Question

Calculating partial pressure in a gas mixture A 8.00 L tank at 3.01 °C is filled with 2.19 g of carbon dioxide gas and 4.83 g of chlorine pentafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Round each of your answers to 3 significant digits. mole fraction: carbon dioxide partial pressure 1x S ? mole fraction: chlorine pentafluoride partial pressure: Total presenta

Answer 1

We know toom ideal gas lawo equation - Pr=nRT . P= Pressure in atru, va volecane of gas ên litre 23 member of nucles, R= gas. Comstock = 0.082069tml neofita Iz Tembus a fase du Kelvin V= 8001, I= 3:01C= (3.01+273.15) K = 276:16K m = Mass in goleme Molas nass Carbon dioxide :- Molar mass of CO2 = 44 g/mol NECO2) = 2.19g = 0.04977 mol 02 44g/mol Parféal pressure of carbon dioxide, Fromo Pico) cºncouxRxT – Jagt977 noflx6208206 Pema 0.04977 mol X 0.08206 atub mol'k' X 276.16k 8.00L 8.00 Picozz 1.428 atm = 0.141 atman Partial Pressure of coz gas = 0.141 atm den Chlorine Pentafluoride (ClEs):-, Molasmess of UES= 120.449|ual es cams MCCLES) = het were = 0.037 mol

Ice V From O. Peceta) = n(CLFS)X RX T HCLES) = Pican 0.037 mol x 0.08206 afub mõll k'X 276-16K 2.00L Picite = 0.030s atm = 0.105 atm - 8.00L Parféal Presscese of elfs = 0-105 atm on Total Pressuse Partial Pressuse of CO2+ Partial Pressure of cete Total Pescase = (0-141+0.40%) atan = 0.246 am As Total numbes of moles= (0.04977+0.037) mol 0 = 0·08677 mol Mole fraclean of casten dioxidera Nurnibes of roles of CO2 – 0-04977 mof -0.6736 Total sentibes rueles 0-08677 malo e Mole fracleenitok Es = 0.037 mol er 0 0 -08677 mot = 0.4264 dhe mScanner o