Question on: JAMB Chemistry - 1990
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 β PbS + 2HNO3 the percentage by volume of hydrogen sulphide in the air is?
(Pb = 207, S = 32, GMV at s.t.p = 22.4dm3)
Pb(NO3)2 + H2S β 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
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Correct Option: C
Pb(NO3)2 + H2S β PbS + 2HNO3
34 g H2S β 239 g pbs
5.02g pbs β (34)/(237g) Ο 5.02 g = 0.714 g
34g β 22.4 dm3
=0.714 β (22.4)/34g) Ο 100 = 4.70
34 g H2S β 239 g pbs
5.02g pbs β (34)/(237g) Ο 5.02 g = 0.714 g
34g β 22.4 dm3
=0.714 β (22.4)/34g) Ο 100 = 4.70
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