In the reaction 2co ( g) + o2( g) → 2co2( g), what is the ratio of moles of oxygen used to moles of co2produced?

In the reaction 2co ( g) + o2( g) → 2co2( g), the ratio of moles of oxygen used to moles of co2produced is 1:2.

When: (CH3)3COOH(CH3)3⇒ C2H6 + 2CH3CoCH3 
                       A                            B                 2C
SO first for A:
When feed rate to reactor = FA(o), and Effluent rate from reactor= FA=Fa(o)(1-x)

Second for B:
When feed rate to reactor = 0, and Effluent rate from reactor FB= FA(o)X

Third for C:
When feed rate tp reactor = 0, and the Effluent rate from reactor FC= 2FA(o)X

So, the total feed rate to reactor FT(o) = FA(o)
and the total Effluent rate from reactor FT = FA(o)(1+2X)
For the CSTR:
V= FA(o) X / γa where -γ = KCA
We can get the total concentration at any point CT and the entrance CT(o) from this equation:
CT= FT/Q= P/ZRT
and CT(o) = FT(o)/ Q(o) = P(o)/ Z(o)RT(o)

We can neglect the changes in Z (the compressibility factor) So:
Q = Q(o) FT/FT(o) P(o)/P T/T(o)
Because there is no pressure drop and for the isothermal system we can assume P & T are constants:
∴ Q = Q(o) FT/ FT(o) 
∴ CA = FA / Q = (FA(o)(1-X)) / (Q(o)(1+2X))
                        = CA(o)((1-X)/(1+2X)) > eq 1
When K (the reaction rate constant) at 127C° can be evaluated with R (the gas constant)  8.314 J/mole.K° = 0.08205 L.atm/mole.°K

When, K2=K1(exp)[E/R ( (1/T1) - (1/T2)]
when T1 by kelvin = 50+273= 323 K° and T2= 127+273 = 400 K° 
and activition energy = 85 KJ/mol = 85000 J/mol
       ∴     K2   =10^-4 (exp) [85000/8314 ((1/323) - (1/400)] = 0.0443 min^-1 
We can get the concentration of (A) at 10 atm and 127C° (400K°) by:

CA(o)= PA(o) /RT = 10 / ( 0.08205 * 400 ) = 0.305 mol / L

∴ The reactor volume of the CSTR is:
V = FA(o)X / KCA---> eq 2 
by substitute eq 1 in eq 2
∴ V=[ FA(o)x * (1+2x)] / [ KCA(o) * (1-X) ] 
      = [ 2.5 * 0.9 * (1+1.8)] / [0.0443 * 0.305* 0.1]
     = 4662.7 L
- The equilibrium conversion:
according to the reaction:
                       A⇔B + 2C
and we have:
-γ = KfCA - Kb CB( C(C)^2) = 0 and KC = Kf / Kb = (CB* CC^2)/ CA= 0.025(as it is agiven)

∴[ ( XCA(o) ) ( 2XCA(o)^2 ) ] / [ (1+2X)^2 * CA(o) * (1-X) ] = 0.025
 
when we have CA(o) (the initial concentration)= 0.305 mol/L, So we have this eq:
[(XCA(o)) (2XCA(o))^2] / [(1+2X)^2 CA(o) (1-X) ] = 0.025
 By the matlab function solve
∴ X= X eq = 0.512 for 90% of the equilibrium conversion
∴ X= 0.9 X eq =  0.4608