`BrO_(3)^(ɵ) + 5Br^(ɵ) rarr Br_(2) + 3 H_(2) O`
IF `50 mL 0.1 M BrO_(3)^(ɵ)` is mixed with `30 mL` of `0.5 M Br^(ɵ)` solution that contains excess of `H^(o+)` ions, the moles of `Br_(2)` formed are

Options

A. `6.0 xx 10^(4)`
B. `1.2 xx 10^(-4)`
C. `9.0 xx 10^(-3)`
D. `1.8 xx 10^(-3)`

Explanation

To solve the problem, we need to determine how many moles of \( Br_2 \) are formed when \( 50 \, \text{mL} \) of \( 0.1 \, \text{M} \, BrO_3^- \) is mixed with \( 30 \, \text{mL} \) of \( 0.5 \, \text{M} \, Br^- \) solution.

### Step-by-Step Solution:

1. **Calculate the moles of \( BrO_3^- \)**:
\[ \text{Moles of } BrO_3^- = \text{Volume (L)} \times \text{Molarity (mol/L)} \]
\[ = 0.050 \, \text{L} \times 0.1 \, \text{mol/L} = 0.005 \, \text{mol} \, (5 \, \text{mmol}) \]

2. **Calculate the moles of \( Br^- \)**:
\[ \text{Moles of } Br^- = \text{Volume (L)} \times \text{Molarity (mol/L)} \]
\[ = 0.030 \, \text{L} \times 0.5 \, \text{mol/L} = 0.015 \, \text{mol} \, (15 \, \text{mmol}) \]

3. **Identify the limiting reagent**:
The balanced equation is:
\[ BrO_3^- + 5Br^- \rightarrow 3Br_2 + 3H_2O \]
From the equation, 1 mole of \( BrO_3^- \) reacts with 5 moles of \( Br^- \).

- We have \( 5 \, \text{mmol} \) of \( BrO_3^- \), which would require:
\[ 5 \, \text{mmol} \times 5 = 25 \, \text{mmol} \, Br^- \]
- However, we only have \( 15 \, \text{mmol} \) of \( Br^- \). Therefore, \( Br^- \) is the limiting reagent.

4. **Calculate the moles of \( Br_2 \) produced**:
From the balanced equation, 5 moles of \( Br^- \) produce 3 moles of \( Br_2 \):
\[ \text{Moles of } Br_2 = \left(\frac{3 \, \text{mol} \, Br_2}{5 \, \text{mol} \, Br^-}\right) \times \text{moles of } Br^- \]
\[ = \left(\frac{3}{5}\right) \times 0.015 \, \text{mol} = 0.009 \, \text{mol} \, (9 \, \text{mmol}) \]

### Final Answer:
The moles of \( Br_2 \) formed are \( 0.009 \, \text{mol} \) or \( 9 \times 10^{-3} \, \text{mol} \).