Hey everyone! Have you ever wondered how much of one chemical you need to react with another? Today, we're diving into a classic chemistry problem: how many grams of ethylene (C₂H₄) react with 50.5 L of O₂ at Room Temperature and Pressure (RTP) to form carbon dioxide and water? This is a common type of question in stoichiometry, which is all about the quantitative relationships between reactants and products in chemical reactions. So, grab your lab coats (metaphorically, of course!) and let's get started!
Understanding the Chemical Equation
Before we jump into calculations, we need to understand the balanced chemical equation. It’s like the recipe for our reaction, telling us exactly how much of each ingredient we need. The equation given is:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
This equation tells us that one molecule of ethylene (C₂H₄) reacts with three molecules of oxygen (O₂) to produce two molecules of carbon dioxide (CO₂) and two molecules of water (H₂O). The coefficients in front of each chemical formula (1, 3, 2, and 2) are crucial. They represent the molar ratios, which are the proportions in which the substances react.
- Ethylene (C₂H₄): This is our main reactant, the gas we want to find the mass of.
- Oxygen (O₂): This is the other reactant, and we know the volume of it at RTP.
- Carbon Dioxide (CO₂): One of the products, a greenhouse gas.
- Water (H₂O): The other product, essential for life.
Why is balancing the equation so important, guys? Well, it ensures that we're adhering to the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. In simple terms, the number of atoms of each element must be the same on both sides of the equation. If the equation isn't balanced, our calculations will be way off!
Step 1: Calculate Moles of Oxygen (O₂)
Okay, so we know the volume of oxygen gas (50.5 L) at RTP. The first step is to convert this volume into moles. To do this, we'll use the molar volume of a gas at RTP. At RTP (which is usually defined as 25°C and 1 atm pressure), one mole of any gas occupies approximately 24.5 L. This is a handy constant to remember!
The formula we'll use is:
Moles = Volume / Molar Volume
So, for oxygen:
Moles of O₂ = 50.5 L / 24.5 L/mol
Let's crunch those numbers:
Moles of O₂ ≈ 2.06 moles
Awesome! We now know that we have approximately 2.06 moles of oxygen gas. This is a crucial piece of information because it links the amount of oxygen to the amount of ethylene needed for the reaction.
Step 2: Determine Moles of Ethylene (C₂H₄) Required
This is where the balanced chemical equation really shines. Remember those coefficients? They tell us the molar ratios of the reactants. The equation shows that 1 mole of ethylene (C₂H₄) reacts with 3 moles of oxygen (O₂).
We can set up a simple ratio to find out how many moles of ethylene are needed:
(Moles of C₂H₄) / (Moles of O₂) = 1 / 3
We already know the moles of O₂ (2.06 moles), so we can plug that in and solve for moles of C₂H₄:
(Moles of C₂H₄) / 2.06 moles = 1 / 3
Now, let's solve for moles of C₂H₄:
Moles of C₂H₄ = (1 / 3) * 2.06 moles
Moles of C₂H₄ ≈ 0.687 moles
Fantastic! We've calculated that approximately 0.687 moles of ethylene are required to react with 2.06 moles of oxygen. We're getting closer to our final answer!
Step 3: Convert Moles of Ethylene to Grams
The final step is to convert the moles of ethylene into grams. To do this, we'll use the molar mass of ethylene. The molar mass is the mass of one mole of a substance, and it's usually expressed in grams per mole (g/mol).
To find the molar mass of ethylene (C₂H₄), we need to add up the atomic masses of all the atoms in the molecule. The atomic mass of carbon (C) is approximately 12.01 g/mol, and the atomic mass of hydrogen (H) is approximately 1.01 g/mol.
So, the molar mass of C₂H₄ is:
(2 * 12.01 g/mol) + (4 * 1.01 g/mol) ≈ 24.02 g/mol + 4.04 g/mol
Molar mass of C₂H₄ ≈ 28.06 g/mol
Now we can use the following formula to convert moles to grams:
Grams = Moles * Molar Mass
Plugging in the values we have:
Grams of C₂H₄ = 0.687 moles * 28.06 g/mol
Let's calculate that:
Grams of C₂H₄ ≈ 19.28 grams
The Final Answer
Alright, guys, we've made it! We've successfully calculated the mass of ethylene required to react with 50.5 L of oxygen at RTP.
Approximately 19.28 grams of ethylene (C₂H₄) will react with 50.5 L of O₂ at RTP to form carbon dioxide and water.
This is a pretty neat example of how stoichiometry works. By understanding the balanced chemical equation and using concepts like molar volume and molar mass, we can figure out the exact amounts of reactants and products involved in a chemical reaction.
Key Concepts Recap
Before we wrap up, let's quickly recap the key concepts we used in this problem:
- Balanced Chemical Equation: The foundation of stoichiometry, showing the molar ratios of reactants and products.
- Molar Volume at RTP: One mole of any gas occupies approximately 24.5 L at RTP.
- Molar Ratio: The ratio of moles of reactants and products in a balanced equation.
- Molar Mass: The mass of one mole of a substance, calculated by adding up the atomic masses of the atoms in the molecule.
- Stoichiometry: The study of the quantitative relationships between reactants and products in chemical reactions.
Practice Problems
Want to test your understanding? Here are a couple of practice problems you can try:
- How many grams of methane (CH₄) are required to react completely with 100 L of oxygen at RTP?
- If 25 grams of hydrogen gas (H₂) react with nitrogen gas (N₂) to form ammonia (NH₃), what volume of nitrogen gas is required at RTP?
Why is Stoichiometry Important?
Stoichiometry isn't just some abstract concept you learn in chemistry class. It has tons of real-world applications! It's used in:
- Industrial Chemistry: To optimize chemical reactions and ensure efficient production of chemicals.
- Pharmaceuticals: To calculate the correct amounts of reactants needed to synthesize drugs.
- Environmental Science: To understand and mitigate pollution, like calculating the amount of pollutants produced in a combustion reaction.
- Cooking: Yes, even cooking! Baking, for instance, relies on precise ratios of ingredients for the best results.
So, understanding stoichiometry can help you in many different fields. It’s a fundamental skill for anyone interested in chemistry and related sciences.
Final Thoughts
I hope this guide has helped you understand how to calculate the mass of ethylene reacting with oxygen at RTP. Stoichiometry can seem tricky at first, but with practice, you'll become a pro in no time! Remember to always start with the balanced chemical equation, and take it step by step. You've got this!
If you have any questions or want to dive deeper into stoichiometry, feel free to ask in the comments. Keep exploring, keep learning, and have fun with chemistry, guys! Chemistry is an amazing subject, and the more you understand it, the more you'll appreciate the world around you. So, keep experimenting, keep asking questions, and most importantly, keep learning. You never know what discoveries you might make along the way! So next time, bye bye guys.