How many mL of 0.125 M `Cr^(3+)` must be rected with 12.00 mL of 0.200 M `MnO_(4)^(-)` if the redox products are `Cr_(2)O_(7)^(2-)` and `Mn^(2+)` ?

Options

A. 8 mL
B. 16 mL
C. 24 mL
D. 32 mL

Explanation

To solve the problem, we need to determine how many mL of 0.125 M `Cr^(3+)` must react with 12.00 mL of 0.200 M `MnO4^(-)`. The products of the reaction are `Cr2O7^(2-)` and `Mn^(2+)`.

### Step-by-Step Solution:

1. **Write the balanced redox reaction**:
The reaction between `Cr^(3+)` and `MnO4^(-)` can be represented as:
\[ 6 \, Cr^{3+} + 5 \, MnO_4^{-} + 14 \, H^{+} \rightarrow 3 \, Cr_2O_7^{2-} + 5 \, Mn^{2+} + 7 \, H_2O \]
From the balanced equation, we can see that 6 moles of `Cr^(3+)` react with 5 moles of `MnO4^(-)`.

2. **Calculate the number of moles of `MnO4^(-)`**:
Using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]
For `MnO4^(-)`:
\[ \text{Moles of } MnO_4^{-} = 0.200 \, \text{M} \times 0.012 \, \text{L} = 0.0024 \, \text{moles} \]

3. **Determine the moles of `Cr^(3+)` needed**:
From the balanced equation, the mole ratio of `Cr^(3+)` to `MnO4^(-)` is 6:5. Therefore, we can set up the proportion:
\[ \frac{6 \, \text{moles of } Cr^{3+}}{5 \, \text{moles of } MnO_4^{-}} = \frac{x \, \text{moles of } Cr^{3+}}{0.0024 \, \text{moles of } MnO_4^{-}} \]
Solving for \( x \):
\[ x = 0.0024 \, \text{moles of } MnO_4^{-} \times \frac{6}{5} = 0.00288 \, \text{moles of } Cr^{3+} \]

4. **Calculate the volume of `Cr^(3+)` solution needed**:
We know the molarity of `Cr^(3+)` is 0.125 M. Using the formula:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.00288 \, \text{moles}}{0.125 \, \text{M}} = 0.02304 \, \text{L} \]
Converting to mL:
\[ \text{Volume (mL)} = 0.02304 \, \text{L} \times 1000 = 23.04 \, \text{mL} \]

### Final Answer:
Approximately **23.04 mL** of 0.125 M `Cr^(3+)` is required to react with 12.00 mL of 0.200 M `MnO4^(-)`.