What volume is produced at STP in the reaction #2"H"_2"S"(g) + 3"O"_2(g) -> 2"SO"_2(g) + 2"H"_2"O"(l)# if we start with #"200 mL H"_2"S"# and #"200 mL O"_2#?

I'm assuming you want the volume of #"SO"_2# at STP. We will assume your definition of STP is #"1 atm"# and #25^@ "C"#.

(If you mean the total volume, then we are ignoring the volume of the liquid water.)

#V_(SO_2) = V_(H_2S)#, since #"H"_2"S"# is the limiting reagent and is #1:1# with #"SO"_2#.

We assume throughout that the gases are ideal.

Your reaction was:

#2"H"_2"S"(g) + 3"O"_2(g) -> 2"SO"_2(g) + 2"H"_2"O"(l)#

Since the reduction of oxygen is a known standard reduction half reaction, it can be confirmed that the water is a liquid. Liquids have small molar volumes, so to a good approximation, the #bb("SO"_2(g))# volume will dominate the volume at STP.

Since the mol ratio of #"H"_2"S":"O"_2# is #2:3#, your reactant gas volumes are #1:1#, and this mol ratio is less than #1:1#, #"O"_2# is in excess and #"H"_2"S"# is the limiting reagent. For example, if you had #"2 mols H"_2"S"#, you would only need #"2 mols O"_2#, and you'd actually have #3#.

Therefore, to calculate the mols of #"SO"_2#, which ARE #1:1# with #"H"_2"S"#, use #"H"_2"S"#, as it would give you the maximum yield of product:

#"mols H"_2"S" = "mols SO"_2#

As we have been assuming all of these gases are ideal gases, which have the same molar volume, #V/n#, at STP, and since #"H"_2"S"# is #1:1# with #"SO"_2#...

we started with #"200 mL H"_2"S"# and produce #color(blue)("200 mL")# of #"SO"_2#.