Equilibrium and Kp: Find HI Mass at 700K

[MeMoses]

Homework Statement

The following reaction has Kp = 55.5 at 700 K:
H2(g) + I2(g) <-> 2HI(g)
You introguce 2.00 moles of H2(g) and 1.00 mole of I2(g) into a previously evacuated 5.00 L flash, and then heat the flask to 700 K. What mass of HI(g) will be present at eqilibrium?

Homework Equations

PV=nRT
Kp

The Attempt at a Solution

The pressure of 3 moles in 5L at 700K is 34.4652atm. So for every atm the reactant pressure decreases the product pressure increases the same. I get x^2 / (34.4652-x) = 55.6 => x=24.0466atm. I covert that to mol using pv=nrt and then multiply by the molecular wight of HI and get around 250 or 260 g *i forgot to wrrite it down). The answer however is 240g and every method I try I always get over by around 20 which doesn't seem right. Can you tell what I'm doing wrong? Thanks for any help.

 

[Borek]

You are trying to cut corners which yielded two mistakes - one in the nominator, one in the denominator. The one in the nominator is a simple simple omission of coefficient, the one in the denominator is an atrocity done to physics. Do it correctly, step by step. Start with the ICE table.

Note that you don't have to use pressures - there is no volume change, so Kc=Kp. Final result you will get using Kc will be the same, but you don't have to convert to p.