Sulphur trioxide is preapared by the following two reactions:
`S_(8)(s)+8O_(2)(g)to 8SO_(2)(g)`
`2SO_(2)(g)+O_(2)(g) to 2SO_(3)(g)`
How many grams of `SO_(3)` are produced from 1 mole fo `S_(8)`?

Options

A. 1280 g
B. 960 g
C. 640 g
D. 320 g

Answer

To solve the problem of how many grams of \( SO_3 \) are produced from 1 mole of \( S_8 \), we can follow these steps:

### Step 1: Analyze the first reaction
The first reaction is:
\[ S_8 (s) + 8 O_2 (g) \rightarrow 8 SO_2 (g) \]

From this reaction, we can see that 1 mole of \( S_8 \) produces 8 moles of \( SO_2 \).

### Step 2: Analyze the second reaction
The second reaction is:
\[ 2 SO_2 (g) + O_2 (g) \rightarrow 2 SO_3 (g) \]

From this reaction, we observe that 2 moles of \( SO_2 \) produce 2 moles of \( SO_3 \). This means that 1 mole of \( SO_2 \) produces 1 mole of \( SO_3 \).

### Step 3: Calculate the total moles of \( SO_3 \) produced
Since 1 mole of \( S_8 \) produces 8 moles of \( SO_2 \), and each mole of \( SO_2 \) produces 1 mole of \( SO_3 \), we can conclude:
\[ 8 \text{ moles of } SO_2 \rightarrow 8 \text{ moles of } SO_3 \]

Thus, 1 mole of \( S_8 \) will produce 8 moles of \( SO_3 \).

### Step 4: Calculate the mass of \( SO_3 \)
Now, we need to find the mass of 8 moles of \( SO_3 \). The molar mass of \( SO_3 \) can be calculated as follows:
- Sulfur (S) has a molar mass of approximately 32 g/mol.
- Oxygen (O) has a molar mass of approximately 16 g/mol.

Thus, the molar mass of \( SO_3 \) is:
\[ 32 \, \text{g/mol} + (3 \times 16 \, \text{g/mol}) = 32 \, \text{g/mol} + 48 \, \text{g/mol} = 80 \, \text{g/mol} \]

Now, we can calculate the total mass of 8 moles of \( SO_3 \):
\[ \text{Mass of } SO_3 = 8 \, \text{moles} \times 80 \, \text{g/mol} = 640 \, \text{grams} \]

### Final Answer
Therefore, the mass of \( SO_3 \) produced from 1 mole of \( S_8 \) is **640 grams**.

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