Ethane Combustion: Oxygen And CO2 Volume Calculation

Hey guys! Let's dive into a cool chemistry problem involving the combustion of ethane. We're going to figure out the volumes of oxygen that reacts and the carbon dioxide that forms when some ethane burns. Specifically, we're working with 2.24 dm³ (at standard conditions, or n. u.) of ethane, and we've got plenty of oxygen to make sure it all burns completely. Sounds fun, right? It's like a little puzzle where we use some basic chemical principles to find our answers. This type of problem is super common in chemistry, and understanding how to solve it will help you with a bunch of other reactions too. So, grab your pencils, and let's get started. We'll break it down step by step to make it super easy to follow. Don't worry, it's not as scary as it sounds. We'll go through the balanced chemical equation, molar volumes, and some simple calculations to get to the answer. By the end, you'll be a pro at these types of calculations! Get ready to impress your friends with your newfound chemistry skills. This is a crucial topic for anyone studying chemistry, and mastering it will give you a solid foundation for more complex problems later on. So, let’s get those numbers crunching!

Understanding the Basics: Combustion and Stoichiometry

Alright, before we get to the nitty-gritty of the calculations, let's make sure we're all on the same page with the basics. First off, what exactly is combustion? Simply put, it's a rapid chemical process that involves a substance reacting with an oxidant to produce heat and light. In our case, the substance is ethane (C₂H₆), and the oxidant is oxygen (O₂). The products of complete combustion for hydrocarbons like ethane are always carbon dioxide (CO₂) and water (H₂O). Pretty straightforward, yeah? Now, here's where stoichiometry comes in. Stoichiometry is the part of chemistry that deals with the relative quantities of reactants and products in chemical reactions. Basically, it's about figuring out how much of each substance is involved in a reaction based on the balanced chemical equation. The balanced chemical equation gives us the molar ratios. For our reaction, the balanced equation is: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. This equation tells us that for every 2 moles of ethane that react, we need 7 moles of oxygen to produce 4 moles of carbon dioxide and 6 moles of water. The coefficients in front of each chemical formula are super important because they show the mole ratios. This knowledge is absolutely crucial for solving our problem. So, always make sure your chemical equations are balanced before you start any calculations. The mole ratios are your key to unlock the problem. Think of it like a recipe: the coefficients are the amounts of each ingredient you need. If you don't use the correct amounts, your 'cake' (or in our case, the reaction) won't work! Understanding the mole ratios is the most important part of this kind of calculation. Once you have this down, the rest is smooth sailing. Understanding these basics will make the calculations much easier to follow, believe me.

The Balanced Chemical Equation

To begin, we need the balanced chemical equation for the combustion of ethane. As we mentioned earlier, the balanced equation is: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. This equation is the foundation for all our calculations. Notice how there's a '2' in front of C₂H₆, a '7' in front of O₂, a '4' in front of CO₂, and a '6' in front of H₂O. These coefficients are super important because they represent the mole ratios of the reactants and products. They tell us exactly how many moles of each substance are involved in the reaction. For every 2 moles of ethane, we need 7 moles of oxygen to make 4 moles of carbon dioxide and 6 moles of water. Make sure you can write and balance equations like this; it's a fundamental skill in chemistry. If you mess up the equation, you mess up everything. To balance the equation, you have to make sure there are the same number of each type of atom on both sides of the equation. If the equation isn't balanced, the law of conservation of mass won't be followed, and your calculations will be wrong. This is the most fundamental step to solving this problem. Master balancing equations, and these problems become much easier. This balanced equation tells us everything we need to know about the relationships between the reactants and products in terms of moles. It's the blueprint of the reaction.

