# 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#?

##### 1 Answer

I'm assuming you want the volume of

(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** **volume will dominate the volume at STP**.

Since the mol ratio of

Therefore, to calculate the mols of

#"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,

we started with

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