The value of `n` in the molecular formula `Be_(n)Al_(2) Si_( 6)O_(18)` is …..

Explanation

To find the value of \( n \) in the molecular formula \( Be_nAl_2Si_6O_{18} \), we need to analyze the charges of each element in the formula and ensure that the overall charge is neutral (which is 0).

### Step-by-Step Solution:

1. **Identify the oxidation states:**
- Beryllium (Be) typically has an oxidation state of +2.
- Aluminum (Al) has an oxidation state of +3.
- Silicon (Si) has an oxidation state of +4.
- Oxygen (O) has an oxidation state of -2.

2. **Set up the equation based on the charges:**
The total positive charge from the cations must equal the total negative charge from the anions for the compound to be neutral.

The contributions to the charge can be expressed as:
- For \( Be_n \): The charge is \( 2n \) (since each Be contributes +2).
- For \( Al_2 \): The charge is \( 3 \times 2 = 6 \) (since each Al contributes +3).
- For \( Si_6 \): The charge is \( 4 \times 6 = 24 \) (since each Si contributes +4).
- For \( O_{18} \): The charge is \( -2 \times 18 = -36 \) (since each O contributes -2).

3. **Write the charge balance equation:**
\[ 2n + 6 + 24 - 36 = 0 \]

4. **Simplify the equation:**
Combine the constant terms:
\[ 2n + 6 + 24 - 36 = 0 \implies 2n - 6 = 0 \]

5. **Solve for \( n \):**
\[ 2n = 6 \implies n = \frac{6}{2} = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).

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