When `B_2H_6` react with `O_2` and `H_2O` independently then product formed are:

Answer

To solve the question regarding the reaction of diborane (B₂H₆) with oxygen (O₂) and water (H₂O), we will analyze each reaction step by step.

### Step 1: Reaction of B₂H₆ with O₂

1. **Write the unbalanced reaction**:
\[ \text{B}_2\text{H}_6 + \text{O}_2 \rightarrow \text{B}_2\text{O}_3 + \text{H}_2\text{O} \]

2. **Identify the products**:
When diborane reacts with oxygen, it forms boron trioxide (B₂O₃) and water (H₂O).

3. **Balance the equation**:
- There are 2 boron atoms in B₂H₆, so we need 1 B₂O₃.
- There are 6 hydrogen atoms in B₂H₆, which will produce 3 H₂O (since each H₂O has 2 hydrogen).
- Now, we need to balance the oxygen. From 1 B₂O₃, we have 3 oxygen atoms, and from 3 H₂O, we have another 3 oxygen atoms, totaling 6 oxygen atoms. Therefore, we need 3 O₂ molecules.

The balanced equation is:
\[ \text{B}_2\text{H}_6 + 3\text{O}_2 \rightarrow \text{B}_2\text{O}_3 + 3\text{H}_2\text{O} \]

### Step 2: Reaction of B₂H₆ with H₂O

1. **Write the unbalanced reaction**:
\[ \text{B}_2\text{H}_6 + \text{H}_2\text{O} \rightarrow \text{B(OH)}_3 + \text{H}_2 \]

2. **Identify the products**:
When diborane reacts with water, it forms boric acid (B(OH)₃) and hydrogen gas (H₂).

3. **Balance the equation**:
- There are 2 boron atoms in B₂H₆, so we need 2 B(OH)₃.
- Each B(OH)₃ contributes 3 hydrogen atoms, giving a total of 6 hydrogen atoms from 2 B(OH)₃.
- From B₂H₆, we have 6 hydrogen atoms, which means we will release 3 H₂ (since each H₂ has 2 hydrogen).

The balanced equation is:
\[ \text{B}_2\text{H}_6 + 6\text{H}_2\text{O} \rightarrow 2\text{B(OH)}_3 + 3\text{H}_2 \]

### Summary of Products

- When B₂H₆ reacts with O₂, the products are:
- Boron trioxide (B₂O₃)
- Water (H₂O)

- When B₂H₆ reacts with H₂O, the products are:
- Boric acid (B(OH)₃)
- Hydrogen gas (H₂)

### Final Answer:
1. From the reaction with O₂: B₂O₃ and H₂O.
2. From the reaction with H₂O: B(OH)₃ and H₂.