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$\text{Hint: Use limiting reagent concept}$\n\n$\text{Step 1: Write the balanced chemical equation and calculate the moles of reactants.}$\n\n$2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$\n\n$\text{Molar mass of } \text{SO}_2 = 32 + 2 \times 16 = 64 \text{ g/mol}$\n\n$\text{Molar mass of } \text{O}_2 = 2 \times 16 = 32 \text{ g/mol}$\n\n$\text{Molar mass of } \text{SO}_3 = 32 + 3 \times 16 = 80 \text{ g/mol}$\n\n$\text{Moles of } \text{SO}_2 = \frac{6.4 \text{ g}}{64 \text{ g/mol}} = 0.1 \text{ mol}$\n\n$\text{Moles of } \text{O}_2 = \frac{3.2 \text{ g}}{32 \text{ g/mol}} = 0.1 \text{ mol}$\n\n$\text{Step 2: Determine the limiting reagent.}$\n\n$\text{From the balanced equation:}$\n\n$2 \text{ moles of } \text{SO}_2 \text{ require } 1 \text{ mole of } \text{O}_2$\n\n$0.1 \text{ mole of } \text{SO}_2 \text{ requires } \frac{0.1 \times 1}{2} = 0.05 \text{ mole of } \text{O}_2$\n\n$\text{We have 0.1 mole of } \text{O}_2\text{, which is more than required, so } \text{SO}_2 \text{ is the limiting reagent.}$\n\n$\text{Alternatively, we can check:}$\n\n$\text{For } \text{O}_2\text{: } 0.1 \text{ mole of } \text{O}_2 \text{ can react with } 0.1 \times 2 = 0.2 \text{ mole of } \text{SO}_2$\n\n$\text{Since we only have 0.1 mole of } \text{SO}_2\text{, } \text{SO}_2 \text{ is the limiting reagent.}$\n\n$\text{Step 3: Calculate the mass of } \text{SO}_3 \text{ formed.}$\n\n$\text{According to the balanced equation:}$\n\n$2 \text{ moles of } \text{SO}_2 \text{ produce } 2 \text{ moles of } \text{SO}_3$\n\n$0.1 \text{ mole of } \text{SO}_2 \text{ will produce } \frac{0.1 \times 2}{2} = 0.1 \text{ mole of } \text{SO}_3$\n\n$\text{Mass of } \text{SO}_3 \text{ formed} = 0.1 \text{ mol} \times 80 \text{ g/mol} = 8 \text{ g}$\n\n$\text{Therefore, the mass of } \text{SO}_3 \text{ formed is 8 g.}$