A mixture containing equimolar amounts of `Ca(OH)_(2)` and `Al(OH)_(3)` requires `0.5` L of `4.0` M HCl to react with it completely. Moles of the mixture are :

Answer

To solve the problem, we need to determine the total moles of the mixture of \( Ca(OH)_2 \) and \( Al(OH)_3 \) that reacts with the given amount of hydrochloric acid (HCl).

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The reaction of \( Ca(OH)_2 \) with HCl is:
\[ Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O \]
This indicates that 1 mole of \( Ca(OH)_2 \) reacts with 2 moles of HCl.

- The reaction of \( Al(OH)_3 \) with HCl is:
\[ Al(OH)_3 + 3HCl \rightarrow AlCl_3 + 3H_2O \]
This indicates that 1 mole of \( Al(OH)_3 \) reacts with 3 moles of HCl.

2. **Define Moles of Each Component**:
- Let the moles of \( Ca(OH)_2 \) be \( A \).
- Since the mixture is equimolar, the moles of \( Al(OH)_3 \) will also be \( A \).

3. **Calculate Total Moles of HCl Required**:
- The total moles of HCl required for the reaction can be expressed as:
\[ \text{Total moles of HCl} = 2A + 3A = 5A \]

4. **Calculate Moles of HCl Provided**:
- The moles of HCl can be calculated using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume} \]
- Given that the molarity (M) is 4.0 M and the volume (V) is 0.5 L:
\[ \text{Moles of HCl} = 4.0 \, \text{M} \times 0.5 \, \text{L} = 2 \, \text{moles} \]

5. **Set Up the Equation**:
- From the previous steps, we have:
\[ 5A = 2 \]

6. **Solve for A**:
- Rearranging the equation gives:
\[ A = \frac{2}{5} = 0.4 \]

7. **Calculate Moles of the Mixture**:
- Since the mixture contains equimolar amounts of \( Ca(OH)_2 \) and \( Al(OH)_3 \), the total moles of the mixture is:
\[ \text{Moles of mixture} = A + A = 2A = 2 \times 0.4 = 0.8 \, \text{moles} \]

### Final Answer:
The total moles of the mixture are \( 0.8 \, \text{moles} \).

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