, we usually measure things by mass. So, we need the molar mass to translate between these two worlds. The molar mass is like a universal translator that allows us to speak both the language of moles and the language of grams.

Again, let's look at the units. We're multiplying moles of nitrogen by grams per mole of nitrogen. The moles cancel out, leaving us with grams of nitrogen, which is exactly what we want. This is another check that we've set up the calculation correctly. If the units didn't work out, we'd know we needed to double-check our work.

So, we've successfully converted moles of nitrogen to grams of nitrogen using the molar mass as our trusty conversion factor. This is a fundamental skill in chemistry, and you'll use it in many different types of calculations. Remember, the molar mass is a key piece of information that allows us to connect the microscopic world of molecules to the macroscopic world of lab measurements.

Final Answer: The Grand Finale

So, to create 9.3 moles of ammonia, you would require approximately 130.393 grams of nitrogen gas. That’s it! We’ve solved the problem step-by-step, using the balanced chemical equation and molar mass as our guides. Wasn’t that fun, guys?

Let's recap the key steps we took to solve this problem. First, we used the balanced chemical equation to determine the molar ratio between nitrogen and ammonia. This gave us the crucial conversion factor we needed to go from moles of ammonia to moles of nitrogen. Then, we used the molar mass of nitrogen to convert from moles of nitrogen to grams of nitrogen. These two steps are the foundation of stoichiometry calculations, and they're used in many different chemical contexts.

It's also important to remember the importance of units in these calculations. We paid close attention to the units at each step, making sure they canceled out correctly. This is a simple but powerful way to check our work and avoid errors. So, always keep an eye on those units, guys – they're your friends!

This problem is a great example of how chemistry connects the microscopic world of atoms and molecules to the macroscopic world of lab measurements. We started with a balanced equation that describes how molecules react with each other, and we ended up with a mass of nitrogen that we could actually weigh out in the lab. This connection between the microscopic and macroscopic is what makes chemistry so fascinating and useful.

Practice Makes Perfect: Further Learning

If you want to get even better at stoichiometry, try practicing more problems like this one. You can find plenty of examples in your textbook or online. The more you practice, the more comfortable you’ll become with the concepts and the calculations. And remember, if you get stuck, don't be afraid to ask for help! Your teacher, classmates, or even online forums can be great resources.

Also, try changing the numbers in this problem and see if you can still solve it. What if we wanted to make 15 moles of ammonia? Or what if we had a different amount of nitrogen to start with? By changing the problem slightly, you can deepen your understanding of the concepts and make sure you really know your stuff.

Stoichiometry is a fundamental topic in chemistry, and it's essential for understanding chemical reactions and calculations. So, take the time to master these concepts, and you'll be well on your way to becoming a chemistry whiz! Keep practicing, keep asking questions, and most importantly, keep having fun with chemistry, guys!

Keywords: Stoichiometry, Ammonia Synthesis, Nitrogen, Moles, Grams, Molar Mass, Chemical Reaction, Balanced Equation

FAQs on Nitrogen and Ammonia Synthesis

1. What is the balanced chemical equation for the synthesis of ammonia?

The balanced chemical equation for the synthesis of ammonia is $N_2 + 3H_2 \rightarrow 2NH_3$ . This equation signifies that one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia.

2. How does the molar mass of nitrogen play a role in this calculation?

The molar mass of nitrogen ( $N_2$ ), which is 28.02 g/mol, acts as a conversion factor between moles and grams. It allows us to convert the amount of nitrogen needed in moles to grams, a unit that can be easily measured in a lab setting.

3. What is the significance of the 1:2 molar ratio between nitrogen and ammonia?

The 1:2 molar ratio between nitrogen and ammonia, derived from the balanced equation, indicates that for every one mole of nitrogen consumed, two moles of ammonia are produced. This ratio is crucial for calculating the amount of nitrogen needed to produce a specific quantity of ammonia.

4. Why is stoichiometry important in chemistry?

Stoichiometry is a fundamental concept in chemistry because it allows us to quantify the relationships between reactants and products in chemical reactions. This is essential for predicting the amount of reactants needed or products formed in a reaction, which has numerous applications in industry and research.

5. Can this calculation be applied to other chemical reactions?

Yes, the principles used in this calculation can be applied to a wide range of chemical reactions. By using the balanced chemical equation and molar masses, you can calculate the amounts of reactants and products involved in any stoichiometric reaction.

6. What if the problem provided grams of ammonia instead of moles? How would the solution differ?

If the problem provided grams of ammonia instead of moles, the first step would be to convert grams of ammonia to moles using the molar mass of ammonia ( $NH_3$ ). After that, the problem would be solved similarly, using the molar ratio to find moles of nitrogen and then converting to grams if needed.

7. What are some real-world applications of ammonia synthesis?

Ammonia synthesis has numerous real-world applications, most notably in the production of fertilizers for agriculture. It is also used in the manufacture of various chemicals, plastics, and explosives. The Haber-Bosch process, which is the industrial method for ammonia synthesis, has had a significant impact on global food production.

8. What is the Haber-Bosch process?

The Haber-Bosch process is an industrial process for the synthesis of ammonia from nitrogen and hydrogen. It is a crucial process for the production of fertilizers and has greatly increased agricultural yields worldwide. The process involves reacting nitrogen and hydrogen gases under high pressure and temperature in the presence of a catalyst.

9. How does temperature and pressure affect the ammonia synthesis reaction?

The ammonia synthesis reaction is exothermic, meaning it releases heat. According to Le Chatelier's principle, lower temperatures favor the formation of ammonia. However, lower temperatures also slow down the reaction rate, so a compromise temperature is used in practice. High pressure favors the formation of ammonia because there are fewer moles of gas on the product side of the equation. Industrial processes typically use high pressures to maximize ammonia yield.

10. What is the role of a catalyst in ammonia synthesis?

A catalyst speeds up the rate of a chemical reaction without being consumed in the reaction itself. In ammonia synthesis, a catalyst, typically iron-based, is used to increase the rate of reaction between nitrogen and hydrogen. The catalyst provides a surface on which the reaction can occur more easily, lowering the activation energy of the reaction.