[Cheistrynoob768]

Question 4: http://imgur.com/Wa4BkSk

My Work: http://imgur.com/MTQNR8M

I scratched a lot of things off by mistake and I bubbled the answer I got which was wrong, I felt like I did everything I could but the correct answer is actually 0.0132 mol.

[magician4]

from what I can decipher from your attempts , your initial error beginns here :
[tex][HI]_{eq.}^2 = \left( \frac {0.120}{0.5} \right)^2 = \color{red}{ 0.0576 \neq 0.21}  [/tex]

... though this is irrelevant, as you can't calculate [HI]_eq. in this manner from c₀(HI)

a much better idea would be to set up a LMA expression, and fill in what you know:

[tex] K_c = \frac {[H_2] \cdot [I_2]}{[HI]^2} = \frac { x \cdot x }{(c_0(HI) - 2x)^2} = 2 \cdot 10^{-2}[/tex]

resolve for x , calculate n(I₂) thereof


regards

Ingo

[Cheistrynoob768]

I tried again but I think I've gone somewhere wrong with my math,

0.0200 = x^2/(0.0576-2x)^2

0.0048 = x / 0.0576-2x

0.0048(0.0576-2x) = x

2.7648E-4 - 0.0096x = x

2.7648E-4 = 1.0096x

x = 2.73851E-4

When I tried to get the moles, I did 2.73851E-4 * 0.5 = 1.369255E-4 moles which is still wrong...

[magician4]

[tex]0.056 \neq c_0(HI)[/tex]

that's the source of your error


regards

Ingo

[Cheistrynoob768]

I am still stuck then, I tried all I could really.

[magician4]

ok, step by step..

what's your initial concentration c₀ for HI due to the original task ?

[Cheistrynoob768]

0.0576 mol/L?

[magician4]

if you have an in intial amount of n = 0.120 moles in V = 500 mL , what concentration does this result in?

[AWK]

This is the reaction with Δn=0 hence you can use moles instead of concentrations. This approach will give you moles of iodine directly.

[Vicktor]

What did you come up with on your final answer? I calculated it out a bit differently using the quadratic formula. I also noticed your x² disappeared?