which of the following is the mass of oxygen produced when 100

Have you ever wondered about chemical reactions and how much stuff they produce? Today, we’re going to tackle a common chemistry problem: which of the following is the mass of oxygen produced when 100 grams of a certain compound breaks down? Don’t worry, it sounds more complicated than it is! We’ll break it down step by step.

Understanding the Question

So, what exactly are we trying to figure out? The question is about a chemical reaction called decomposition. Decomposition is when a single substance breaks down into two or more simpler substances. Specifically, we’re focusing on how much oxygen gas (O₂) is created when potassium chlorate (KClO₃) decomposes. We need to use stoichiometry to solve this problem. We are going to use a balanced chemical equation and molar masses of the compounds.

Step-by-Step Solution: Calculating Oxygen Production

Let’s work through this problem together. Here’s how we’ll find the mass of oxygen produced:

1. Write the Balanced Chemical Equation: Potassium chlorate (KClO₃) decomposes to produce potassium chloride (KCl) and oxygen gas (O₂). The balanced equation is:
   2 KClO₃ (s) → 2 KCl (s) + 3 O₂ (g)

2. Find the Molar Masses:

   • Molar mass of KClO₃: (39.1 g/mol K) + (35.5 g/mol Cl) + 3(16.0 g/mol O) = 122.6 g/mol
   • Molar mass of O₂: 2(16.0 g/mol O) = 32.0 g/mol

3. Convert grams of KClO₃ to moles: Using the molar mass, we convert 100 g of KClO₃ to moles:

   • Moles of KClO₃ = 100 g / 122.6 g/mol = 0.816 mol

4. Use the Mole Ratio: From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂. Therefore:

   • Moles of O₂ produced = 0.816 mol KClO₃ * (3 mol O₂ / 2 mol KClO₃) = 1.224 mol O₂

5. Convert moles of O₂ to grams: Multiply the moles of O₂ by its molar mass to find the mass in grams:

   • Grams of O₂ = 1.224 mol * 32.0 g/mol = 39.17 g

Final Answer

So, the mass of oxygen produced when 100 grams of potassium chlorate decomposes is approximately:

39.17 grams

Why This Answer is Correct

We found the answer by carefully following the stoichiometry of the reaction. The balanced chemical equation provided the mole ratios, which allowed us to convert between the amount of reactant (KClO₃) and the amount of product (O₂).

Alternative Methods

An alternative method could involve using the concept of percent yield if you’re given additional information, such as the actual yield of oxygen produced. However, in this case, we have all the information needed to calculate the theoretical yield.

Common Mistakes

Here are some common pitfalls to avoid:

• Forgetting to balance the equation: This messes up the mole ratios!
• Using the wrong molar masses: Always double-check your values!
• Incorrectly setting up the conversion factors: Make sure your units cancel out correctly.
• Not considering the mole ratios: this is the key to stoichiometry problems

Conclusion

Congratulations! You’ve successfully calculated the mass of oxygen produced in a chemical reaction. Remember to use balanced equations, molar masses, and mole ratios, and you’ll be well on your way to mastering stoichiometry. The key is practicing these types of problems.

FAQ

1. What if I’m asked about a different reactant or product? The process is similar! You’ll adjust the balanced equation and molar masses accordingly.
2. Why is balancing the equation important? It ensures that the law of conservation of mass is followed.
3. Can I use this method for other chemical reactions? Yes, as long as you have a balanced chemical equation and molar masses!