10.0dm³ of air containing H₂S as an impurity was passed through…
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10.0dm³ 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
According to the equation: Pb(NO₃)₂ + H₂S → PbS + 2HNO₃
1 mole of H₂S will produce 1 mole of PbS
Or 1 mole of H₂S will produce 1 molar mass of PbS = 207 + 32 = 239g
Therefore, x mole of H₂S will produce 5.02g of PbS
Cross multiplying to find x, we have:
1(5.02) = x(239)
x = 5.02/239 = 0.021mole
now, according to Avogadro’s law, the standard volume of 1 mole of any gas is = 22.4dm³
so, 1 mole of H₂S = 22.4dm³ therefore,
0.021 mole of H₂S = y
Cross multiplying to find y, we have:
1(y) = 0.021(22.4)
y = 0.021 x 22.4 = 0.47dm³
what this simply mean is that the fraction of H₂S in air = 0.47dm³ out of a total of 10dm³
so, the percentage of 0.47dm³ in 10dm³ = (0.47/10) × 100 = 4.7
Now for the right answer to the above question:
- Option A is incorrect.
- Option B is incorrect.
- C is correct.
- D is not the correct answer. This is the exact volume of H2S in air and not the percentage.
You may please note these/this:
- Always use cross multiplication to solve for the equivalents, as it makes your calculation easy and straight forward.
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/ culled from 2015 JAMB-UTME Chemistry past question 13 /