How do you find the number of molecules in #"2 L CO"_2#?

You can either use the density of #"CO"_2# at STP or assume that #"CO"_2# is an ideal gas.

USING THE DENSITY

This is the more accurate way, since #"CO"_2# is not quite ideal. Wikipedia gives the density as #"1.977 g/L"# at STP. We can use the molar mass and density to get the number of mols and hence the number of molecules.

#2 cancel("L CO"_2) xx ("1.977 g CO"_2)/cancel("L CO"_2) = "3.954 g CO"_2#

Therefore, the mols are:

#3.954 cancel("g CO"_2) xx "1 mol CO"_2/(44.009 cancel("g CO"_2)) = "0.0898 mols CO"_2#

(How can you get the molar mass of #"CO"_2#?)

Therefore, the number of molecules is:

#0.0898 cancel("mols CO"_2) xx (6.0221413 xx 10^(23) "anything ever")/cancel("1 mol anything ever")#

#= color(blue)(5.41 xx 10^(22))# #color(blue)("molecules CO"_2)#

USING THE IDEAL GAS LAW

Of course, we could also use the ideal gas law by assuming #"CO"_2# is ideal. In that case, we would assume that it has a molar volume of #"22.414 L/mol"# at #0^@ "C"# and #"1 atm"#. Therefore:

#2 cancel("L CO"_2) xx cancel"1 mol ideal gas at STP"/(22.414 cancel"L")#

#= "0.0892 mols CO"_2#

Therefore, the number of molecules is:

#0.0892 cancel("mols CO"_2) xx (6.0221413 xx 10^(23) "anything ever")/cancel("1 mol anything ever")#

#= color(blue)(5.37 xx 10^(22))# #color(blue)("molecules CO"_2)#

There is a small error (about #0.7%#), but not a bad one. It's a rather good assumption that #"CO"_2# is ideal at STP. At higher pressures and lower temperatures though, the approximation fails (why?).