Stoichiometry How Many Moles Of Oxygen React With 24 Moles Of Ethane

Hey everyone! Today, we're diving into the fascinating world of stoichiometry, where we explore the quantitative relationships between reactants and products in chemical reactions. In simpler terms, we're going to figure out how much of one substance we need to react completely with another. Let's tackle a problem together that involves figuring out the amount of oxygen needed to react with a specific amount of ethane. This is a classic chemistry problem, and by the end of this article, you'll be a pro at solving similar questions. So, buckle up, and let's get started!

The Chemical Equation Our Roadmap

At the heart of every stoichiometry problem is a balanced chemical equation. Think of it as a recipe for a chemical reaction. It tells us exactly how many 'units' of each substance are needed to make the reaction happen. In our case, we have the following balanced equation:

2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

This equation tells us that two moles of ethane (C₂H₆) react with seven moles of oxygen (O₂) to produce four moles of carbon dioxide (CO₂) and six moles of water (H₂O). These mole ratios are the key to solving our problem. It's super important to have a balanced equation because it ensures that we're working with the correct proportions of each substance. Without it, our calculations would be way off, and we might end up with the wrong answer. Balancing chemical equations is a fundamental skill in chemistry, so if you're not quite comfortable with it yet, it's definitely worth practicing!

Deciphering the Balanced Equation

Let's break down this equation a little further. The coefficients in front of each chemical formula represent the number of moles of that substance involved in the reaction. So, the '2' in front of C₂H₆ means two moles of ethane, and the '7' in front of O₂ means seven moles of oxygen. These coefficients are like the ingredients list in a recipe; they tell us the precise amounts we need. The arrow (→) indicates the direction of the reaction, showing that the reactants (ethane and oxygen) are transformed into the products (carbon dioxide and water). Now, the cool thing about this balanced equation is that it gives us a mole ratio. A mole ratio is like a conversion factor that helps us switch between the amounts of different substances in the reaction. In this case, we can see that for every 2 moles of ethane that react, we need 7 moles of oxygen. This 2:7 ratio is our golden ticket to solving the problem. We're going to use this ratio to figure out how many moles of oxygen are needed for 24 moles of ethane. It's like saying, 'If I'm baking a cake and the recipe calls for 2 eggs for every cup of flour, how many eggs do I need if I'm using 3 cups of flour?' The same principle applies here, just with chemical substances instead of baking ingredients!

The Problem How Much Oxygen Do We Need?

Okay, now let's state the problem clearly. We want to find out how many moles of O₂ are needed to react completely with 24 moles of C₂H₆. This is a classic stoichiometry question, and it's all about using the mole ratio from the balanced equation. The problem gives us a starting amount (24 moles of C₂H₆) and asks us to find the corresponding amount of another substance (O₂). This is where the mole ratio comes in handy. It allows us to 'convert' from moles of one substance to moles of another. Think of it like a bridge that connects the amount of ethane to the amount of oxygen needed. We know that 2 moles of ethane react with 7 moles of oxygen. So, if we have 24 moles of ethane, we can use this ratio to calculate how many moles of oxygen we need. It's all about setting up the calculation correctly, making sure our units cancel out, and arriving at the right answer. We're going to walk through the steps together, so you'll see exactly how it's done. By breaking down the problem into smaller parts and using the mole ratio as our guide, we can solve even the trickiest stoichiometry questions. So, let's move on to the next section and see how to set up the calculation!

Visualizing the Stoichiometric Relationship

To really grasp what's going on, imagine you have 24 tiny boxes, each containing one mole of ethane. Now, for each pair of ethane boxes, you need seven boxes of oxygen to make the reaction happen completely. The balanced equation tells us this fundamental relationship: 2 ethane boxes require 7 oxygen boxes. Our task is to figure out the total number of oxygen boxes needed for all 24 ethane boxes. It's like scaling up a recipe. If you know the amount of ingredients for one batch, you can easily calculate the amounts needed for multiple batches. In our case, the 'recipe' is the balanced chemical equation, and we're scaling it up from 2 moles of ethane to 24 moles of ethane. Visualizing the problem in this way can make it less abstract and more intuitive. You can see the direct connection between the amounts of the reactants and how the mole ratio acts as the bridge between them. This is the essence of stoichiometry – understanding and applying these quantitative relationships in chemical reactions. So, with this visual in mind, let's move on to setting up the calculation and finding the answer!

