Question

A gas mixture contains each of the following gases at the indicated partial pressures: N2, 216 torr ; O2, 123 torr ; and He, 122 torr .

What is the total pressure of the mixture?

What mass of each gas is present in a 1.30 −L−L sample of this mixture at 25.0 ∘C∘C?

Answer 1

Poatial poressure of N2 = Partial Pressure gl O2 = Partial Pressure of He = 216 boor 123 took 122 tooor Pltolal) - 216+123+122 - 461 toe total. Using the equation, PV=nRT, we calculate the total number of moles- = 461 toode * 1:30 L (6236367 L Torockmel JX (25.0+ 273.1537 = 599.3 = 0.0286 mol gases 20921.45 Nihogen (N2 = 28.01g/mol) = 0.0286 mol x(216 boor N2, 461 toga * (28.01g Namo1) = 0.375g N2 Oxygen CO2 = 32g/mol) = 0·0286 mol % (123 toode 02) « (32 g Oz/mol) 461 toga ) = 0.2449 02 Helium ( He - 4g/mol) = 0.0286mol x (12 hoa fle)x (AgImol) = O. Ozbate 1 461twa / 122 tooer h ).OZXhaal 'o