A mixture of 1.380 g of H2 and 66.80 g of Br2 is heated in a 2.00-L vessel at 700 K. These substances react as follows.

H2(g) + Br2(g) equilibrium reaction arrow 2 HBr(g)
At equilibrium the vessel is found to contain 0.605 g of H2.
(a) Calculate the equilibrium concentrations of H2, Br2, and HBr.
(b) Calculate Kc.
So I found the molarity of each one but im really confused on how to do this. Do I try an ICE box or is it something else completely? Any help is appreciated thanks!

Do the ICE box although that isn't completely necessary. Post your work and I'll be glad to check it.

Okay im actually a bit confused on how to do the ICE box in all honesty, we just learned this today. From my calculation the molarity for H2 is 0.3423 and the molarity for Br2 is 0.2090. ans the molarity for H2 at equilibrium is 0.150.

Assuming your numbers are right the ICE box will just about do them all.

........H2 + Br2 ==> 2HBr
I....0.342.0.209.......0
C......-x.....-x.......2x
E..0.342-x..0.209-x....2x

Read the problem again; it tells you that the Equilibrium concn (H2) = 0.150M. That tells you that 0.342-x = 0.150 which lets you solve for x, then knowing x you can calculate the E values for the other two. Know those you can calculate Kc.

ohhhhh I see okay, thanks so much for your help!so x= 0.193 therefore the equilibrium for H2 is 0.15 and Br is 0.017 and HBr is 0.384 right?

All of that is right except x is 0.192. However, the other numbers are right so that must be just a typo.

Yeah it is, sorry about that. Thank you so much for your help, I don't think I could have done it without you!

One more thing Im, sorry so would Kc be [0.384]2/[0.15][0.017]?

Yes, that's right. Good work. You did all of this; I just got you started. By the way, I would not type in a last name on the net (unless of course it's a phony one). We recommend no identifying information on Jiskha. Just good safe sense.

And thank you. You're quite welcome.

oh okay thank you for the advice!(: