Question

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`?

Answer 1

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.

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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`.