Setting Up the Calculation The Correct Molar Ratio

The key to solving this problem is using the correct mole ratio from the balanced equation as a conversion factor. We know that 2 moles of C₂H₆ react with 7 moles of O₂. This gives us two possible conversion factors:

• (7 mol O₂ / 2 mol C₂H₆)
• (2 mol C₂H₆ / 7 mol O₂)

The first conversion factor tells us how many moles of O₂ are needed per mole of C₂H₆, while the second tells us how many moles of C₂H₆ react per mole of O₂. We need to choose the conversion factor that will cancel out the units we want to get rid of (moles of C₂H₆) and leave us with the units we want (moles of O₂). This is a crucial step in stoichiometry, and it's all about dimensional analysis – making sure our units work out correctly. Think of it like converting from inches to feet. You need to use the conversion factor (1 foot / 12 inches) to cancel out the inches and get the answer in feet. The same principle applies here, just with moles of different substances. So, let's see how we apply this to our problem.

Choosing the Right Path

We start with 24 moles of C₂H₆, and we want to end up with moles of O₂. To do this, we'll multiply our starting amount by the conversion factor that has moles of O₂ in the numerator (the top part) and moles of C₂H₆ in the denominator (the bottom part). This will allow the "mol C₂H₆" units to cancel out, leaving us with "mol O₂". The correct setup looks like this:

24 mol C₂H₆ × (7 mol O₂ / 2 mol C₂H₆)

Notice how the "mol C₂H₆" units appear in both the numerator and the denominator, so they cancel each other out. This is exactly what we want! It's like having a fraction where the same number appears on the top and bottom – you can cancel them out. This process ensures that we're using the mole ratio correctly and that our answer will be in the desired units. Now, all that's left is to do the math. We multiply 24 by 7 and then divide by 2. This will give us the number of moles of oxygen needed to react with 24 moles of ethane. It's a simple calculation, but it's based on the fundamental principles of stoichiometry. By understanding how to set up the calculation correctly, you can solve a wide range of similar problems.

The Importance of Unit Cancellation

Pay close attention to unit cancellation! This is a fundamental technique in chemistry and physics. It's like a safety check to make sure you're setting up the problem correctly. If your units don't cancel out as expected, it's a sign that you've chosen the wrong conversion factor or made a mistake in your setup. In our case, if we had used the incorrect conversion factor (2 mol C₂H₆ / 7 mol O₂), our units wouldn't have canceled out properly, and we would have ended up with a meaningless answer. So, always double-check your units and make sure they cancel out to give you the units you're looking for. This simple step can save you from making costly mistakes and ensure that you get the right answer. It's like having a built-in error detector in your calculations. By mastering the art of unit cancellation, you'll become a more confident and accurate problem solver in chemistry.

The Solution Crunching the Numbers

Now that we have the correct setup:

24 mol C₂H₆ × (7 mol O₂ / 2 mol C₂H₆)

We can do the arithmetic. Multiply 24 by 7, which gives us 168. Then, divide 168 by 2, and we get 84. So, the answer is 84 moles of O₂. This means that 84 moles of oxygen are needed to react completely with 24 moles of ethane. We've successfully used the mole ratio from the balanced equation to solve the problem! It's like following a recipe and getting the desired result. By applying the principles of stoichiometry, we were able to determine the exact amount of oxygen needed for the reaction. This is a powerful skill in chemistry, as it allows us to predict and control the outcomes of chemical reactions. Whether you're working in a lab, designing a chemical process, or simply trying to understand the world around you, stoichiometry is an essential tool.

