To solve the given reaction and find the values of x, y, and z, we need to balance the chemical equation for the decomposition of potassium chlorate (KClO₃) in the presence of manganese dioxide (MnO₂) when heated.
The unbalanced reaction is:
\[ x \text{KClO}_3(s) \xrightarrow{\text{MnO}_2, \text{Heat}} y \text{KCl}(s) + z \text{O}_2(g) \]
### Step-by-Step Solution:
1. **Write the unbalanced equation**:
\[
KClO_3(s) \xrightarrow{\text{MnO}_2, \text{Heat}} KCl(s) + O_2(g)
\]
2. **Count the number of atoms on both sides**:
- Reactants: 1 K, 1 Cl, 3 O (from KClO₃)
- Products: 1 K, 1 Cl, and O from O₂
3. **Balance the oxygen atoms**:
- On the product side, O₂ has 2 oxygen atoms. To balance the 3 oxygen atoms from KClO₃, we need to adjust the coefficients.
- Place a coefficient of 2 in front of KClO₃ to get 6 oxygen atoms:
\[
2 \text{KClO}_3 \xrightarrow{\text{MnO}_2, \text{Heat}} y \text{KCl}(s) + z \text{O}_2(g)
\]
4. **Now, balance the oxygen atoms**:
- Since we have 6 oxygen atoms from 2 KClO₃, we can place a coefficient of 3 in front of O₂ to balance the oxygen:
\[
2 \text{KClO}_3 \xrightarrow{\text{MnO}_2, \text{Heat}} y \text{KCl}(s) + 3 \text{O}_2(g)
\]
5. **Balance potassium and chlorine**:
- Since we have 2 KClO₃, we will need 2 KCl on the product side:
\[
2 \text{KClO}_3 \xrightarrow{\text{MnO}_2, \text{Heat}} 2 \text{KCl}(s) + 3 \text{O}_2(g)
\]
6. **Final balanced equation**:
\[
2 \text{KClO}_3(s) \xrightarrow{\text{MnO}_2, \text{Heat}} 2 \text{KCl}(s) + 3 \text{O}_2(g)
\]
7. **Identify the coefficients**:
- From the balanced equation, we find:
- \( x = 2 \) (for KClO₃)
- \( y = 2 \) (for KCl)
- \( z = 3 \) (for O₂)
### Conclusion:
The values of x, y, and z are:
- \( x = 2 \)
- \( y = 2 \)
- \( z = 3 \)
