Question

Consider the reaction represented by the following chemical equation.

2SO2(g)+O2(g)⟶2SO3(g)
3.24mol of sulfur dioxide reacts with an excess of oxygen gas to form sulfur trioxide (SO3). Assuming the reaction occurs at a temperature of 300.0K and a pressure of 2.50atm what volume of sulfur trioxide will be produced in L?
Use 0.08206 L atm/mol K for the ideal gas constant.

Answer 1

To calculate the volume of SO3 produced in the reaction of SO2 with oxygen, the ideal gas law PV = nRT is used. Given the reaction temperature of 300.0 K, a pressure of 2.50 atm, and that 3.24 moles of SO2 yield the same number of moles of SO3, the volume of SO3 produced is 31.742688 L.

Step-by-step explanation:

The student is asking how to calculate the volume of sulfur trioxide (SO3) produced from a reaction between sulfur dioxide (SO2) and oxygen (O2) given the number of moles of sulfur dioxide, the temperature, and the pressure of the reaction. To find this volume, we will use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we must recognize that the reaction is balanced with a stoichiometry of 2 moles of SO2 producing 2 moles of SO3. Therefore, 3.24 moles of SO2 will produce 3.24 moles of SO3, because they are in a 1:1 molar ratio.

Second, using the ideal gas constant R = 0.08206 L atm/mol K, the pressure P = 2.50 atm, and the temperature T = 300.0 K, we can rearrange the ideal gas law to solve for the volume (V) as follows:

V = (nRT)/P

Inserting the values we have:

V = (3.24 moles * 0.08206 L atm/mol K * 300.0 K) / 2.50 atm

V = (3.24 * 0.08206 * 300.0) / 2.50

V = (79.35672) / 2.50

V = 31.742688 L

Therefore, 3.24 moles of SO2 will produce 31.742688 L of SO3 at 300 K and 2.50 atm.