A gas mixture contains each of the following gases at the indicated partial pressures: N2, 315 torr; O2, 134 torr; and He, 219 torr?

Question

What is the total pressure of the mixture? What mass of each gas is present in a 2.15 L sample of this mixture at 25.0 C?

Hermina Ferry · Accepted Answer

total pressure = partial pressure due to N2 + partial pressure due to O2 + partial pressure due to He = 315 + 134 + 219 = 668 torr or 668/760 = 0.879 atm ( as 1 atm = 760 torr)now PV = nRT in order to calculate total mole of gasP = 0.879 atm V = 2.15 Ln = total no.of moles of all three gases = ?R = 0.0821 L atm/K/moleT = 25 + 273 = 298 K0.879 X 2.15 = n X 0.0821 X 298 1.889 = n X 24.466n = 1.889/24.466 = 0.077now partial pressure = mole fraction X total pressure partial pressure of N2 = mole fraction of N2 X total pressure315 = no.of moles of N2/total no.of moles X 668315 = no.of moles of N2/ 0.077 X 668 no.of moles of N2 = 315 X 0.077 / 668 = 0.036molar mass of N2 = 28 g/moleso amount of N2 = 28 X 0036 = 1.008 gsimilalrly ...partial pressure of O2 = mole fraction of O2 X total pressure134 = no.of moles of O2/0.077 X 668 no.of moles of O2 = 134 X 0.077 / 668 = 0.015molar mass of O2 = 32 g/moleso amount of O2 = 32 X 0.015 = 0.48 gand finally....partial pressure of He = mole fracion of He X total pressure219 = no.of moles of He/0.077 X 668no.of moles of He = 219 X 0.077/668 = 0.025 molar mass of He = 4 g/moleamount of He = 0.025 X 4 = 0.1 g

Hermina Ferry · Answer

(315 torr) + (134 torr) + (219 torr) = 668 torr total n = PV / RT = (668 torr) x (2.15 L) / ((62.36367 L torr/K mol) x (25.0 + 273.15) K) = 0.07724 mol total (0.07724 mol) x (315 torr N2 / 668 torr total) x (28.01344 g N2/mol) = 1.02 g N2(0.07724 mol) x (134 torr O2 / 668 torr total) x (31.99886 g O2/mol) = 0.496 g O2(0.07724 mol) x (219 torr He) / 668 torr total) x (4.002602 g He/mol) = 0.101 g He

Kelsie Sawayn

A gas mixture contains each of the following gases at the indicated partial pressures: N2, 315 torr; O2, 134 torr; and He, 219 torr?

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