In the following equation
`CrO_(4)^(2-)+S_(2)O_(3)^(2-)+overset(ɵ)(O)Hto[Cr(OH)_(4)]^(-1)+SO_(4)^(2-)`
What volume of 0.2 M `Na_(2)S_(2)O_(3)` solution require to completely react with 0.016 mole of chromate ion.

Options

A. 30 mL
B. 80 mL
C. 20 mL
D. 60mL

Explanation

To solve the problem, we need to determine the volume of a 0.2 M Na₂S₂O₃ solution required to completely react with 0.016 moles of chromate ions (CrO₄²⁻) based on the given reaction.

### Step-by-Step Solution:

1. **Identify the Reaction and Write the Balanced Equation:**
The reaction given is:
\[ \text{CrO}_4^{2-} + \text{S}_2\text{O}_3^{2-} + \text{OH}^- \rightarrow [\text{Cr(OH)}_4]^{-1} + \text{SO}_4^{2-} \]
We need to find the stoichiometry of the reaction to understand how many moles of S₂O₃²⁻ are needed for the reaction with CrO₄²⁻.

2. **Calculate the n-factor for CrO₄²⁻:**
- The oxidation state of chromium in CrO₄²⁻ is +6.
- In the product [Cr(OH)₄]⁻, the oxidation state of chromium is +3.
- The change in oxidation state is:
\[ 6 - 3 = 3 \]
- Since there is 1 chromium atom, the n-factor for CrO₄²⁻ is:
\[ n_{\text{CrO}_4^{2-}} = 1 \times 3 = 3 \]

3. **Calculate the n-factor for S₂O₃²⁻:**
- The oxidation state of sulfur in S₂O₃²⁻ is +2.
- In the product SO₄²⁻, the oxidation state of sulfur is +6.
- The change in oxidation state is:
\[ 6 - 2 = 4 \]
- Since there are 2 sulfur atoms in S₂O₃²⁻, the n-factor for S₂O₃²⁻ is:
\[ n_{\text{S}_2\text{O}_3^{2-}} = 2 \times 4 = 8 \]

4. **Set Up the Equivalence Equation:**
According to the stoichiometry of the reaction:
\[ \text{Equivalence of CrO}_4^{2-} = \text{Equivalence of S}_2\text{O}_3^{2-} \]
Therefore, we can write:
\[ n_{\text{CrO}_4^{2-}} \times \text{moles of CrO}_4^{2-} = n_{\text{S}_2\text{O}_3^{2-}} \times \text{moles of S}_2\text{O}_3^{2-} \]

5. **Calculate the Equivalents of CrO₄²⁻:**
- Moles of CrO₄²⁻ = 0.016
- Equivalents of CrO₄²⁻:
\[ \text{Equivalents of CrO}_4^{2-} = n_{\text{CrO}_4^{2-}} \times \text{moles of CrO}_4^{2-} = 3 \times 0.016 = 0.048 \]

6. **Set Up the Equation for S₂O₃²⁻:**
Let the moles of S₂O₃²⁻ be \( x \).
\[ \text{Equivalents of S}_2\text{O}_3^{2-} = n_{\text{S}_2\text{O}_3^{2-}} \times x = 8 \times x \]
Since the equivalents are equal:
\[ 0.048 = 8x \]
Solving for \( x \):
\[ x = \frac{0.048}{8} = 0.006 \text{ moles of S}_2\text{O}_3^{2-} \]

7. **Calculate the Volume of 0.2 M Na₂S₂O₃ Solution:**
Using the formula:
\[ \text{Molarity (M)} = \frac{\text{moles}}{\text{volume (L)}} \]
Rearranging gives:
\[ \text{Volume (L)} = \frac{\text{moles}}{\text{Molarity}} = \frac{0.006}{0.2} = 0.03 \text{ L} \]
Converting to mL:
\[ 0.03 \text{ L} = 30 \text{ mL} \]

### Final Answer:
The volume of 0.2 M Na₂S₂O₃ solution required to completely react with 0.016 moles of chromate ion is **30 mL**.