Question

Table of Contents > Test 3 > Module 8 > After Class: Knewton HW and quiz > 8.4 Gas Laws: Ideal Gas Laws 18.4 Gas Laws: Ideal Gas Laws ---- Consider the reaction represented by the following chemical equation. 250,(g) + O2(g) 250, (g) 3.24 mol of sulfur dioxide reacts with an excess of oxygen gas to form sulfur trioxide (SO,)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 L? • Use 0.08206 batm for the ideal gas constant. • Round the answer to three significant figures. Provide your answer below: FEEDBACK MORE INSTRUCTION SUBMIT Content attribution

3.24 mol of surfer dioxide reacts with an excess of oxygen gas to phone sulfur trioxide assuming the reaction occurs as a temperature of 300 K and a pressure of 2.50 ATM what volume of sulfur trioxide want your list in L

Answer 1

To find the volume of sulfur trioxide Given that 3.24 mol of sulfur dioxide (SO) 250, (8) +0,(8) +250 (8) First find the moles of sulfur tri oxide (SO3) According to reaction SO, and SO, is 1:1 mole ratio Therefore moles of sulfur trioxide(SO3) = 3.24 mol To find the volume of sulfur trioxide (SO3) Use the ideal gas law PV = nRT p = pressure V = volume n=n moles of any gas R = gas constant T = temperature p=2.50atm n = 3.24 moles R = 0.08206 L atm /mol. K T = 300K V=? PV = nRT V = nRT 3.24 molesx 0.08206L - atm/mol. K x 300K 2.50atm = 31.90L Hence the volume of sulfur trioxide = 31.90L