What is the percent yield for the reaction below when 364 g SO2 and 42.0 g O2 produce 408 g SO3?2SO2(g) + O2(g) 2SO3(g)A.89.7%B.100%C.97.1%D.51.5%SUBMITarrow_backPREVIOUS

To solve this problem, we need to follow these steps:

1. First, we need to find the limiting reactant. To do this, we need to convert the mass of each reactant to moles using their molar masses. The molar mass of SO2 is approximately 64 g/mol and the molar mass of O2 is approximately 32 g/mol.

   Moles of SO2 = 364 g / 64 g/mol = 5.69 mol Moles of O2 = 42.0 g / 32 g/mol = 1.31 mol

2. According to the balanced chemical equation, the mole ratio of SO2 to O2 is 2:1. This means we need twice as many moles of SO2 as O2 for the reaction. However, we have more than twice as many moles of SO2 as O2, so O2 is the limiting reactant.

3. Next, we need to find the theoretical yield of SO3. According to the balanced chemical equation, the mole ratio of O2 to SO3 is 1:2. This means we can produce twice as many moles of SO3 as we have of our limiting reactant, O2.

   Theoretical yield of SO3 = 1.31 mol O2 * 2 mol SO3/1 mol O2 = 2.62 mol SO3

4. We then convert this amount to grams using the molar mass of SO3, which is approximately 80 g/mol.

   Theoretical yield of SO3 = 2.62 mol * 80 g/mol = 209.6 g

5. Finally, we find the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100%.

   Percent yield = (408 g / 209.6 g) * 100% = 194.5%

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