[lizardo5]

An equilibrium mixture contains 3.100 mol HI and 1.258 moles each of H2 and I2.  HOw many moles of I2 must be removed to obtain an equilibrium mixture with 2.343 mol H2?

H2(g) + I2(g) -> 2HI (g)

______ mol I2 must be removed

I set up a reaction table and used the moles given to solve for the I2 removed but can't seem to get it right?  Can someone please help?

[Borek]

Show your work, hard to help without knowing what you did.

[lizardo5]

first i set up a reaction table
1)     H2              + I2 =              2HI
i       1.258 mol     1.258mol          3.100 mol

delt   + 1.085       +1.085             + 2.170

F     2.343            2.343                5.270

im confused since you use 1-1 of h2 and i2 why would you remove any to reach an equilibrium of 2.343 h2
                                         

[Yggdrasil]

If you remove I₂ then by Le'Chatalier's principle, the reaction will proceed backward to produce more H₂ and I₂ until it reaches a new equilibrium with a higher [H₂].

Now, this is a very complex problem (at least to me it looks like this).  Here's how I would approach the problem.  Since you can remove only I₂, that means the amount of hydrogen atoms in the mixture (whether in H₂ or HI) will remain constant.  Therefore, if you have 2.343 mol of H_{2 at equilibrium, you should be able to calculate how much HI you have at equilibrium.  From those two values, you can get the number of moles of I2 present at equilibrium.  Then, you can work back to find out how much I2 you removed from the system.}