Expressing the Final Answer

Therefore, 24 moles of C₂H₆ will react completely with 84 moles of O₂. This is our final answer, and it tells us a lot about the reaction. It tells us the precise amount of oxygen needed to consume all 24 moles of ethane, leaving no reactants leftover. This is important in many chemical processes, where we want to ensure that reactions go to completion and that we get the maximum yield of products. Our answer also highlights the importance of the mole ratio. The 2:7 ratio between ethane and oxygen dictates the exact proportions in which they react. Changing the amount of one reactant will affect the amount of the other reactant needed. Stoichiometry allows us to quantify these relationships and make accurate predictions. So, the next time you see a chemical equation, remember that it's not just a list of chemicals; it's a recipe that tells you exactly how much of each substance is needed. And with the power of stoichiometry, you can unlock the secrets of these recipes and master the art of chemical calculations.

Double-Checking Our Work

It's always a good idea to double-check your work, especially in chemistry. One way to do this is to think about whether the answer makes sense intuitively. In our case, we started with 24 moles of ethane, and the balanced equation tells us that we need 7 moles of oxygen for every 2 moles of ethane. So, we would expect the amount of oxygen needed to be significantly larger than the amount of ethane. Our answer of 84 moles of oxygen is indeed much larger than 24 moles of ethane, which gives us confidence that we're on the right track. Another way to check is to work backward. If 84 moles of oxygen react, how much ethane would we expect to react with? Using the mole ratio, we can calculate that 84 moles of O₂ would react with (2/7) * 84 = 24 moles of C₂H₆, which is exactly what we started with. This confirms that our answer is consistent with the given information and the balanced equation. By using these double-checking techniques, you can minimize errors and ensure the accuracy of your calculations. It's like having a safety net that catches any potential mistakes. So, always take a few extra moments to review your work and make sure everything adds up.

Conclusion Stoichiometry Success!

Great job, guys! We've successfully determined that 84 moles of O₂ are needed to react with 24 moles of C₂H₆. This is a classic example of how stoichiometry can be used to solve real-world chemistry problems. By understanding the balanced chemical equation and using the mole ratio as a conversion factor, we were able to calculate the exact amount of oxygen needed for the reaction. This is a powerful tool that can be applied to a wide range of chemical reactions. Remember, the key to stoichiometry is to start with a balanced equation, identify the mole ratio, and set up the calculation correctly, paying close attention to unit cancellation. With practice, you'll become a stoichiometry master! So, keep practicing, keep exploring, and keep unlocking the secrets of the chemical world.

Mastering Stoichiometry for Future Success

Stoichiometry is not just a topic in chemistry; it's a fundamental skill that will serve you well in many areas of science and engineering. Whether you're designing new materials, developing new drugs, or analyzing environmental samples, stoichiometry will be an essential tool in your toolkit. It's the language of chemical reactions, and by mastering it, you'll be able to communicate effectively with the chemical world. So, don't be intimidated by stoichiometry; embrace it as a challenge and an opportunity to grow your understanding of chemistry. Practice solving problems, ask questions when you're unsure, and celebrate your successes along the way. With dedication and perseverance, you'll become a confident and competent problem solver in stoichiometry. And who knows, maybe one day you'll be using these skills to make groundbreaking discoveries and improve the world around us. The possibilities are endless when you have a solid foundation in stoichiometry!

Keep Exploring the World of Chemistry

Chemistry is a vast and fascinating field, full of exciting discoveries and challenges. Stoichiometry is just one piece of the puzzle, but it's a crucial piece that connects many other concepts together. As you continue your journey in chemistry, be sure to explore other topics like chemical kinetics, thermodynamics, equilibrium, and electrochemistry. Each of these areas builds upon the fundamental principles of stoichiometry and will give you a deeper understanding of how chemical reactions work. And don't forget to stay curious and ask questions. The more you explore, the more you'll discover, and the more you'll appreciate the beauty and complexity of the chemical world. So, keep learning, keep experimenting, and keep pushing the boundaries of your knowledge. The world needs more scientists and problem solvers, and you have the potential to make a real difference. So, go out there and make your mark on the world of chemistry!