Question

A 20.0 L container at 303 K holds a mixture of two gases with a total pressure of 5.00 atm. If there are 2.00 mol of Gas A in the mixture, how many moles of Gas B are present?

Answer 1

PV = nRT —> n = PV/RT
P = 5.00 atm
V = 20.0 L
R = 0.08206 L atm / mol K
T = 303 K

n = (5.00 • 20.0)/(0.08206 • 303)
= 4.02 mol = A + B
A = 2.00 mol
B = 2.02 mol