Question

A2B3 --> 2A + 3B


I 0.001 0 0
C –x 2x 3x
E 0.001-x 2x 3x
3x = 0.006, x = 0.002
A = 0.004
B = 0.006
A2B3 = 0.001-0.002 =
(0.004)2(0.006)3/


Am I close?

Answer 1

I think we agree on A and B but not A2B3. I answered it below. A2B3 = 0.01-0.002 = 0.008
Don't forget that you must convert mols to M before calculating Kc.

Answer 2

3x = 0.006, x = 0.002
A = 0.004
B = 0.006
A2B3 = 0.01-0.002 = 0.008
(0.004)2(0.006)3/(0.008) = 4.32x10-10

Answer 3

No. I didn't calculate it but I know it isn't right becasue you didn't convert mols to M. M = mols/L so you must divide each of the mols by 0.200 L to convert to M. M is what goes into K expression. I thinkl 200 mL is correct although it isn't posted in this post.