The trick here is to realize that because you know the **equilibrium concentration** of sulfur trioxide, #"SO"_3#, you practically know the value of #x#, so you don't *need* to use #x# anymore.

In fact, we only use #x# because we don't know the equilibrium concentration of one or more chemical species that take part in the reaction.

So, your equilibrium reaction looks like this

#color(red)(2)"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(red)(2)"SO"_ (3(g))#

Now, you know that the **equilibrium concentration** of #"SO"_3# is equal to #"0.04 M"#.

Assuming that you started with #"0 M"# of #"SO"_3# in the reaction vessel, you can say that the reaction produced **twice as many** moles of #"SO"_3# than moles of #"O"_2# consumed #-># #"O"_2# and #"SO"_3# are in a #1:color(red)(2)# **mole ratio**.

So if the reaction produced #"0.04 M"# of #"SO"_3#, it follows that it must have consumed #"0.02 M"# of #"O"_2#.

Likewise, the reaction produced #"0.04 M"# of #"SO"_3#, so it must have consumed #"0.04 M"# of #"SO"_2# #-># #"SO"_2# and #"SO"_3# are in a #color(red)(2):color(red)(2)# **mole ratio**.

Therefore, the **ICE table** would look like this

#" " color(red)(2)"SO"_ (2(g)) " "+" " "O"_ (2(g)) " "rightleftharpoons " " color(red)(2)"SO"_ (3(g))#

#color(purple)("I")color(white)(aaaaaacolor(black)(0.06)aaaaaaaacolor(black)(0.05)aaaaaaaaaaacolor(black)(0)#

#color(purple)("C")color(white)(aaaacolor(black)(-0.04)aaaaaacolor(black)(-0.02)aaaaaaaaacolor(black)(+0.04)#

#color(purple)("E")color(white)(aaaaacolor(black)(0.02)aaaaaaaacolor(black)(0.03)aaaaaaaaaaacolor(black)(0.04)#

This is **exactly** what you get if you use #x# to represent the number of moles of oxygen gas that are consumed by the reaction

#" " color(red)(2)"SO"_ (2(g)) " "+" " "O"_ (2(g)) " "rightleftharpoons " " color(red)(2)"SO"_ (3(g))#

#color(purple)("I")color(white)(aaaaaacolor(black)(0.06)aaaaaaaacolor(black)(0.05)aaaaaaaaaaacolor(black)(0)#

#color(purple)("C")color(white)(aaaaacolor(black)(-color(red)(2)x)aaaaaaaacolor(black)(-x)aaaaaaaaaacolor(black)(+color(red)(2)x)#

#color(purple)("E")color(white)(aaacolor(black)(0.06-color(red)(2)x)aaaaacolor(black)(0.05-x)aaaaaaaacolor(black)(color(red)(2)x)#

But since the equilibrium concentration of #"SO"_3# is #"0.04 M"#, it follows that

#color(red)(2)x = "0.04 M" implies x = "0.02 M"#

Once again, the ICE table takes the form given to you by your teacher.

So remember, don't get too hung up on #x#, that's just a way we have of figuring out the equilibrium concentrations of the chemical species when we *don't know what they are*.

When you **do** know what they are, there's no need to use #x#, just use the stoichiometric coefficients of the reaction, i.e. the **mole ratios**.