I got
This is just asking you to:
Since the volume of the container is shared, we can use just the mols of everything instead of the . The first equilibrium constant will stay the same, and thus the starting equilibrium configuration has
$K_c = (["CO"_2]["H"_2])/(["CO"]["H"_2"O"]) = (0.400^2)/(0.200^2) = 4$
We want to add
Your initial ICE table should look like this (after reaching the first equilibrium configuration already, and adding the
$"CO"(g) " "+" " "H"_2"O"(g) " "rightleftharpoons" " "CO"_2(g) + "H"_2(g)$
$"I"" "0.200" "" "" "0.200" "" "" "" "0.400 + y" "" "0.400$
$"C"" "+x" "" "" "+x" "" "" "" "" "" "-x" "" "-x$
$"E"" "0.200 + x" "0.200 + x" "0.400 + y - x" "0.400 - x$
Now, we want the final mols of
$n_(CO) = 0.200 + x = 0.300$
Thus,
$((0.300 + y)(0.300))/((0.300)(0.300)) = 4$
Solve for the amount of
$(4(0.300^2))/(0.300) = 0.300 + y$
$=> color(blue)(y) = (4(0.300^2))/(0.300) - 0.300 = color(blue)("0.900 mols CO"_2 " added")$