# 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?

Aug 31, 2014

The concentration H₂S is 45 % (v/v).

Step 1. Write the balanced chemical equation.

2H₂S + O₂ → 2H₂O + 2S

Step 2. Convert grams of S → moles of S → moles of H₂S

3.2 g S × $\left(1 \text{mol S")/(32.06"g S") × (2"mol H"_2"S")/(2"mol S}\right)$ = 0.0998 mol H₂S (2 significant figures + 1 guard digit)

Step 3. Use the Ideal Gas Law to calculate the volume of H₂S at STP.

$P V = n R T$

$V = \frac{n R T}{P} = \left(0.0998 \text{mol" × 8.314 "kPa·L·K"^-1"mol"^-1 × 273.15"K")/(100"kPa}\right)$ = 2.27 L

Step 3. Calculate the concentration of the H₂S.

Percent by volume = ("volume of H"_2"S")/"volume of air" × 100 % = $\left(2.27 \text{L")/(5.0"L}\right)$ × 100 % = 45 % (v/v)

Note: The definition of Standard Pressure was changed to 100 kPa in 1982.