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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

A. 50.2

B. 47.0

C. 4.70

D. 0.47

QUICK ANSWER…

C

DETAILS…  

According to the equation: Pb(NO₃)₂ + H₂S → PbS + 2HNO

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:

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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 HS 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:

  1. Option A is incorrect.
  2. Option B is incorrect.
  3. C is correct.
  4. D is not the correct answer. This is the exact volume of H2S in air and not the percentage.

KEY-POINTS…

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 /

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