Three solutions eaach of 100 mL containing 0.4 M `As_(2)S_(3),5M` NaOH and `6MH_(2)O_(2)`, respectively were mixed to form `AsO_(4)^(3-)` and `SO_(4)^(2-)` as products.
Q. Percentage strength of the `H_(2)O_(2)` solution left after reaction is

Options

A. `0.017%`
B. `0.113%`
C. `0.51%`
D. `0.68%`

Explanation

To solve the problem, we need to determine the percentage strength of the remaining H₂O₂ solution after the reaction. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the moles of each reactant

1. **Calculate moles of As₂S₃**:
- Volume = 100 mL = 0.1 L
- Molarity = 0.4 M
- Moles of As₂S₃ = Molarity × Volume = 0.4 mol/L × 0.1 L = 0.04 moles

2. **Calculate moles of NaOH**:
- Volume = 100 mL = 0.1 L
- Molarity = 5 M
- Moles of NaOH = 5 mol/L × 0.1 L = 0.5 moles

3. **Calculate moles of H₂O₂**:
- Volume = 100 mL = 0.1 L
- Molarity = 6 M
- Moles of H₂O₂ = 6 mol/L × 0.1 L = 0.6 moles

### Step 2: Identify the reaction and stoichiometry

The reaction involves As₂S₃, NaOH, and H₂O₂ producing AsO₄³⁻ and SO₄²⁻. The stoichiometry of the reaction will determine how much H₂O₂ is consumed.

### Step 3: Determine the limiting reagent

Assuming the reaction consumes H₂O₂ completely, we need to find out how much H₂O₂ is required for the reaction.

1. **Assuming a balanced reaction** (for simplicity, let's assume a 1:1 stoichiometry for H₂O₂ to As₂S₃):
- For 0.04 moles of As₂S₃, we would need approximately 0.04 moles of H₂O₂.

### Step 4: Calculate remaining moles of H₂O₂

1. **Moles of H₂O₂ left after the reaction**:
- Initial moles of H₂O₂ = 0.6 moles
- Moles of H₂O₂ consumed = 0.04 moles
- Remaining moles of H₂O₂ = 0.6 - 0.04 = 0.56 moles

### Step 5: Calculate the normality of the remaining H₂O₂

1. **Convert moles of H₂O₂ to normality**:
- Total volume of the mixed solution = 100 mL + 100 mL + 100 mL = 300 mL = 0.3 L
- Normality (N) = Moles of solute / Volume of solution in L
- Normality of remaining H₂O₂ = 0.56 moles / 0.3 L = 1.87 N

### Step 6: Convert normality to percentage strength

1. **Convert normality to percentage strength**:
- 1 Normal H₂O₂ = 1 equivalent per liter = 5.6 volume strength
- Therefore, 1.87 N corresponds to a volume strength of 1.87 × 5.6 = 10.47 volume strength.
- To convert volume strength to percentage strength, we use the formula:
\[ \text{Percentage Strength} = \frac{\text{Volume Strength}}{10} \approx 10.47 / 10 = 1.047\% \]

### Final Answer

The percentage strength of the H₂O₂ solution left after the reaction is approximately **1.05%**.

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