When `B_2H_6` react with`Cl_2` it produce:

Answer

To solve the question regarding the reaction of \( B_2H_6 \) with \( Cl_2 \), we will follow these steps:

### Step 1: Write the Reactants
We start with the reactants given in the question:
- \( B_2H_6 \) (Diborane)
- \( Cl_2 \) (Chlorine gas)

### Step 2: Identify the Products
When \( B_2H_6 \) reacts with \( Cl_2 \), it produces:
- \( BCl_3 \) (Boron trichloride)
- \( HCl \) (Hydrochloric acid)

### Step 3: Write the Unbalanced Reaction
The unbalanced chemical equation for the reaction can be written as:
\[ B_2H_6 + Cl_2 \rightarrow BCl_3 + HCl \]

### Step 4: Balance the Equation
Now, we need to balance the equation.
1. Count the number of atoms for each element on both sides.
- Left side:
- Boron (B): 2
- Hydrogen (H): 6
- Chlorine (Cl): 2
- Right side:
- Boron (B): 1 (from \( BCl_3 \))
- Hydrogen (H): 1 (from \( HCl \))
- Chlorine (Cl): 4 (3 from \( BCl_3 \) and 1 from \( HCl \))

2. To balance boron, we place a coefficient of 2 in front of \( BCl_3 \):
\[ B_2H_6 + Cl_2 \rightarrow 2BCl_3 + HCl \]

3. Now, recalculate:
- Left side:
- B: 2, H: 6, Cl: 2
- Right side:
- B: 2, H: 1 (from \( HCl \)), Cl: 6 (6 from \( 2BCl_3 \))

4. To balance hydrogen, we need 6 \( HCl \):
\[ B_2H_6 + 6Cl_2 \rightarrow 2BCl_3 + 6HCl \]

5. Finally, we balance chlorine:
- Now, we have 6 \( Cl_2 \) which gives us 12 Cl atoms, and on the right side, we have 6 from \( 2BCl_3 \) and 6 from \( 6HCl \), totaling 12 Cl atoms.

### Step 5: Final Balanced Equation
The final balanced equation is:
\[ B_2H_6 + 6Cl_2 \rightarrow 2BCl_3 + 6HCl \]

### Step 6: Additional Reaction
When \( B_2H_6 \) reacts with \( HCl \), it produces:
\[ B_2H_5Cl + H_2 \]
This can also be balanced similarly.

### Final Answer
When \( B_2H_6 \) reacts with \( Cl_2 \), it produces \( BCl_3 \) and \( HCl \).

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