# 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?

Jun 10, 2017

I'm assuming you want the volume of ${\text{SO}}_{2}$ at STP. We will assume your definition of STP is $\text{1 atm}$ and ${25}^{\circ} \text{C}$.

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

${V}_{S {O}_{2}} = {V}_{{H}_{2} S}$, since $\text{H"_2"S}$ is the limiting reagent and is $1 : 1$ with ${\text{SO}}_{2}$.

We assume throughout that the gases are ideal.

$2 \text{H"_2"S"(g) + 3"O"_2(g) -> 2"SO"_2(g) + 2"H"_2"O} \left(l\right)$

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 $\boldsymbol{{\text{SO}}_{2} \left(g\right)}$ volume will dominate the volume at STP.

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

Therefore, to calculate the mols of ${\text{SO}}_{2}$, which ARE $1 : 1$ with $\text{H"_2"S}$, use $\text{H"_2"S}$, as it would give you the maximum yield of product:

${\text{mols H"_2"S" = "mols SO}}_{2}$

As we have been assuming all of these gases are ideal gases, which have the same molar volume, $\frac{V}{n}$, at STP, and since $\text{H"_2"S}$ is $1 : 1$ with ${\text{SO}}_{2}$...

we started with $\text{200 mL H"_2"S}$ and produce $\textcolor{b l u e}{\text{200 mL}}$ of ${\text{SO}}_{2}$.