A 5.0-L air sample containing H2S at STP is treated with a catalyst to promote the reaction,
H2S + O2 = H2O + S(solid). If 3.2 g of solid S was collected, what is the volume percentage of H2S in the original sample?

Question

Noah Aldrich · Accepted Answer

45 percent (v/v) of H2O is present.

First, write the chemical equation that is balanced.

2H₂S → 2H₂O + 2S via O₂

Convert grams of S to moles of S to moles of H₂S in step two.

3.2 g S × #(1"mol S")/(32.06"g S") × (2"mol H"_2"S")/(2"mol S")# = 0.0998 mol H₂S (2 significant figures + 1 guard digit)

Step 3: Determine the volume of H2S at STP using the Ideal Gas Law.

#PV = nRT#

#V = (nRT)/P = (0.0998"mol" × 8.314 "kPa·L·K"^-1"mol"^-1 × 273.15"K")/(100"kPa")# = 2.27 L

Step 3: Determine the H2S concentration.

Percent by volume = #("volume of H"_2"S")/"volume of air"# × 100 % = #(2.27"L")/(5.0"L")# × 100 % = 45 % (v/v)

Notably, in 1982, the definition of Standard Pressure was modified to 100 kPa.

Mateo Aguilar · Answer

undefined The molar mass of sulfur (S) is 32.06 g/mol. From the reaction, 1 mole of H2S produces 1 mole of S. Therefore, 3.2 g of S corresponds to 0.1 moles. According to the ideal gas law, at STP, 1 mole of any gas occupies 22.4 L. So, the volume of 0.1 moles of H2S is 2.24 L. The initial volume of the air sample is 5.0 L. Thus, the volume percentage of H2S in the original sample is $ \frac{2.24}{5.0} \times 100 $, which is 44.8%.

A 5.0-L air sample containing H2S at STP is treated with a catalyst to promote the reaction, H2S + O2 = H2O + S(solid). If 3.2 g of solid S was collected, what is the volume percentage of H2S in the original sample?