The reaction
`Cl_(2) + S_(2)O_(3)^(2-) + OH^(-) to SO_(4)^(2-) + Cl^(-) + H_(2)O`
Starting with 0.15 mole `Cl_(2)` , 0.010 mole `S_(2)O_(3)^(2-)` and 0.30 mole `OH^(-)` mole of `Cl_(2)` left in solution will be

Options

A. 0.11
B. 0.01
C. 0.04
D. 0.09

Answer

To solve the problem, we need to analyze the given redox reaction and determine how much Cl₂ remains after the reaction has occurred. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the balanced chemical equation
The balanced reaction is:
\[ \text{Cl}_2 + \text{S}_2\text{O}_3^{2-} + \text{OH}^- \rightarrow \text{SO}_4^{2-} + \text{Cl}^- + \text{H}_2\text{O} \]

### Step 2: Identify the initial amounts of reactants
From the question, we have:
- Moles of Cl₂ = 0.15 moles
- Moles of \( \text{S}_2\text{O}_3^{2-} \) = 0.010 moles
- Moles of OH⁻ = 0.30 moles

### Step 3: Determine the stoichiometry of the reaction
From the balanced equation, we see that:
- 1 mole of \( \text{S}_2\text{O}_3^{2-} \) reacts with 4 moles of Cl₂.

### Step 4: Calculate the amount of Cl₂ required for the reaction
Using the stoichiometry:
- Moles of Cl₂ required for 0.010 moles of \( \text{S}_2\text{O}_3^{2-} \):
\[ 0.010 \, \text{moles of } S_2O_3^{2-} \times 4 \, \text{moles of Cl}_2/\text{mole of } S_2O_3^{2-} = 0.040 \, \text{moles of Cl}_2 \]

### Step 5: Identify the limiting reagent
We compare the moles of Cl₂ required (0.040 moles) with the available moles (0.15 moles). Since we have more than enough Cl₂, \( \text{S}_2\text{O}_3^{2-} \) is the limiting reagent.

### Step 6: Calculate the remaining moles of Cl₂
Now we subtract the moles of Cl₂ that reacted from the initial amount:
- Initial moles of Cl₂ = 0.15 moles
- Moles of Cl₂ that reacted = 0.040 moles

Remaining moles of Cl₂:
\[ 0.15 \, \text{moles} - 0.040 \, \text{moles} = 0.11 \, \text{moles} \]

### Final Answer
The moles of Cl₂ left in the solution will be **0.11 moles**.

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