Equilibrium Calculation for Fe3O4-CO Reaction at 600°C and 5.00 atm

[cheme2019]

Homework Statement

The Reaction
Fe₃O₄(s) + CO(g) ↔ 3 FeO(s) + CO₂(g)
took place at 600°C and at a constant total pressure of 5.00 atm. Determine the amount of each substance at equilibrium if there were 1 mole of Fe₃O₄, 2 moles of CO, .5 moles of FeO, and .3 moles of CO₂ originally present in the mixture and knowing that K_p = 1.15.

Homework Equations

K_p = K_c* (R* T)^Δn

The Attempt at a Solution

So I've written out my RICE table. I assumed since Δn = 0 that K_c = K_p that I would be able to use Kc in my equation. Because the activities of solids is 1, I can assume they won't impact the equilibrium equation. My solution looked like
1.15 = (.3+x)/(2-x) which goes to x = .482. However the solution says that x should be .93. I'm not sure where I went wrong. Thank you in advance!

 

[Borek]

Check your math.