Answer: The balanced chemical equation for the reaction between SO2 and O2 is:
2 SO2 + O2 → 2 SO3
According to the equation, 1 mole of O2 reacts with 2 moles of SO2 to form 2 moles of SO3.
To determine which substance is the limiting reactant, we need to calculate the amount of product that can be formed by each reactant and choose the reactant that produces the least amount of product.
First, we need to determine the number of moles of SO2 and O2 present:
Number of moles of SO2 = mass / molar mass = 20 g / 64.06 g/mol = 0.312 moles
Number of moles of O2 = mass / molar mass = 20 g / 32.00 g/mol = 0.625 moles
Now we can calculate the amount of product formed by each reactant:
Amount of SO3 formed from SO2 = 0.312 moles × (2 moles SO3 / 2 moles SO2) × (80.06 g/mol SO3) = 12.48 g SO3
Amount of SO3 formed from O2 = 0.625 moles × (2 moles SO3 / 1 mole O2) × (80.06 g/mol SO3) = 100.10 g SO3
From the calculations, we can see that the limiting reactant is SO2, as it produces the least amount of product (12.48 g of SO3). Therefore, 12.48 g of SO3 is formed when 20 g of both SO2 and O2 are reacted.
Step-by-step explanation: