Step 1: Calculate moles of SO₂. Given 64 g SO₂ and molar mass 64 g/mol: \[ \text{Moles of } \text{SO}_2 = \frac{64 \, \text{g}}{64 \, \text{g/mol}} = 1 \, \text{mol}. \]
Step 2: Apply reaction stoichiometry. The balanced equation \( 2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \) indicates a 1:1 mole ratio between SO₂ and SO₃.
Step 3: Determine moles of SO₃ produced. With a 1:1 ratio, 1 mole of SO₂ yields 1 mole of SO₃: \[ \text{Moles of } \text{SO}_3 = 1 \, \text{mol}. \]
Step 4: Compute mass of SO₃. Using the molar mass of SO₃ (80 g/mol): \[ \text{Mass of } \text{SO}_3 = 1 \, \text{mol} \cdot 80 \, \text{g/mol} = 80 \, \text{g}. \]
Step 5: Confirm the result. The 1:1 mole ratio and consistent molar masses validate the calculation.