1. Write the Balanced Chemical Equation:

The reaction between sulfur (S8) and oxygen (O2) produces sulfur dioxide (SO2):

S8 + 8 O2 → 8 SO2

2. Calculate the Moles of Each Reactant:

* Moles of S8:

- Molar mass of S8 = 256.52 g/mol

- Moles of S8 = (31.5 g) / (256.52 g/mol) = 0.123 mol

* Moles of O2:

- Molar mass of O2 = 32.00 g/mol

- Moles of O2 = (8.65 g) / (32.00 g/mol) = 0.270 mol

3. Determine the Limiting Reactant:

The limiting reactant is the one that gets used up first, limiting the amount of product that can be formed. To find this, compare the mole ratios from the balanced equation:

* From the balanced equation: 1 mole of S8 reacts with 8 moles of O2.

* In our reaction mixture: We have 0.123 moles of S8 and 0.270 moles of O2.

* Ratio comparison: (0.123 moles S8) / (1 mole S8) = 0.123

(0.270 moles O2) / (8 moles O2) = 0.03375

Since the ratio for O2 is smaller, O2 is the limiting reactant. This means the amount of SO2 produced will be determined by how much O2 is available.

4. Calculate the Moles of SO2 Produced:

From the balanced equation, 8 moles of O2 produce 8 moles of SO2. Since O2 is the limiting reactant, we'll use its moles to calculate the moles of SO2:

* Moles of SO2 = (0.270 moles O2) * (8 moles SO2 / 8 moles O2) = 0.270 moles SO2

5. Calculate the Mass of SO2 Produced:

* Molar mass of SO2 = 64.07 g/mol

* Mass of SO2 = (0.270 moles SO2) * (64.07 g/mol) = 17.3 g

Therefore, 17.3 g of SO2 is produced in this reaction.