In the following equation:
`CIO_(3)^(-)+6H^(+)+.XrarrCl^(-)+3H_(2)O`, then `X` is

Options

A. `O`
B. `6e^(-)`
C. `O_(2)`
D. `6e^(-)`

Explanation

To solve the equation \( \text{ClO}_3^{-} + 6\text{H}^{+} + X \rightarrow \text{Cl}^{-} + 3\text{H}_2\text{O} \) and find the value of \( X \), we will follow these steps:

### Step 1: Identify the Reactants and Products
We have the following reactants:
- \( \text{ClO}_3^{-} \) (Chlorate ion)
- \( 6\text{H}^{+} \) (Hydrogen ions)
- \( X \) (unknown species)

And the products are:
- \( \text{Cl}^{-} \) (Chloride ion)
- \( 3\text{H}_2\text{O} \) (Water)

### Step 2: Balance the Chlorine Atoms
In the reactants, we have 1 chlorine atom in \( \text{ClO}_3^{-} \) and in the products, we have 1 chlorine atom in \( \text{Cl}^{-} \). Thus, chlorine is balanced.

### Step 3: Balance the Oxygen Atoms
In the reactants, there are 3 oxygen atoms in \( \text{ClO}_3^{-} \). In the products, we have 3 oxygen atoms from \( 3\text{H}_2\text{O} \) (since each water molecule contains 1 oxygen atom). Thus, oxygen is also balanced.

### Step 4: Balance the Hydrogen Atoms
In the reactants, we have 6 hydrogen ions (\( 6\text{H}^{+} \)). In the products, we have \( 3\text{H}_2\text{O} \), which contains 6 hydrogen atoms (2 from each water molecule). Thus, hydrogen is balanced.

### Step 5: Balance the Charges
Now, let's look at the charges:
- The charge on the left side is \( -1 + 6 = +5 \) (from \( \text{ClO}_3^{-} \) and \( 6\text{H}^{+} \)).
- The charge on the right side is \( -1 \) (from \( \text{Cl}^{-} \)).

To balance the charges, we need to account for the transfer of electrons. The left side has a total charge of \( +5 \) and the right side has \( -1 \), which means we need to add electrons to the left side to balance the charge.

### Step 6: Calculate the Number of Electrons
To balance the charges:
- The difference in charge is \( +5 - (-1) = +6 \).
- Therefore, we need to add 6 electrons (\( 6e^{-} \)) to the left side to balance the charge.

### Conclusion
Thus, the value of \( X \) in the reaction is \( 6e^{-} \) (6 electrons).

### Final Answer
\( X = 6e^{-} \)

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