Chemical equilibrium help

Chemical equilibrium help!!!!

For a reaction 2SO2(g) + O2(g) ↔ 2 SO3(g), 0.1mol of each SO2 and SO3 are mixed in a 2.0L flask at 27 degrees Celsius. After Equilibrium total pressure is 2.78atm.
Calculate a) The mole fraction of O2 at equilibrium
b) The value of Kp

I dont know how to find out the mole fraction of O2...
I did the problem finding Kp.. but I got a large number of 19.31....
And my Nt at equilibrium= n(4-alpha)

Any ideas??

 

Re: Chemical equilibrium help!!!!

Here is my work:

2SO2(g) + O2(g) ↔ 2 SO3(g)
2n 0 ↔ 2n
-2n@ -n@ ↔ +2n@

2n-2n@ -n@ ↔ 2n +2n@ Where Nt= 2n-2n@-n@+2n+2n@
2n(1-@) -n@ ↔ 2n(1+@) Nt=4n-n@ Nt= n(4-@)

2n(1-@)/n(4-@) -n@/(n4-@) ↔ 2n(1+@)/ (n (4-@))

2(1-@)/(4-@) Pt 1/4 Pt ↔ 2(1+@)/ (4-@))

Where Kp= (P SO3(g))^2 / ((P SO2)^2 * (P O2))

Kp= ((4(1+@)^2) / ((1-@)^2 ) ) 1/Pt

Where Nt at equilibrium= ( 2.78atm*2L) /( 0.08206 atm dm3 mol-1 k-1)(301.15)
Nt= .2249 moles

solving for @
.2449= .1(4-@)
@= 1.751

Plugging in @ to find Kp.... I got 19.31....... but i'ts wrong it's supposed to be 0.356