Answer: 9.3 mol of SO3 at STP occupies 20.2 liters.
Explanation: To convert from moles of a gas to liters, we need to use the ideal gas law:
PV = nRT
where:
P = pressure in atm
V = volume in liters
n = number of moles
R = gas constant (0.08206 L·atm/mol·K)
T = temperature in Kelvin
We can rearrange this equation to solve for V:
V = (nRT)/P
First, let's calculate the volume of 9.3 mol of an ideal gas at standard temperature and pressure (STP). STP is defined as 0°C (273.15 K) and 1 atm.
V = (9.3 mol * 0.08206 L·atm/mol·K * 273.15 K) / 1 atm
V = 20.2 L
Therefore, 9.3 mol of SO3 at STP occupies 20.2 liters.
Note that this assumes SO3 is an ideal gas, which may not be the case in reality.