One molecule of aluminium hydroxide will require how many molecules of dil . HCl . For complete neutralisation ?

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To determine how many molecules of dilute HCl are required to completely neutralize one molecule of aluminum hydroxide (Al(OH)₃), we can follow these steps:

### Step 1: Write the chemical equation for the reaction.
The reaction between aluminum hydroxide and hydrochloric acid (HCl) can be represented as follows:

\[ \text{Al(OH)}_3 + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2\text{O} \]

### Step 2: Balance the chemical equation.
To balance the equation, we need to ensure that the number of each type of atom is the same on both sides of the equation.

1. **Aluminum (Al)**: There is 1 aluminum atom on both sides.
2. **Chlorine (Cl)**: On the product side, there are 3 chlorine atoms in AlCl₃, so we need 3 HCl on the reactant side.
3. **Oxygen (O)**: There are 3 hydroxide (OH) groups in Al(OH)₃, which means we need 3 water (H₂O) molecules on the product side.
4. **Hydrogen (H)**: We have 3 from the 3 HCl and 3 from the 3 water molecules, giving a total of 6 hydrogen atoms on both sides.

The balanced equation will be:

\[ \text{Al(OH)}_3 + 3\text{HCl} \rightarrow \text{AlCl}_3 + 3\text{H}_2\text{O} \]

### Step 3: Determine the number of HCl molecules required.
From the balanced equation, we can see that 1 molecule of aluminum hydroxide (Al(OH)₃) reacts with 3 molecules of hydrochloric acid (HCl) for complete neutralization.

### Final Answer:
One molecule of aluminum hydroxide will require **3 molecules of dilute HCl** for complete neutralization.

---