Finding error in ideal gas law calculation for argon volume

[aboredperson]

Homework Statement

Given three rigid 1.00-Liter containers at 25o C filled with 1.00 mole of helium gas, 2.00 moles of neon gas, and 3.00 moles of argon gas respectively. When all three gases are pumped into a fourth 1.00-Liter container, what is the volume occupied by the neon gas in the final mixture?
a) 1.00L b) 2.00L c) 3.00L d) 0.167L. E) 0.333L

Relevant Equations

PV=nRT
P=pressure(atm) V= volume (L) n= Moles R= 0.0821 T= Temperature(K)

PV=nRT
P*1L = 6 moles * 0.0821*298 (I added up all the moles and solved for pressure)
P =146.79 atm
146.79 atm * V = 2 moles Ar * 0.0821* 298 (I plugged in the moles for argon and solved for volume)
V= 0.333 L
Answer key says the answer is 1 Liter. Where did I go wrong?

 

[Hill]

Looks like you have assumed that partial pressure of the neon gas is the same as the total pressure in the mixture.

 

[Chestermiller]

The 3 gases don't occupy distinct portions of the receiving container separately. They all occupy the complete receiving container. Each gas in an ideal gas mixture behaves as if the other gases are not even present.