Question #d2eaf

I got #80.7%# reacted #"I"_2#.

What you've effectively asked for is, "what is the percent of #"I"_2(g)# that has reacted?". #"H"_2(g)# and #"I"_2(g)# are mixed in a 1:1 molar ratio here.

Let #n# represent the mols of #"H"_2(g)# or #"I"_2(g)# in unit volume (#"1 L"#). Numerically, the mols will be equal to the concentrations here, because molarity is the mols divided by #1# #"L"#. We assume the container volume is constant, and we start with no #"HI"(g)# at all.

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 #x# in terms of #n#:

#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 #"I"_2# converted is the amount the #"I"_2# quantity decreased by (#x#), divided by its starting amount (#n#):

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