[Rutherford]

CO is combustioned in oxygen and a mixture of gasses is made. The density of the mixture is 8% higher than the density of air (80% N₂ and 20% O₂). The K_eq of this reaction is very high so the reaction can be used as irreversible. What was the ratio of CO and oxygen moles at the beggining of the reaction? n(CO)/n(O₂)=?

How to calculate the ratio if only percents are given?
The molar mass of air is 29g/mol but how to calculate the density?

[Borek]

If a gas of molar mass M occupies volume V, what is its density? If you have n moles of gas of molar mass M, what is its mass? What is its volume for a given P,T?

[Rutherford]

I figured out that the density of air is 29/V and of the mixture is it 1.08*29/V, so the mass is 1.08*29/V*V=31.32g but what to do now?

[Borek]

What gases are present in the mix? What must be their ratio for this density (or rather for this average molar mass)?

[Rutherford]

I suppose CO₂ and O₂.
m(CO₂)+m(O₂)=31.32
n(CO₂)=n(CO)
n(CO)/n(O₂)=[(m-31.32)/28]/(m/32)
And I have an unknown.

[Borek]

What about oxidation stoichiometry?

[Rutherford]

Still can't solve.
xCO+x/2O₂-->xCO₂
n of CO is x
n of O₂ before the reaction is [32*x/2+(31.32-44x)]/32
and again there is "-" so I can't reduce the unknown.

[Borek]

Sorry, I have no idea what your problems, please elaborate. Write exactly what equations do you have and what are you going to do with them.

[Rutherford]

I write the equation:
CO+1/2O₂-->CO₂
I marked the number of CO moles with x. From the reaction I see that x/2 moles of O₂ react.
I calculated that the obtained mixture of gasses (CO₂ and O₂) has a mass of 31.32g. The mass of CO₂ produced in the reaction is 44*x. the mass of O₂ that left unreacted is 31.32-44*x. So the beggining mass of O₂ (before the reaction) is x/2+31.32-44x.
I need to calculate n(CO)/n(O₂) before the reaction. n of O2 is n=(x/2+31.32-44x)/32 and n of CO I marked as x. When I put those into the equation (n(CO)/n(O₂)) I have 1 unknown so I can't calculate n(CO)/n(O₂).

[AWK]

Problem is solvable only if this density (1.08 of air) concerns a mixture of CO an O₂ before combustion.

[Rutherford]

"a mixture of gasses is made. The density of the mixture is 8% higher than the density of air" I don't know then, I translated it correct. The answer is n(CO)/n(O₂)=10.3

[AWK]

Through combustion we usually understand complete reaction with an excess of oxygen. In this case we have the oxidation of CO with oxygen as limiting reagent. Hence calculate mean molecular mass of mixture CO + CO₂ which is 1.08 times greater than mean molecular mass of air.
Sum of moles CO + CO₂ = moles CO before reaction.
1/2 moles of CO₂ = moles of O₂  before reaction.

This gives ratio CO/O₂ sligtly greater than 10

[Rutherford]

Can't solve again 
Number of moles of oxygen would be x/2. That is ok.
Now the number of CO moles: (31.32-44x)/28+x.44x is the mass of CO₂. Again it can't be solved.

[AWK]

31.32= 44x + 28(1-x)
           CO2    CO

[Rutherford]

31.32= 44x + 28(1-x)
           CO2    CO

Can you explain me how did you get (1-x)