[Cheistrynoob768]

Question 6: http://imgur.com/4Mz6FBR

My Work: http://imgur.com/wDj6O8y

I just did an ice table and I did some calculations on my calculator based on this work and I got a wrong answer, the answer should be 0.71 mol, I got 0.68 mol...

[magician4]

again, this question can only be answered by setting up a decent LMW expression first:

[tex]K_c = \frac {[CO_2]^2}{[CO]^2 \ \cdot \ [O_2]}[/tex]

what's a given , what are we looking for?

- we start with carbon monoxide, oxygen exclusively: no carbondioxide whatsoever
- therefore, all carbondioxide in equilibrium must have come from carbonmonoxide
- as per one carbondioxide gained one carbon monoxide is lost, it follows: c_eq.(CO) = c₀(CO) - c_eq.(CO₂)
- we know the conc. of two of those gases at eq., as we know those moles and the volume that contains them

let's go on by putting numbers in the equation:

[tex]4.2 = \frac {(0.34)^2}{[CO]^2 \ \cdot \ 0.2}[/tex]

this you can resolve for c_eq.(CO) , and calculate c₀(CO) from that value


... and I agree with approx. 0.71 moles

regards
Ingo

[Cheistrynoob768]

I tried doing it again but I got a worse answer,

4.2 = 0.34^2 / (y - 0.68)^2 * 0.20

.916515139 = 0.34/y - 0.68

.916515139y - 0.6232302945 = 0.34

y = 1.05 mol/L

1.05 * 1L = 1.05 mol

[magician4]

[tex]c_eq(CO) \neq c_0(CO) - 0.68 [/tex]

that's the source of your error


regards

Ingo