Calculating the Volume of Reacted Oxygen

Now, let's get down to the calculation for the volume of oxygen. We know that we have 2.24 dm³ of ethane at standard conditions (n. u.). We're going to use this information, along with the balanced chemical equation, to figure out the volume of oxygen that reacts. Here's the game plan: We'll first convert the volume of ethane to moles, then use the mole ratio from the balanced equation to find the moles of oxygen needed, and finally convert the moles of oxygen back to volume. Easy, right? First, we need to know the molar volume of a gas at standard conditions, which is approximately 22.4 dm³/mol. This is a constant value and an important piece of information to remember. This means that one mole of any gas at standard temperature and pressure occupies 22.4 dm³. Step 1: Convert the volume of ethane to moles. Use the ideal gas law at standard temperature and pressure to convert the volume of ethane to moles. Then, use the mole ratio from the balanced chemical equation to find the moles of oxygen. According to the balanced equation (2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O), the mole ratio of ethane to oxygen is 2:7. This means that for every 2 moles of ethane that react, 7 moles of oxygen are needed. Step 2: Calculate the moles of ethane. Then we can use the formula: moles = volume / molar volume. So, moles of ethane = 2.24 dm³ / 22.4 dm³/mol = 0.1 mol. Step 3: Calculate the moles of oxygen. Now, using the mole ratio (2:7) from the balanced equation, we can calculate the moles of oxygen. The calculation is: moles of O₂ = (0.1 mol C₂H₆) * (7 mol O₂ / 2 mol C₂H₆) = 0.35 mol O₂. Step 4: Convert moles of oxygen to volume. Finally, we convert the moles of oxygen to volume using the molar volume at standard conditions again. Volume of O₂ = moles * molar volume, so Volume of O₂ = 0.35 mol * 22.4 dm³/mol = 7.84 dm³. So, the volume of oxygen that reacted is 7.84 dm³. Congratulations, guys, we made it!

Calculating the Volume of Produced Carbon Dioxide

Okay, now that we've found the volume of oxygen, let's figure out the volume of carbon dioxide (CO₂) produced during the combustion. The approach is similar to what we did for oxygen, but this time, we'll use the mole ratio of ethane to carbon dioxide from the balanced equation. Remember, in the equation 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, the mole ratio of ethane to CO₂ is 2:4 (which simplifies to 1:2). This means that for every 1 mole of ethane that reacts, 2 moles of CO₂ are produced. Here's the plan: We'll start with the moles of ethane we calculated earlier, use the mole ratio to find the moles of CO₂, and then convert the moles of CO₂ to volume. Let's do this! Step 1: Use the moles of ethane calculated previously. We already know from the previous calculation that we have 0.1 mol of ethane. Now let's calculate the moles of CO₂. Step 2: Use the mole ratio to find the moles of CO₂. Using the mole ratio from the balanced equation (2:4 or 1:2), we calculate the moles of CO₂. Moles of CO₂ = (0.1 mol C₂H₆) * (4 mol CO₂ / 2 mol C₂H₆) = 0.2 mol CO₂. Step 3: Convert moles of CO₂ to volume. Now, we convert the moles of CO₂ to volume using the molar volume at standard conditions, 22.4 dm³/mol. Volume of CO₂ = moles * molar volume. Volume of CO₂ = 0.2 mol * 22.4 dm³/mol = 4.48 dm³. Therefore, the volume of carbon dioxide produced is 4.48 dm³. We've solved the problem. Awesome, right? Keep practicing these types of problems, and you'll become a pro in no time.

Summary of Results

Alright, let's wrap things up with a quick recap of our findings. We started with 2.24 dm³ of ethane and burned it in an excess of oxygen. Here's what we found:

• Volume of Oxygen (O₂) that reacted: 7.84 dm³.
• Volume of Carbon Dioxide (CO₂) produced: 4.48 dm³.

These results were obtained using the balanced chemical equation, mole ratios, and the molar volume of a gas at standard conditions (22.4 dm³/mol). Understanding these concepts and how to apply them is key to solving this type of problem. We first converted the volume of ethane to moles, used the mole ratios from the balanced chemical equation to find the moles of oxygen and carbon dioxide, and finally converted the moles back to volumes. Remember, stoichiometry is a fundamental concept in chemistry. It is crucial for understanding how chemical reactions work. Keep practicing these types of problems to solidify your understanding. Now that you've got the hang of it, you can apply these principles to other combustion reactions. Well done, guys! You tackled this problem like a champ! With enough practice, you’ll be able to solve these types of problems in your sleep. Keep up the good work and enjoy the journey of learning chemistry!