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)
A. 50.2 B. 47.0 C. 4.70 D. 0.47
Correct Answer: C
Explanation
Pb(NO3)2 + H2S → PbS + 2HNO3 34 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
