10.0 dm3 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 weigh 5.02 g. According to the equation:
Pb(NO3)2 + H2S → PbS + 2HNO3 the percentage by volume of hydrogen sulphide in the air is?
(Pb = 207, S = 32, GMV at s.t.p = 22.4dm3)
- 50.2
- 47.0
-
4.70
- 0.47
The correct answer is: C
Explanation
Pb(NO3)2 + H2S → PbS + 2HNO334 g H2S → 239 g pbs
5.02g pbs → (34)/(237g) χ 5.02 g = 0.714 g
34g → 22.4 dm3
=0.714 → (22.4)/34g) χ 100 = 4.70