Starting with 0.3500mol Co(g) and 0.05500mol COCL2(g) in a 3.050 L flask at 668K, how many moles of Cl2 are present at equilibrium?

Question

Co(g) = Cl2(g) <-> COCL2(g) Kc = 1.2x10^3

Answer 1

Do you mean Co or CO. Surely you meant CO. You must understand that CO is carbon monoxide and Co is cobalt. Also, I assume the = sign should be a + sign.

Convert to concns in molarity.
(CO) = 0.3500/3.050 = ??
(COCl2) = 0.05500/3.050 = yy
.............CO + Cl2 ==> COCl2
initial..0...??....0.......yy
change.....+x......+x......-x
equil....0.??+x...x........yy-x

Substitute the ICE chart I've prepared and substitute into Kc expression, solve for x. That will give you (Cl2) in moles/L, then multiply by L to convert to moles.

Answer 2

When I converted everything I got

1.2x10^3 = (0.02 + x) / (0.12 - x)(x)
= (0.02 + x) / (0.12x - x^2)

1.2x10^3(0.12x - x^2) = 0.02 + x
144x - 1.2x10^3x^2 = 0.02 + x

Then I brought x to the other side and I got

145x - 1.2x10^3x^2 = 0.02

Then I get confused with the x's and don't know how to continue solving.

Thanks so much for the previous help though

Answer 3

You're rounding far too early. I may make an error by carrying too many places when doing these quadratic equations but I like to carry more than I'm allowed and round at the end.
0.3500/3.050 = 0.11475M = (CO)
0.05500/3.050 = 0.01803M = (COCl2)

First, note that K = 1.29E3 or 1290 and not 1200.
............CO + Cl2 ==> COCl2
initial.0.11475...0.......0.01803
change.....+x......+x......-x
equil..0.11475+x...x.......0.01803-x

Kc = 1290 = (COCl2)/(CO)(Cl2)
1290 = (0.01803-x)/(0.11475+x)(x)
0.01803-x = 1290(0.11475+x)(x)
0.01803-x = 1290x^2 + 148.03x
1290x^2 +149.03x -0.01803 = 0
x = 0.0001208 M = (Cl2)
Then convert to moles and round to the appropriate number of s.f.

Answer 4

OK thanks so much, I appreciate it.