How To Calculate Propene Mass: Burning With 728L Of Oxygen

Hey everyone! Today, we're diving into a cool chemistry problem: figuring out the mass of propene that burns when you've got 728 liters of oxygen to work with. Don't worry, it sounds trickier than it is! We'll break it down into easy-to-follow steps, so you'll be a pro at this in no time. This problem is a classic example of stoichiometry, which is basically the relationship between the amounts of reactants and products in a chemical reaction. Understanding this is super important, so let's get started!

We need to determine the mass of propene (C₃H₆) that will react completely with 728 liters of oxygen (O₂) under standard conditions (n.у., which stands for normal conditions - 0°C and 1 atm pressure). This involves using the balanced chemical equation for the combustion of propene, Avogadro's law, and molar masses to convert between volume, moles, and mass. It is a common problem in chemistry and helps to understand the quantitative relationships in chemical reactions.

First things first, we need the balanced chemical equation for the combustion of propene. Combustion is essentially the rapid reaction between a substance with the oxidant, usually oxygen, to produce heat and light. In the case of propene, it reacts with oxygen to produce carbon dioxide and water. The balanced equation looks like this:

2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O

This equation is crucial because it tells us the ratio in which the reactants and products react. For every 2 moles of propene, we need 9 moles of oxygen. Always balance your equations, guys! It is the cornerstone for all stoichiometric calculations. Incorrectly balanced equations lead to incorrect results. Now, let's start solving the problem.

Step 1: Convert Oxygen Volume to Moles

We know we have 728 liters of oxygen. But we need to convert this volume into moles, as the balanced equation deals with moles. To do this, we'll use the molar volume of a gas at standard conditions (n.у.). At standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atm, one mole of any ideal gas occupies 22.4 liters. So, here's how we convert:

Moles of O₂ = Volume of O₂ / Molar Volume Moles of O₂ = 728 L / 22.4 L/mol Moles of O₂ ≈ 32.5 moles

So, we have approximately 32.5 moles of oxygen. The molar volume is a constant derived from the ideal gas law and is a key factor in relating the volume of a gas to the amount of substance (in moles). It simplifies calculations, enabling direct conversions between volume and moles at STP. Remember, this step is essential for bridging the gap between the experimental measurement (volume) and the theoretical calculations (moles).

Step 2: Use the Balanced Equation to Find Moles of Propene

Now we know how many moles of oxygen we have, we can use the balanced equation to figure out how many moles of propene reacted. Looking back at our balanced equation (2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O), we see that 9 moles of O₂ react with 2 moles of C₃H₆. We can set up a proportion here:

(Moles of C₃H₆ / Moles of O₂) = (2 / 9)

Let’s rearrange this to solve for moles of C₃H₆:

Moles of C₃H₆ = (Moles of O₂ * 2) / 9 Moles of C₃H₆ = (32.5 mol * 2) / 9 Moles of C₃H₆ ≈ 7.22 moles

So, approximately 7.22 moles of propene reacted. This step highlights the importance of the stoichiometric coefficients in the balanced equation. They provide the precise molar ratios needed for the calculation. Without them, we would not be able to accurately relate the amount of oxygen consumed to the amount of propene reacted.

Step 3: Convert Moles of Propene to Mass

The final step is to convert the moles of propene into grams (or the mass). To do this, we need the molar mass of propene (C₃H₆). The molar mass is the mass of one mole of a substance, which is measured in grams per mole (g/mol). To find the molar mass of propene, we add up the atomic masses of all the atoms in the molecule:

• Carbon (C): 12.01 g/mol, and there are 3 atoms, so 3 * 12.01 = 36.03 g/mol
• Hydrogen (H): 1.01 g/mol, and there are 6 atoms, so 6 * 1.01 = 6.06 g/mol

Total Molar Mass of Propene = 36.03 g/mol + 6.06 g/mol = 42.09 g/mol

Now, to find the mass of propene:

Mass of Propene = Moles of Propene * Molar Mass Mass of Propene = 7.22 mol * 42.09 g/mol Mass of Propene ≈ 303.9 g

So, approximately 303.9 grams of propene will burn when 728 liters of oxygen are consumed. Remember, understanding the concept of molar mass is very important. It is the bridge between the microscopic world of molecules and the macroscopic world where we measure mass. Without this knowledge, we cannot convert between moles and grams.

Wrapping it Up!

And there you have it, folks! We've successfully calculated the mass of propene. Remember to always start with a balanced equation, convert any volumes to moles, use the mole ratios from the balanced equation, and then convert back to mass using molar masses. It might seem like a lot of steps, but trust me, with a little practice, you'll become a pro at these calculations. Keep practicing, and you'll ace these problems in no time. If you have any more questions, feel free to ask! Understanding stoichiometry is not just about solving problems; it gives you an insight into how chemical reactions work on a molecular level. It's a fundamental concept in chemistry that opens up a world of understanding about how substances interact and transform!

This entire process, from balancing the equation to calculating the final mass, is a testament to the power of chemistry. It showcases how a deep understanding of fundamental principles enables us to predict and quantify chemical reactions. Keep practicing, and you'll find that these calculations become second nature. You've got this!

Tips and Tricks for Success

Let's add some extra tips to help you succeed in these types of problems. Here's a quick rundown of helpful strategies and things to keep in mind:

• Always Balance Your Equation First: This cannot be stressed enough! A balanced chemical equation provides the mole ratios that are absolutely crucial for your calculations. If your equation is unbalanced, all subsequent calculations will be incorrect. Take your time with this step, and double-check your work.
• Units are Your Friends: Pay close attention to your units throughout the calculation. Make sure your units cancel out correctly so you arrive at the right final unit (in this case, grams). Writing the units with each number helps you track whether you are performing the calculations correctly. It is a simple step, but a very effective method to avoid errors.
• Significant Figures: Be mindful of significant figures. Your final answer should reflect the number of significant figures in your initial measurements. This ensures that you are reporting your result with appropriate precision.
• Molar Mass is Key: Learn to calculate molar masses quickly and accurately. This skill is fundamental to stoichiometry. Refer to the periodic table for atomic masses, and always double-check your calculations.
• Practice Makes Perfect: The more you practice, the easier these problems will become. Work through multiple examples, and try different variations. The more exposure you get, the more comfortable and confident you'll become.
• Check Your Work: Always re-examine your calculations. Simple mistakes can often be avoided by taking a few moments to review each step. Consider doing the problem again to verify your initial answer. It’s also helpful to have a friend or classmate check your work.

Common Mistakes to Avoid

Let's also look at some common pitfalls to avoid when working through these problems. Being aware of these errors can help you prevent them:

• Forgetting to Balance the Equation: As mentioned earlier, this is the most common mistake. Always start by balancing the equation before you do anything else.
• Using Incorrect Mole Ratios: Make sure you are using the correct mole ratios from your balanced equation. This is where incorrect balancing will really hurt you.
• Mixing Up Units: Be careful with units. Make sure all your values are in the correct units before you start the calculation. For example, if you are given a volume in liters, make sure you convert it to moles using the correct molar volume (22.4 L/mol at STP).
• Incorrect Molar Mass Calculation: Double-check that you have calculated the molar mass of each compound correctly, adding the atomic masses accurately.
• Not Accounting for Limiting Reactants: If the problem provides information on the amount of multiple reactants, make sure you identify the limiting reactant first. The limiting reactant will determine how much product can be formed. We did not cover this here, but keep it in mind.

By keeping these tips and common mistakes in mind, you will be well on your way to mastering these kinds of problems. Remember, practice and attention to detail are key!