10.0dm3 of air containing H2S as an impurity was passed through a solution of Pb(NO3)2 until all the H2S had reacted. The precipitate of PbS was found to weight 5.02g. According to the equation: Pb(NO3)2 + H2S→PbS+2HNO3 the percentage by volume of hydrogen sulphide in the air is

10.0dm3 of air containing H2S as an impurity was passed through a solution of Pb(NO3)2 until all the H2S had reacted. The precipitate of PbS was found to weight 5.02g. According to the equation:

Pb(NO3)2 + H2S→PbS+2HNO3 the percentage by volume of hydrogen sulphide in the air is

A. 50.2

B. 47.0

C. 4.70

D. 0.47

[Pb(NO3)2=207, S=32, GMV at s.t.p.=22.4dm3]

The correct is C. 4.70

Calculation:

First, find the amount of PbS formed.

Molar mass of PbS
= Pb + S = 207 + 32 = 239 g mol⁻¹

Moles of PbS formed
5.02/{239 = 0.021 mol

From the equation:
Pb{NO}3)2 + {H2S}→{PbS} + 2{HNO}3

The mole ratio of H₂S : PbS = 1 : 1, so:

Moles of H₂S = 0.021 mol

Volume of H₂S at STP

Volume = 0.021 * 22.4 = 0.470 dm^3

Percentage by volume of H₂S in air

{0.470}/{10.0} * 100 = 4.70%

✅ Correct Answer: C. 4.70

NUMBER THIRTEEN