2. KCIO3 _KCI +_O2 a. Balance the reaction b. What is the reaction type?. C. What is the theoretical yield in moles of oxygen that are produced by the reaction of 6 moles c potassium chlorate?

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**Chemical Reaction Problem**

2. ____ KClO₃ → ____ KCl + ____ O₂

a. Balance the reaction

b. What is the reaction type? ____________________

c. What is the theoretical yield in moles of oxygen that are produced by the reaction of 6 moles of potassium chlorate?

d. If 11 moles of oxygen are actually produced, what is the percent yield?

e. What is the percent error?

---

**Explanation and Solution Steps:**

1. **Balancing the Reaction:**
   - The unbalanced equation is: KClO₃ → KCl + O₂
   - Balance each element:
     - K: 1 K on both sides
     - Cl: 1 Cl on both sides
     - O: There are 3 O on the left and 2 O on the right, so you need to adjust to make them equal.

   - Balanced Equation: 2 KClO₃ → 2 KCl + 3 O₂

2. **Reaction Type:**
   - It is a decomposition reaction because KClO₃ breaks down into KCl and O₂.

3. **Theoretical Yield Calculation:**
   - From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂.
   - Calculate the yield for 6 moles of KClO₃ using the ratio (6 moles KClO₃ x 3 moles O₂ / 2 moles KClO₃ = 9 moles O₂).

4. **Percent Yield:**
   - If actually 11 moles of O₂ are produced: 
     - Percent Yield = (Actual Yield / Theoretical Yield) x 100%
     - Percent Yield = (11 moles / 9 moles) x 100% = 122.22%

5. **Percent Error:**
   - Percent Error = | (Actual - Theoretical) / Theoretical | x 100%
   - Percent Error = | (11 moles - 9 moles) / 9 moles | x 100% = 22.22%
Transcribed Image Text:**Chemical Reaction Problem** 2. ____ KClO₃ → ____ KCl + ____ O₂ a. Balance the reaction b. What is the reaction type? ____________________ c. What is the theoretical yield in moles of oxygen that are produced by the reaction of 6 moles of potassium chlorate? d. If 11 moles of oxygen are actually produced, what is the percent yield? e. What is the percent error? --- **Explanation and Solution Steps:** 1. **Balancing the Reaction:** - The unbalanced equation is: KClO₃ → KCl + O₂ - Balance each element: - K: 1 K on both sides - Cl: 1 Cl on both sides - O: There are 3 O on the left and 2 O on the right, so you need to adjust to make them equal. - Balanced Equation: 2 KClO₃ → 2 KCl + 3 O₂ 2. **Reaction Type:** - It is a decomposition reaction because KClO₃ breaks down into KCl and O₂. 3. **Theoretical Yield Calculation:** - From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂. - Calculate the yield for 6 moles of KClO₃ using the ratio (6 moles KClO₃ x 3 moles O₂ / 2 moles KClO₃ = 9 moles O₂). 4. **Percent Yield:** - If actually 11 moles of O₂ are produced: - Percent Yield = (Actual Yield / Theoretical Yield) x 100% - Percent Yield = (11 moles / 9 moles) x 100% = 122.22% 5. **Percent Error:** - Percent Error = | (Actual - Theoretical) / Theoretical | x 100% - Percent Error = | (11 moles - 9 moles) / 9 moles | x 100% = 22.22%
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