Diborane reacts violently with water and releases hydrogen. Write a balanced equation of this reaction.

Answer

To write a balanced equation for the reaction of diborane (B2H6) with water (H2O) that produces ortho boric acid (H3BO3) and hydrogen gas (H2), follow these steps:

### Step 1: Write the unbalanced equation
The initial unbalanced equation for the reaction is:
\[ \text{B}_2\text{H}_6 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{BO}_3 + \text{H}_2 \]

### Step 2: Identify the number of atoms for each element
- On the left-hand side (LHS):
- Boron (B): 2 (from B2H6)
- Hydrogen (H): 6 (from B2H6) + 2 (from H2O) = 8
- Oxygen (O): 1 (from H2O)

- On the right-hand side (RHS):
- Boron (B): 1 (from H3BO3)
- Hydrogen (H): 3 (from H3BO3) + 2 (from H2) = 5
- Oxygen (O): 3 (from H3BO3)

### Step 3: Balance the boron atoms
Since there are 2 boron atoms on the LHS and only 1 on the RHS, we need to multiply H3BO3 by 2:
\[ \text{B}_2\text{H}_6 + \text{H}_2\text{O} \rightarrow 2 \text{H}_3\text{BO}_3 + \text{H}_2 \]

### Step 4: Update the atom count after adjusting boron
- LHS:
- B: 2
- H: 6 + 2 = 8
- O: 1

- RHS:
- B: 2 (from 2 H3BO3)
- H: 6 (from 2 H3BO3) + 2 (from H2) = 8
- O: 6 (from 2 H3BO3)

### Step 5: Balance the oxygen atoms
Now we have 6 oxygen atoms on the RHS (from 2 H3BO3) and only 1 on the LHS (from H2O). To balance the oxygen, we need to have 6 water molecules on the LHS:
\[ \text{B}_2\text{H}_6 + 6 \text{H}_2\text{O} \rightarrow 2 \text{H}_3\text{BO}_3 + \text{H}_2 \]

### Step 6: Update the hydrogen count
Now, recalculate the hydrogen atoms:
- LHS:
- H: 6 (from B2H6) + 12 (from 6 H2O) = 18
- RHS:
- H: 6 (from 2 H3BO3) + 2 (from H2) = 8

### Step 7: Balance the hydrogen atoms
To balance the hydrogen, we need to adjust the number of H2 produced. Since we have 18 H on the LHS, we need to produce 18 H on the RHS:
\[ \text{B}_2\text{H}_6 + 6 \text{H}_2\text{O} \rightarrow 2 \text{H}_3\text{BO}_3 + 6 \text{H}_2 \]

### Final Balanced Equation
The final balanced equation for the reaction of diborane with water is:
\[ \text{B}_2\text{H}_6 + 6 \text{H}_2\text{O} \rightarrow 2 \text{H}_3\text{BO}_3 + 6 \text{H}_2 \]