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#?

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#?

Ellie Andrews · Accepted Answer

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 by volume you mean total volume, then the volume of the liquid water is not being considered.)

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

Throughout, we will assume that the gases are perfect.

Your response 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#.

Cameron Andrews · Answer

undefined To find the volume of gas produced at STP, we first need to calculate the moles of each reactant using the ideal gas law. Then, we determine the limiting reactant and use stoichiometry to find the volume of gas produced.

Given:
- $2\, \text{H}_2\text{S(g)} + 3\, \text{O}_2\text{(g)} \rightarrow 2\, \text{SO}_2\text{(g)} + 2\, \text{H}_2\text{O(l)}$
- Initial volume of $H_2S(g) = 200\, \text{mL}$
- Initial volume of $O_2(g) = 200\, \text{mL}$

Using the ideal gas law: $PV = nRT$
$n = \frac{PV}{RT}$

For $H_2S(g)$:
$n_{\text{H}_2S} = \frac{(200\, \text{mL})(1\, \text{atm})}{(0.0821\, \text{atm}\cdot \text{L/mol}\cdot \text{K})(273\, \text{K})}$
$n_{\text{H}_2S} ≈ 7.61\, \text{mol}$

For $O_2(g)$:
$n_{\text{O}_2} = \frac{(200\, \text{mL})(1\, \text{atm})}{(0.0821\, \text{atm}\cdot \text{L/mol}\cdot \text{K})(273\, \text{K})}$
$n_{\text{O}_2} ≈ 7.61\, \text{mol}$

The stoichiometric ratio of $H_2S$ to $O_2$ is 2:3. Since both $H_2S$ and $O_2$ produce the same number of moles of $SO_2$, $H_2S$ is the limiting reactant.

Given that 2 moles of $H_2S$ produce 2 moles of $SO_2$, the volume of $SO_2$ produced is the same as the volume of $H_2S$ consumed.

Therefore, the volume of $SO_2$ produced is $200\, \text{mL}$.

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

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#?