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

Correct answer is C

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

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