Consider the reaction `2Al (g) + 3Cl_(2) (g) rArr 2Al Cl_(3) (g)`. The approximate volume of chlorine that would react with 324 g of aluminium at STP is :

Options

A. 112 L
B. 134 L
C. 260 L
D. 403 L

Explanation

To solve the problem of finding the approximate volume of chlorine that would react with 324 g of aluminium at STP, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The balanced chemical reaction is:
\[ 2Al (g) + 3Cl_2 (g) \rightarrow 2AlCl_3 (g) \]

2. **Calculate the Moles of Aluminium**:
Given the mass of aluminium (Al) is 324 g, and the molar mass of aluminium is 27 g/mol, we can calculate the number of moles of aluminium using the formula:
\[ \text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar Mass of Al}} = \frac{324 \text{ g}}{27 \text{ g/mol}} = 12 \text{ moles} \]

3. **Determine the Moles of Chlorine Needed**:
From the balanced equation, we see that 2 moles of Al react with 3 moles of Cl2. Therefore, we can find the moles of Cl2 required for 12 moles of Al:
\[ \text{Moles of Cl2} = \left(\frac{3 \text{ moles Cl2}}{2 \text{ moles Al}}\right) \times 12 \text{ moles Al} = 18 \text{ moles Cl2} \]

4. **Calculate the Volume of Chlorine at STP**:
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of 18 moles of Cl2 can be calculated as:
\[ \text{Volume of Cl2} = \text{Moles of Cl2} \times \text{Volume per mole at STP} = 18 \text{ moles} \times 22.4 \text{ L/mole} = 403.2 \text{ L} \]

5. **Final Answer**:
The approximate volume of chlorine that would react with 324 g of aluminium at STP is approximately:
\[ \text{Volume of Cl2} \approx 403 \text{ L} \]