Question

Balance the following equation:
KOH+Cl2 = KCl+KClO+H2O
Select the coefficient that would go in front of KClO

Answer 1

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**