Ethylene And Chlorine Reaction: Yield And Volume Calculations
Hey guys! Let's dive into a cool chemistry problem involving the reaction between ethylene and chlorine. We'll figure out the relative yield of the product and the volumes of the gases that participated in the reaction. This is super interesting, so let's get started. We have ethylene with a mass of 60 g that reacts with chlorine and forms 198 g of dichloroethane. Let's figure out what's going on! This type of problem is pretty common in chemistry, and it's all about understanding how reactions work and how to apply some basic calculations.
The Reaction and the Chemistry Behind It
First off, let's look at the actual reaction. Ethylene (Câ‚‚Hâ‚„) reacts with chlorine (Clâ‚‚) to produce dichloroethane (Câ‚‚Hâ‚„Clâ‚‚). The reaction equation is:
C₂H₄ + Cl₂ → C₂H₄Cl₂
This is an addition reaction where chlorine molecules add to the double bond in ethylene, forming a new molecule. Now, let's break down the problem step by step to find the relative yield and volumes.
Calculating the Moles and Theoretical Yield
To begin, we need to calculate the number of moles of ethylene. The molar mass of ethylene (Câ‚‚Hâ‚„) is 28 g/mol (2 * 12 g/mol for carbon + 4 * 1 g/mol for hydrogen). We know the mass of ethylene is 60 g, so:
Moles of C₂H₄ = mass / molar mass = 60 g / 28 g/mol ≈ 2.14 mol.
According to the stoichiometry of the reaction, 1 mole of ethylene reacts with 1 mole of chlorine to produce 1 mole of dichloroethane. Therefore, if all the ethylene reacted, we'd expect 2.14 moles of dichloroethane to be formed. Now, we'll calculate the theoretical yield of dichloroethane. The molar mass of dichloroethane (Câ‚‚Hâ‚„Clâ‚‚) is 99 g/mol (2 * 12 g/mol for carbon + 4 * 1 g/mol for hydrogen + 2 * 35.5 g/mol for chlorine).
Theoretical yield = moles * molar mass = 2.14 mol * 99 g/mol ≈ 211.86 g.
This is the amount of dichloroethane we should get if the reaction goes perfectly, which is also called the theoretical yield. But in reality, reactions don't always go perfectly, and the actual yield might be less than this.
Calculating the Relative Yield
Next up, we need to find the relative yield of the product. The relative yield, often expressed as a percentage, tells us how efficient our reaction was. The formula for relative yield is:
Relative yield = (actual yield / theoretical yield) * 100%
We were told that the actual yield of dichloroethane is 198 g. We calculated the theoretical yield to be about 211.86 g. Plugging the numbers into the formula:
Relative yield = (198 g / 211.86 g) * 100% ≈ 93.46%.
So, the relative yield of the reaction is approximately 93.46%. That means this reaction is pretty efficient; it produced nearly all of the dichloroethane it could have theoretically made!
Volume Calculations of Reactants
Now, let's find the volumes of the gases that participated. We need to find out how many moles of chlorine reacted and then use the ideal gas law to calculate the volume. First, let's assume that the reaction goes to completion, and all the ethylene has reacted.
Since 2.14 moles of ethylene reacted, and the reaction ratio is 1:1 for ethylene and chlorine, then 2.14 moles of chlorine reacted. Now let's calculate the volume of chlorine at standard conditions (n.u. – normal conditions). At n.u., one mole of any gas occupies 22.4 liters. Thus:
Volume of Cl₂ = moles * molar volume = 2.14 mol * 22.4 L/mol ≈ 47.94 L.
Also, the volume of ethylene, which is our 2.14 moles of ethylene, can be found similarly. Its volume at normal conditions is:
Volume of C₂H₄ = moles * molar volume = 2.14 mol * 22.4 L/mol ≈ 47.94 L.
So, at normal conditions, the volume of both gases is approximately 47.94 L.
Summary of Results
Let's wrap up our calculations.
The relative yield of dichloroethane is approximately 93.46%. The volume of ethylene that reacted at normal conditions (n.u.) is approximately 47.94 L. The volume of chlorine that reacted at normal conditions (n.u.) is approximately 47.94 L.
Final Thoughts and Why It Matters
This problem showed us how to do some important calculations in chemistry. We learned how to find the relative yield of a product, which is super useful for figuring out how efficient a chemical reaction is. We also practiced with stoichiometry and the ideal gas law to calculate volumes of gases at normal conditions. Understanding the amounts of reactants and products, the volumes of gases, and the efficiency of reactions is vital in both lab and industrial settings.
These calculations help chemists optimize reactions, making more of the desired products and reducing waste. So, whether you're working in a lab or just curious about chemistry, these skills are fundamental. Keep practicing and keep asking questions, and you'll be a pro in no time! Keep up the good work, guys! This is a good basis for understanding stoichiometry and reaction yields. Remember that the accuracy of our results is dependent on the conditions of the reaction. Keep an eye out for how this can affect the outcomes. Understanding these basics is critical for success in chemistry!
Further Exploration and Next Steps
Now that you've got a handle on this, here are some things you can do to dig deeper:
- Vary the conditions: Try redoing the calculations with different initial masses of reactants or under non-standard conditions. How does this affect the relative yield? The volume calculations? This lets you explore the impact of reaction conditions.
- Explore limiting reactants: What if you didn't have enough chlorine to react with all the ethylene? What would the calculations look like then? This is a super important concept in chemistry.
- Real-world applications: Find out where dichloroethane is used. How is it produced industrially? Understanding the practical applications of these reactions can make chemistry even more interesting.
By following these steps, you're not just solving a chemistry problem; you're building a solid foundation in chemical principles! Keep up the great work, and don't hesitate to ask if you have any more questions.