Suppose a 250.mL flask is filled with 0.70mol of H2 and 1.6mol of HCl. The following reaction becomes possible:

H2 (g) + Cl2 (g) <---> 2HCl (g)

The equilibrium constant K for this reaction is 0.262 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places.

6.97 M

Explanation:

The reaction that takes place is

H₂(g) + Cl₂(g) ↔ 2HCl (g)

With a equilibrium constant ke =   = 0.262

The original concentrations for each species are:

250 mL ⇒ 250 / 1000 = 0.25 L

The ICE table thus is:

                      H₂(g) + Cl₂(g) ↔ 2HCl (g)

Initial                 2.8      0           6.4

Change            -x          -x             +2x

Equilibium        2.8-x      0-x        6.4 + 2x

So the equilibrium molarity of HCl is 6.4 + 2*0.285 = 6.97 M

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

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