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The hydrogen sulfide in 80.0 gram crude petroleum was removed by distillation and uncollected in a solution of CdCl2. The precipitated CdS was filtered, washed, and ignited to CdSO4. Calculate the percentage of H2S in the sample if 0.125 g of CdSO4 was recovered

Sagot :

barramedasm

Answer:

The percentage of H₂S in the sample is 0.0255%.

Explanation:

This problem is an example of gravimetry. Gravimetry is one of the basic analytical methods where you study the composition of raw materials or substances. It deals with the quantitative determination of an analyte based on the mass of the solid compositions in a sample.

In solving gravimetric problems, stoichiometry is vital. For us to use stoichiometry, we first need a balanced equation, then we proceed to dimensional analysis.

The balanced chemical equation for the problem is:

CdSO₄ + H₂S  → CdS + H₂SO₄

Getting the molar mass of the compounds needed, we have:

Molar mass (MM) of CdSO₄ = 208.472 g/mol

MM of H₂S = 34.081 g/mol

Solving for the mass of H₂S in the sample using stoichiometry:

0.125 g CdSO₄ × [tex](\frac{1 mol CdSO4}{208.472 g}) x (\frac{1 mol H2S}{1 mol CdSO4}) x (\frac{34.081 g H2S}{1 mol})[/tex]

= 0.020435 grams H₂S

From the dimensional analysis, we can see that the units will cancel until the unit grams H₂S will remain.

Solving for percentage of H₂S in the sample:

  • Mass of sample = 80 g
  • Mass of H₂S in the sample = 0.020435 g
  • % H₂S = [tex]\frac{mass H2S}{mass sample}[/tex] × 100%
  • % H₂S = [tex]\frac{0.020435 g H2S}{80 g sample}[/tex] × 100%
  • % H₂S = 0.0255%

Therefore, the percentage of H₂S in the sample is 0.0255%.

Read more about balancing equations here: https://brainly.ph/question/543874

Learn more about stoichiometry here:

https://brainly.ph/question/443932

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