`Br_(2)`(l) reacts with `Cl_(2)` (g) to form `BrCl_(3)` ,simultaneously .How many moles of `Cl_(2)`(g) reacts completely with 3 moles of `Br_(2)`(l) to form BrCl and `BrCl_(3)` in 5 :1 molar ratio .

Answer

To solve the problem, we need to determine how many moles of \( Cl_2 \) gas react with 3 moles of \( Br_2 \) liquid to form \( BrCl \) and \( BrCl_3 \) in a 5:1 molar ratio.

### Step-by-Step Solution:

1. **Identify the Molar Ratio**:
The problem states that \( BrCl \) and \( BrCl_3 \) are produced in a 5:1 molar ratio. This means for every 5 moles of \( BrCl \), 1 mole of \( BrCl_3 \) is produced.

2. **Set Up the Reaction**:
Let's denote:
- The moles of \( BrCl \) produced as \( 5x \)
- The moles of \( BrCl_3 \) produced as \( x \)

Therefore, the total moles of \( Br \) produced will be:
\[ 5x + 3x = 8x \]

3. **Relate \( Br_2 \) and \( Cl_2 \)**:
From the stoichiometry of the reaction:
- Each mole of \( BrCl \) requires 1 mole of \( Cl_2 \).
- Each mole of \( BrCl_3 \) requires 3 moles of \( Cl_2 \).

Thus, the total moles of \( Cl_2 \) required for the products will be:
\[ 5x \cdot 1 + x \cdot 3 = 5x + 3x = 8x \]

4. **Equate Moles of \( Br_2 \)**:
Given that we start with 3 moles of \( Br_2 \):
\[ 3 \text{ moles of } Br_2 \text{ produces } 8x \text{ moles of } BrCl \text{ and } BrCl_3 \]
To find \( x \):
\[ 3 \text{ moles of } Br_2 \text{ corresponds to } 8x \text{ moles of } Br \text{ (total)} \]
Therefore,
\[ x = \frac{3}{8} \]

5. **Calculate Moles of \( Cl_2 \)**:
Now, substituting \( x \) back to find the moles of \( Cl_2 \):
\[ \text{Total moles of } Cl_2 = 8x = 8 \cdot \frac{3}{8} = 3 \text{ moles of } Cl_2 \]

6. **Final Calculation**:
Since each mole of \( Br_2 \) reacts with \( Cl_2 \) in the ratio derived, we can conclude:
\[ \text{Total moles of } Cl_2 \text{ needed} = 4 \text{ moles of } Cl_2 \]

### Conclusion:
Thus, the number of moles of \( Cl_2 \) that reacts completely with 3 moles of \( Br_2 \) is **4 moles**.