What is the theoretical yield of SO3 produced by 3.15 g of S?
The balanced chemical equation for the reaction of S with O2 to produce SO3 is:
2S + 3O2 -> 2SO3
From the equation, we can see that 2 moles of S will react with 3 moles of O2 to produce 2 moles of SO3.
First, we need to determine the number of moles of S present in 3.15 g of S:
molar mass of S = 32.06 g/mol
number of moles of S = mass / molar mass
number of moles of S = 3.15 g / 32.06 g/mol
number of moles of S = 0.0984 mol
Now, using the mole ratio from the balanced equation, we can calculate the theoretical yield of SO3 produced:
moles of SO3 = moles of S x (2 moles SO3 / 2 moles S)
moles of SO3 = 0.0984 mol x 1
moles of SO3 = 0.0984 mol
Finally, we can convert the moles of SO3 to grams using the molar mass of SO3:
molar mass of SO3 = 80.06 g/mol
theoretical yield of SO3 = moles of SO3 x molar mass of SO3
theoretical yield of SO3 = 0.0984 mol x 80.06 g/mol
theoretical yield of SO3 = 7.87 g
Therefore, the theoretical yield of SO3 produced by 3.15 g of S is 7.87 g.