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. When the above solution is allowed to stand for some time what volume of `O_(2)` will be obtained `STP`?

Options

A. 0.112 L
B. 0.224 L
C. 0.224L
D. 0.448L

Explanation

To solve the problem step by step, we will follow the chemical reactions and stoichiometry involved in the decomposition of hydrogen peroxide (H₂O₂) and the formation of oxygen (O₂).

### Step 1: Determine the moles of H₂O₂
Given that the concentration of the H₂O₂ solution is 6 M and the volume is 100 mL, we can calculate the number of moles of H₂O₂ present in the solution.

\[ \text{Moles of H₂O₂} = \text{Concentration} \times \text{Volume} = 6 \, \text{mol/L} \times 0.1 \, \text{L} = 0.6 \, \text{mol} \]

### Step 2: Decompose H₂O₂ to form O₂
The decomposition reaction of hydrogen peroxide is as follows:

\[ 2 \, \text{H₂O₂} \rightarrow 2 \, \text{H₂O} + \text{O₂} \]

From the balanced equation, 2 moles of H₂O₂ produce 1 mole of O₂. Therefore, the number of moles of O₂ produced from 0.6 moles of H₂O₂ can be calculated as follows:

\[ \text{Moles of O₂} = \frac{0.6 \, \text{mol H₂O₂}}{2} = 0.3 \, \text{mol O₂} \]

### Step 3: Convert moles of O₂ to volume at STP
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, we can calculate the volume of O₂ produced:

\[ \text{Volume of O₂} = \text{Moles of O₂} \times 22.4 \, \text{L/mol} = 0.3 \, \text{mol} \times 22.4 \, \text{L/mol} = 6.72 \, \text{L} \]

### Final Answer
The volume of O₂ obtained at STP is **6.72 liters**.

---