How many grams of Xe gas are present in a 21.00 L cylinder held at a pressure of 21.00 atm and 67.0°C?
A) 2074 g
B) 514.7 g
C) 7361 g
D) 8043 g

Question

Quizplus · Accepted Answer

We can use the ideal gas law to solve for the number of moles of Xe gas present and then convert to grams using the molar mass of Xe.
PV = nRT
n = PV/RT
n = (21.00 atm)(21.00 L) / [(0.08206 L*atm/mol*K)(340.2 K)]
n = 2.064 mol Xe
m = nM
m = (2.064 mol)(131.3 g/mol)
m = 271.5 g Xe (rounded to 3 significant figures)
Therefore, the correct answer is A) 2074 g.

How Many Grams of Xe Gas Are Present in a 21.00