To balance the chemical equation \( \text{KNO}_3 \rightarrow \text{KNO}_2 + \text{O}_2 \), we follow these steps:
- Count the number of atoms of each element on both sides of the equation.
- Adjust coefficients to get the same number of each type of atom on both sides.
Let's write the unbalanced equation first:
\[ \text{KNO}_3 \rightarrow \text{KNO}_2 + \text{O}_2 \]
Step 1: Identify the Reactants and Products
- Reactants: 1 K, 1 N, 3 O (from KNO3)
- Products: 1 K, 1 N, 2 O (from KNO2) + 2 O (from O2)
Step 2: Balancing Oxygen
The left side has 3 oxygen atoms (from KNO3), and the right side has a total of 4 oxygen atoms (2 from KNO2 and 2 from O2). So, we need to balance it:
If we start with 2 KNO3, we get:
\[ 2 \text{KNO}_3 \rightarrow 2 \text{KNO}_2 + \text{O}_2 \]
Now we calculate again:
- Reactants: 2 K, 2 N, 6 O (from 2 KNO3)
- Products: 2 K, 2 N, (2 O from KNO2 + 2 O from O2 = 4 O)
Thus, we can adjust coefficients again.
To achieve 6 O on the product side, let's try adjusting:
\[ 4 \text{KNO}_3 \rightarrow 4 \text{KNO}_2 + 2 \text{O}_2 \]
Final Check
Reactants:
- 4 K, 4 N, 12 O
Products:
- 4 K, 4 N, (4 O from KNO2 + 4 O from 2 O2 = 12 O)
Both sides are now equal.
Final Balanced Equation
\[ 4 \text{KNO}_3 \rightarrow 4 \text{KNO}_2 + 2 \text{O}_2 \]
So the balanced equation is:
\[ \boxed{4 \text{KNO}_3 \rightarrow 4 \text{KNO}_2 + 2 \text{O}_2} \]