Question

Part A Three gases (8.00 g of methane, CH4, 18.0 g of ethane, CH, and an unknown amount of propane, CH.) were added to the same 10.0-L container. At 23.0 °C, the total pressure in the container is 4.50 atm. Calculate the partial pressure of each gas in the container. Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane. View Available Hint(s) 10 AED Owa ? atm Submit Previous Answers

Answer 1

I Given. Total volume- 10.OL Total pressure = 4.50 atm. Temperature = 23.0°(= 23.0 (+273=296k using Ideal Ideal-gal equation To find Total moles PV = nRT pv 4.50 atm x 10.0L 1.852 moles 0.0821 Latmktmol x 296k RT moles of (Hy = 8.00g 0.5 moles 16 g/mol moles of Ca Ho = 18.09 0.6 moles 30glmol Total = Mehty + Vesty the Hy NC ang = 1.852 moles 0.5 moles - 0.6 moles nez Hg = 0.75 moles Now, partial pressure of CHy = totalt total 4.50 atm x 0. 5 mol 1.22 atm 7.85 mol Partial pressure of C2H6 = 4.50 atmx 0.6 mol = 1.46 atm CHY 1.85 mol Partial pressure of C3 Hg = 4.50 4tm x 0.75 mol = 1.83 atm. 1.85 mol Hence, 1.22 , 1.46, 1.83 atm