Complete and balance the following chemical equations.
(i)`BrO_(3)^(-)+F_(2)+H_(2)Oto`
(ii)`F_(2)+NaOH ("dilute & conc.) to`
(iii)`NaOH (aq) ("hot and cold")+Cl_(2)(aq)to`

Answer

Let's break down the solutions to each of the chemical equations step by step.

### (i) Reaction of `BrO3^(-) + F2 + H2O`

1. **Identify Reactants and Products**:
- Reactants: `BrO3^(-)`, `F2`, `H2O`
- Products: `F^-`, `BrO4^(-)`, `H^+`

2. **Write the Unbalanced Equation**:
\[ BrO3^- + F2 + H2O \rightarrow F^- + BrO4^- + H^+ \]

3. **Balance the Equation**:
- Start with balancing the number of atoms for each element.
- For `F`, we have 2 `F` atoms on the left, so we need 2 `F^-` on the right.
- For `Br`, we have 1 `Br` on both sides.
- For `O`, we have 3 `O` from `BrO3^-` and 1 `O` from `H2O`, totaling 4 `O` on the left, and we have 4 `O` from `BrO4^-` on the right.
- For `H`, we have 2 `H` from `H2O` and 2 `H^+` on the right.

The balanced equation is:
\[ BrO3^- + 2F2 + 2H2O \rightarrow 2F^- + BrO4^- + 2H^+ \]

### (ii) Reaction of `F2 + NaOH (dilute)`

1. **Identify Reactants and Products**:
- Reactants: `F2`, `NaOH (dilute)`
- Products: `OF2`, `NaF`, `H2O`

2. **Write the Unbalanced Equation**:
\[ F2 + NaOH \rightarrow OF2 + NaF + H2O \]

3. **Balance the Equation**:
- We have 2 `F` on the left and 2 `F` in `OF2` and `NaF` on the right.
- For `Na`, we have 1 `Na` on both sides.
- For `H`, we have 1 `H` from `NaOH` and 1 `H` in `H2O` on the right.
- For `O`, we have 1 `O` from `NaOH` and 1 `O` in `OF2`.

The balanced equation is:
\[ F2 + 2NaOH \rightarrow OF2 + 2NaF + H2O \]

### (iii) Reaction of `F2 + NaOH (conc.)`

1. **Identify Reactants and Products**:
- Reactants: `F2`, `NaOH (conc.)`
- Products: `O2`, `NaF`, `H2O`

2. **Write the Unbalanced Equation**:
\[ F2 + NaOH \rightarrow O2 + NaF + H2O \]

3. **Balance the Equation**:
- For `F`, we have 2 `F` on the left and 2 `F` in `NaF` on the right.
- For `Na`, we have 1 `Na` on both sides.
- For `H`, we have 1 `H` from `NaOH` and 1 `H` in `H2O`.
- For `O`, we have 1 `O` from `NaOH` and 1 from `O2`.

The balanced equation is:
\[ 2F2 + 4NaOH \rightarrow O2 + 4NaF + 2H2O \]

### Final Balanced Equations:
1. \( BrO3^- + 2F2 + 2H2O \rightarrow 2F^- + BrO4^- + 2H^+ \)
2. \( F2 + 2NaOH \rightarrow OF2 + 2NaF + H2O \) (dilute)
3. \( 2F2 + 4NaOH \rightarrow O2 + 4NaF + 2H2O \) (conc.)