Question

Consider the reaction represented by the following chemical equation.

2SO₂(g) + O₂(g) →→→ 2SO3(g)
3.24 mol of sulfur dioxide reacts with an excess of oxygen gas to form sulfur trioxide (SO3). Assuming the reaction
occurs at a temperature of 300.0 K and a pressure of 2.50 atm what volume of sulfur trioxide will be produced in La
• Use 0.08206 Latm for the ideal gas constant.
mol K
Round the answer to three significant figures.

Answer 1

To solve this problem, we can use the information given to calculate the number of moles of sulfur trioxide produced and then use the ideal gas law to calculate the volume. The volume of sulfur trioxide produced will be 9.79 liters.

Step-by-step explanation:

To solve this problem, we can use the information given to calculate the number of moles of sulfur trioxide produced. First, we need to determine the number of moles of sulfur dioxide that reacts. We are given that 3.24 mol of sulfur dioxide reacts, so we can use the balanced chemical equation to determine the amount of sulfur trioxide produced. According to the equation, 2 moles of sulfur dioxide react to produce 2 moles of sulfur trioxide, so the amount of sulfur trioxide produced will be equal to the amount of sulfur dioxide reacted, which is 3.24 mol.

Next, we can use the ideal gas law to calculate the volume of sulfur trioxide produced. The ideal gas law equation 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. Rearranging the equation, we have V = (nRT) / P. Plugging in the values given in the problem, we get:

V = (3.24 mol)(0.08206 Latm/mol K)(300.0 K) / (2.50 atm) = 9.79 L

Therefore, the volume of sulfur trioxide produced will be 9.79 liters.