Final answer:
The reaction between 96g of SO2 and 32g of O2 produces 120 g of SO3, with SO2 being the limiting reactant.
Step-by-step explanation:
To determine the mass of product formed from the reaction of 96g of SO2 with 32g of O2, we first need to look at the balanced chemical equation for the reaction:
2SO2(g) + O2(g) → 2SO3(g)
From the balanced equation, we can see that two moles of SO2 react with one mole of O2 to produce two moles of SO3. The molar mass of SO2 is approximately 64.07 g/mol and the molar mass of O2 is 32.00 g/mol.
Thus, 96g of SO2 is equivalent to 96g / 64.07g/mol = 1.50 mol of SO2. Similarly, 32g of O2 is equivalent to 32g / 32.00g/mol = 1.00 mol of O2.
Since the reaction ratio is 2:1, the limiting reactant is SO2. Therefore, 1.50 mol of SO2 fully reacts with 0.75 mol of O2.
The molar mass of SO3 is 80.07 g/mol, so the mass of SO3 produced by 1.50 mol of SO2 is 1.50 mol × 80.07 g/mol = 120.105g of SO3
Our final answer, rounded to three significant figures, is 120 g of SO3 produced when 96g of SO2 reacts with 32g of O2.