In the reaction `X^(-)+XO_(3)^(-)+H^(+)rarrX_(2)+H_(2)O` , the molar ratio in which `X^(-)` and `XO_(3)^(-)` react is :

Options

A. `1:5`
B. `5:1`
C. `2:3`
D. `3:2`

Answer

To determine the molar ratio in which \( X^- \) and \( XO_3^- \) react in the given redox reaction, we will follow these steps:

### Step 1: Identify the oxidation and reduction half-reactions
In the reaction:
\[ X^- + XO_3^- + H^+ \rightarrow X_2 + H_2O \]

- **Oxidation half-reaction:** \( X^- \) is oxidized to \( X_2 \).
- **Reduction half-reaction:** \( XO_3^- \) is reduced to \( X_2 \).

### Step 2: Write the half-reactions
1. **Oxidation half-reaction:**
\[ 2X^- \rightarrow X_2 + 2e^- \]
Here, each \( X^- \) loses 1 electron, so 2 \( X^- \) will lose 2 electrons.

2. **Reduction half-reaction:**
\[ XO_3^- + 10e^- + 12H^+ \rightarrow X_2 + 6H_2O \]
Here, \( XO_3^- \) gains 10 electrons to be reduced to \( X_2 \).

### Step 3: Balance the electrons
To balance the overall reaction, we need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

- From the oxidation half-reaction, we have 2 electrons lost.
- From the reduction half-reaction, we have 10 electrons gained.

To equalize the number of electrons, we can multiply the oxidation half-reaction by 5:
\[ 5(2X^- \rightarrow X_2 + 2e^-) \Rightarrow 10X^- \rightarrow 5X_2 + 10e^- \]

### Step 4: Combine the half-reactions
Now we can combine the balanced half-reactions:
\[ 10X^- + XO_3^- + 12H^+ \rightarrow 5X_2 + 6H_2O \]

### Step 5: Determine the molar ratio
From the balanced equation:
- \( 10 \) moles of \( X^- \) react with \( 1 \) mole of \( XO_3^- \).

Thus, the molar ratio of \( X^- \) to \( XO_3^- \) is:
\[ \text{Molar ratio} = 10:1 \]

### Final Answer
The molar ratio in which \( X^- \) and \( XO_3^- \) react is \( 10:1 \).

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