Question

In the reaction \( 2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \), if 64 g of \(\text{SO}_2\) reacts completely, how many grams of \(\text{SO}_3\) are produced? (Molar mass of \(\text{SO}_2 = 64 \, \text{g/mol}, \text{SO}_3 = 80 \, \text{g/mol}\))

• A. 80 g
• B. 64 g
• C. 100 g
• D. 128 g

Hint:

Use the mole ratio from the balanced equation and convert moles to mass using the molar mass for stoichiometry problems.

Answer 1

The Correct Option is A

Step 1: Determine the number of moles of \(\text{SO}_2\). The molar mass of \(\text{SO}_2\) is 64 g/mol, and 64 g is given: \[ \text{Moles of } \text{SO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{64}{64} = 1 \, \text{mol}. \]
Step 2: Use the stoichiometry of the reaction. The balanced equation is \( 2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \), showing a 2:2 (1:1) mole ratio between \(\text{SO}_2\) and \(\text{SO}_3\).
Step 3: Calculate the moles of \(\text{SO}_3\) produced. Since 1 mole of \(\text{SO}_2\) produces 1 mole of \(\text{SO}_3\): \[ \text{Moles of } \text{SO}_3 = 1 \, \text{mol}. \]
Step 4: Compute the mass of \(\text{SO}_3\). The molar mass of \(\text{SO}_3\) is 80 g/mol: \[ \text{Mass of } \text{SO}_3 = \text{moles} \cdot \text{molar mass} = 1 \cdot 80 = 80 \, \text{g}. \]
Step 5: Verify the result. The 1:1 mole ratio and consistent molar masses confirm the calculation.