Question

Hydrogen sulfide gas is a flammable gas. The gas burns with

a blue flame according to the following equation:
2 H₂S (g) + 3 O2 (g) => 2 SO2 (g) + 2 H₂O(g)
5.00 L of hydrogen sulfide gas and 7.00 L of oxygen were
allowed to react.
Calculate the theoretical yield, in L (liters), of sulfur dioxide.
All the gases were at the same temperature and pressure.

Answer 1

4.67 liters of SO2 with sig figs

Step-by-step explanation:

To do this reaction, we first must find the limiting agent.
in this problem, 5 liters of H2S are reacted with 7 liters of O2, and each two moles of H2S corresponds to 3 moles of O2. Assuming ideal behavior, 1 liter of H2S has the same number of moles as 1 liter of O2 has. So, 2 liters of H2S react with 3 liters of O2. To find the limiting reagent, focus on one reagent and calculate how much of the other reagents would be needed to fully consume that one reagent.
Focus on H2S:
5 liters of H2S would require 7.5 liters of O2 (since each 2 liters of H2S requires 3 liters of O2)

Focus on O2:
7 liters of O2 would require 14/3, or 4.666... liters of H2S.

Since we don't have 7.5 liters of O2 (we only have 7 liters) to react with all 5 liters of H2S, we would say that O2 is the limiting reagent, and that we have 7 liters of it.

Next, figure out how much product can be produced per mole of that limiting reagent. 3 moles of O2 and 2 moles of H2S turns into 2 moles of SO2 and 2 moles of H2O. Thus, the ratio of O2 to SO2 is 3:2
We have 7 liters of O2, and since the ratio of O2 to SO2 is 3:2, we can set up the equation:
7:x = 3:2 where x is the number of liters of SO2
cross multiply (or solve with another method) to get:
x = 14/3, or 4.666... liters of SO2