Final answer:
To find the amount of sulfur trioxide that could be formed, we need to determine which reactant is limiting. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. We compare the moles of the limiting reactant to the mole ratio from the balanced equation to find the theoretical yield. Finally, we convert the moles of product to grams using its molar mass to get the answer in grams.
Step-by-step explanation:
To find the amount of sulfur trioxide that could be formed, we need to determine which reactant is limiting.
The balanced equation for the reaction is:
2S + 3O2 -> 2SO3
From the equation, we can see that 2 moles of sulfur react with 3 moles of oxygen to produce 2 moles of sulfur trioxide.
First, we need to convert the given masses of sulfur and oxygen to moles using their molar masses: sulfur (32.07 g/mol) and oxygen (16.00 g/mol).
11.3 g S * (1 mol S / 32.07 g S) = 0.352 mol S
14.0 g O2 * (1 mol O2 / 32.00 g O2) = 0.438 mol O2
Next, we compare the mole ratios from the balanced equation:
0.352 mol S / (2 mol S / 2 mol SO3) = 0.352 mol SO3
0.438 mol O2 / (3 mol O2 / 2 mol SO3) = 0.292 mol SO3
Since the amount of sulfur trioxide that can be formed is limited by the amount of oxygen, the theoretical yield of sulfur trioxide is 0.292 moles.
Finally, we convert the moles of sulfur trioxide to grams using its molar mass (80.06 g/mol):
0.292 mol SO3 * (80.06 g SO3 / 1 mol SO3) = 23.37 g SO3 (rounded to two decimal places)
Therefore, the correct answer is B) 22.0 g.