To balance the equation \( \text{KOH} + \text{Cl}_2 \rightarrow \text{KCl} + \text{KClO} + \text{H}_2\text{O} \), we need to ensure that the number of each type of atom is equal on both sides of the equation.
Let's start by writing down the unbalanced equation:
\[ \text{KOH} + \text{Cl}_2 \rightarrow \text{KCl} + \text{KClO} + \text{H}_2\text{O} \]
Counting the number of each type of atom on both sides, we have:
On the left:
- K: 1 (from KOH)
- O: 1 (from KOH)
- H: 1 (from KOH)
- Cl: 2 (from Cl2)
On the right:
- K: 2 (1 from KCl and 1 from KClO)
- O: 1 (from KClO)
- H: 2 (from H2O)
- Cl: 2 (1 from KCl and 1 from KClO)
Next, we adjust the coefficients to balance the equation. We can start by balancing potassium (K), since we have 1 K on the left and 2 on the right.
To balance K, we place a coefficient of 2 in front of KOH:
\[ 2\text{KOH} + \text{Cl}_2 \rightarrow \text{KCl} + \text{KClO} + \text{H}_2\text{O} \]
Now, recounting the atoms:
On the left:
- K: 2 (from 2 KOH)
- O: 2 (from 2 KOH)
- H: 2 (from 2 KOH)
- Cl: 2 (from Cl2)
On the right:
- K: 2 (1 from KCl and 1 from KClO)
- O: 1 (from KClO)
- H: 2 (from H2O)
- Cl: 2 (1 from KCl and 1 from KClO)
Now, the potassium and chlorine are balanced, while oxygen is not. We have 2 O on the left and 1 O on the right.
To balance oxygen, we can place a coefficient of 2 in front of \( \text{KClO} \):
\[ 2\text{KOH} + \text{Cl}_2 \rightarrow \text{KCl} + 2\text{KClO} + \text{H}_2\text{O} \]
Now recounting:
On the left:
- K: 2
- O: 2
- H: 2
- Cl: 2
On the right:
- K: 2 (1 from KCl and 2 from KClO)
- O: 2 (2 from 2 KClO and 1 from H2O)
- H: 2
- Cl: 2 (1 from KCl and 2 from KClO)
Now that everything is balanced, we can conclude that the coefficient that would go in front of \( \text{KClO} \) is 2.
So, the answer is:
2
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