Ap chem question help needed

A 6.19g sample of pcl5 is placed in an evacuated 2.00 l flasi and is completely vaporized at 252 c.

a) calculate the pressure if no chem reaction were to occur. (I got the right answer for this, it is 487mm hg)
b) actually @ 252 c, pcl5 is partially dissociated according to the following equation:
Pcl5 (g) <---> pcl3 (g) + cl2
The observed pressure is found to be 1 atm. In view of this observation, calculate partial pressure of pcl5 and pcl3 in the flask at 252 c.

I need help with part b because I'm not entirely sure how to find partial pressure from that...if someone could help that would be great. Thanks!

6.19 / 208.2 g per mole = 0.0297 mole PCl5

moles of PCl5 + PCl3 + Cl2 = n = PV/RT
n = 1 am x 2.00 L / (0.082057 x 525 K) = 0.0464 mole
Let X = moles of PCl5 that dissociate
PCl5 = 0.0297 - X
PCl3 = X
Cl2 = X
Total moles = 0.0297 - X + X + X = 0.0297 + X = 0.0464
X = 0.0464 - 0.0297 = 0.0167
Moles of PCl5 = 0.0297 - 0.0167 = 0.013
Moles of PCl3 = 0.0167
Moles of Cl2 = 0.0167
Moles are directly proportional to pressure (at the same temp and volume)
Partial pressure of PCl5 = 1.00 atm * (0.013 / 0.0464) = 0.280 atm

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