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)

Logicalugo

2 years ago

To find the percentage by volume of hydrogen sulfide in the air, we need to first calculate the volume of H2S that reacted.

1. Calculate the molar mass of PbS:
Molar mass of Pb = 207 g/mol
Molar mass of S = 32 g/mol
Molar mass of PbS = 207 + 32 = 239 g/mol

2. Calculate the number of moles of PbS formed:
Mass of PbS formed = 5.02 g
Number of moles of PbS = Mass / Molar mass = 5.02 g / 239 g/mol = 0.021 moles

3. According to the balanced chemical equation, 1 mole of PbS is formed from 1 mole of H2S. So, 0.021 moles of PbS were formed from 0.021 moles of H2S.

4. Calculate the volume of H2S at STP (standard temperature and pressure):
1 mole of any gas at STP occupies 22.4 dm³
Volume of H2S at STP = 0.021 moles × 22.4 dm³/mol = 0.4704 dm³

5. Calculate the volume percentage of H2S in the air:
Total volume of air = 10.0 dm³
Volume percentage of H2S = (Volume of H2S / Total volume of air) × 100
= (0.4704 dm³ / 10.0 dm³) × 100
= 4.704%

So, the percentage by volume of hydrogen sulfide in the air is approximately 4.704%.

Spectrojamz

6 years ago

impeccable explanation,Thanks

cent_dave

3 months ago

please NO 39

Omahzosia

1 year ago

thanks

pablosavage76

3 years ago

that's what i also got