# Question #d2eaf

##### 1 Answer

I got

What you've effectively asked for is, "what is the percent of

Let

The ICE table becomes:

#"H"_2(g) " "+" " "I"_2(g) rightleftharpoons 2"HI"(g)#

#"I"" "n" "" "" "" "" "n" "" "" "0#

#"C"" "-x" "" "" "-x" "" "+2x#

#"E"" "n-x" "" "n-x" "" "2x#

Remember that the amount of reactant lost or product gained is *weighted by the stoichiometric coefficient* of that substance. It also shows up in the exponent.

So, the equilibrium constant is:

#K_c = 70.0 = (["HI"]^2)/(["H"_2]["I"_2])#

#= (2x)^2/((n-x)(n-x))#

That means this is easily solved without using the quadratic formula to find

#sqrt(70.0) = (2x)/(n-x)#

#nsqrt(70.0) - sqrt(70.0)x = 2x#

#(2 + sqrt(70.0))x = nsqrt(70.0)#

#=> x = sqrt(70.0)/(2 + sqrt(70.0))n#

#~~ 0.807n#

The amount of

#color(blue)(%"I"_2) = (x)/(n) = (0.807n)/(n) ~~ color(blue)(80.7%)#