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A gas mixture contains each of the following gases at the indicated partial pressures: N2, 236 torr ; O2, 159 torr ; and He, 131 torr .?
What mass of each gas is present in a 1.10-L sample of this mixture at 25.0 degrees Celsius?
Mass of N2=
Mass of O2=
Mass of He=
1 Answer
- Roger the MoleLv 75 years ago
236 torr + 159 torr + 131 torr = 526 torr total
n = PV / RT = (526 torr) x (1.10 L) / ((62.36367 L torr/K mol) x (25.0 + 273.15) K) = 0.031118 mol total
(0.031118 mol) x (236 torr / 526 torr) x (28.01344 g N2/mol) = 0.391 g N2
(0.031118 mol) x (159 torr / 526 torr) x (31.99886 g O2/mol) = 0.301 g O2
(0.031118 mol) x (131 torr / 526 torr) x (4.002602 g He/mol) = 0.0310 g He
