How do you balance #CO(g) + I_2O_5(s) -> I_2(s) + CO_2(g)#?

You're actually dealing with a redox reaction used to convert carbon monoxide, #"CO"#, to carbon dioxide, #"CO"_2#, by using what is known as the Schutze reagent, which is iodine pentoxide, #"I"_2"O"_5#, on silica gel.

Iodine pentoxide acts as an oxidizing agent here and oxidizes carbon monoxide to carbon dioxide while being reduced to iodine in the process.

Assign oxidation numbers to the atoms that take part in the reaction - for the sake of simplicity, I will not add the states of the chemical species involved in the reaction

#stackrel(color(blue)(+2))("C") stackrel(color(blue)(-2))("O") + stackrel(color(blue)(+5))("I")_2 stackrel(color(blue)(-2))("O")_5 -> stackrel(color(blue)(0))("I")_2 + stackrel(color(blue)(+4))("C")stackrel(color(blue)(-2))("O")_2#

You can balance this equation by assuming that the reaction takes place in acidic conditions (you will get the same result if you assume basic conditions).

So, carbon's oxidation state goes from #color(blue)(+2)# on the reactants' side, to #color(blue)(+4)# on the products' side, which of course means that it's being oxidized.

The oxidation half-equation will look like this

#stackrel(color(blue)(+2))("C")"O" -> stackrel(color(blue)(+4))("C")"O"_2 + 2"e"^(-)#

Here each carbon atoms loses two electrons. Balance the oxygen atoms by adding water molecules to the side that needs oxygen, and the hydrogen atoms by adding protons, #"H"^(+)#, to the side that needs hydrogen.

#"H"_2"O" + stackrel(color(blue)(+2))("C")"O" -> stackrel(color(blue)(+4))("C")"O"_2 + 2"e"^(-) + 2"H"^(+)#

The oxidation state of iodine goes from #color(blue)(+5)# on the reactant's side, to #color(blue)(0)# on the products' side, which means that it's being reduced.

The reduction half-reaction will be

#stackrel(color(blue)(+5))("I")_2"O"_5 + 10"e"^(-) -> stackrel(color(blue)(0))"I"_2#

Here each iodine atom gains five electrons, so two atoms will gain a total of ten electrons. Once again, balance the oxygen and hydrogen to get

#10 "H"^(+) + stackrel(color(blue)(+5))("I")_2"O"_5 + 10"e"^(-) -> stackrel(color(blue)(0))"I"_2 + 5"H"_2"O"#

As you know, in any redox reaction, the number of electrons lost in the oxidation half-equation must be equal to the number of electrons gained in the reduction half-equation.

In order to get these two balanced, multiply the oxidation half-equation by #5#. Add the two half-equations to get

#{ ("H"_2"O" + stackrel(color(blue)(+2))("C")"O" -> stackrel(color(blue)(+4))("C")"O"_2 + 2"e"^(-) + 2"H"^(+) | xx 5), (10 "H"^(+) + stackrel(color(blue)(+5))("I")_2"O"_5 + 10"e"^(-) -> stackrel(color(blue)(0))"I"_2 + 5"H"_2"O") :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaa)#

#color(red)(cancel(color(black)(5"H"_2"O"))) + 5"CO" + color(red)(cancel(color(black)(10"H"^(+)))) + "I"_2"O"_5 + color(red)(cancel(color(black)(10"e"^(-)))) -> 5"CO"_2 + color(red)(cancel(color(black)(10"e"^(-)))) + "I"_2 + color(red)(cancel(color(black)(10"H"^(+)))) + color(red)(cancel(color(black)(5"H"_2"O")))#

The balanced chemical equation for this reaction will thus be - state symbols included!

#color(green)(|bar(ul(color(white)(a/a)color(black)(5"CO"_ ((g)) + "I"_ 2"O"_ (5(s)) -> "I"_ (2(s)) + 5"CO"_ (2(g)))color(white)(a/a)|)))#

As you can see, the reaction must be balanced in neutral conditions, so the result will be the same regardless if you pick acidic or basic conditions.