Using the chemical equation: 2S + 3O3 -> 2SO3, we can see that 2 moles of sulfur react with 3 moles of oxygen to form 2 moles of sulfur trioxide.
To find the amount of oxygen required to produce 64.5 grams of sulfur trioxide, we need to first calculate the number of moles of sulfur trioxide produced.
The molar mass of sulfur trioxide is 80.06 g/mol (32.07 g/mol for sulfur + 3(16.00 g/mol) for oxygen).
Number of moles of SO3 = mass of SO3 / molar mass of SO3
Number of moles of SO3 = 64.5 g / 80.06 g/mol
Number of moles of SO3 = 0.805 mol
Since 2 moles of sulfur trioxide are produced for every 3 moles of oxygen, we can set up a proportion to find the number of moles of oxygen required to produce 0.805 mol of sulfur trioxide:
2 moles SO3 / 3 moles O2 = 0.805 moles SO3 / x moles O2
Solving for x, we get:
x = (3 moles O2 * 0.805 moles SO3) / 2 moles SO3
x = 1.21 moles O2
Finally, we can convert moles of oxygen to milliliters using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Assuming standard temperature and pressure (STP), which is 0°C and 1 atm, we can use the following values:
P = 1 atm
V = x (unknown)
n = 1.21 mol
R = 0.08206 L atm / mol K
T = 273 K
Solving for V, we get:
V = nRT / P
V = (1.21 mol * 0.08206 L atm / mol K * 273 K) / (1 atm)
V = 27.1 L
Finally, we need to convert liters to milliliters:
27.1 L * 1000 mL / L = 27,100 mL
So, 27,100 milliliters of oxygen are