Question

balance the below equations by adding appropriate coefficients
a. LiCl+Br2~LiBr+Cl2

Answer 1

To balance the equation \( \text{LiCl} + \text{Br}_2 \rightarrow \text{LiBr} + \text{Cl}_2 \), we will follow these steps:

1. List the number of each type of atom on both sides of the equation.
2. Adjust coefficients until both sides have the same number of each type of atom.

### Unbalanced Equation:
\[
\text{LiCl} + \text{Br}_2 \rightarrow \text{LiBr} + \text{Cl}_2
\]

### Counting Atoms:
- **Reactants:**
- Li: 1 (from LiCl)
- Cl: 1 (from LiCl)
- Br: 2 (from Br₂)

- **Products:**
- Li: 1 (from LiBr)
- Cl: 2 (from Cl₂)
- Br: 1 (from LiBr)

### Balancing:
1. From the counting, we notice there are 2 Cl atoms on the products side and only 1 on the reactants side. We can place a coefficient of 2 in front of LiCl.

\[
2 \text{LiCl} + \text{Br}_2 \rightarrow \text{LiBr} + \text{Cl}_2
\]

2. Now let's update the equation and recount:

- **Reactants:**
- Li: 2 (from 2 LiCl)
- Cl: 2 (from 2 LiCl)
- Br: 2 (from Br₂)

- **Products:**
- Li: 1 (from LiBr)
- Cl: 2 (from Cl₂)
- Br: 1 (from LiBr)

3. Since we have 2 Li in reactants, we need to put a coefficient of 2 in front of LiBr on the product side:

\[
2 \text{LiCl} + \text{Br}_2 \rightarrow 2 \text{LiBr} + \text{Cl}_2
\]

### Final Counting:
- **Reactants:**
- Li: 2
- Cl: 2
- Br: 2

- **Products:**
- Li: 2
- Cl: 2
- Br: 2

### Balanced Equation:
The balanced equation is:

\[
\boxed{2 \text{LiCl} + \text{Br}_2 \rightarrow 2 \text{LiBr} + \text{Cl}_2}
\]