Chemistry - Salt Analysis Question with Solution | TestHub

ChemistrySalt AnalysisTEST OF CATIONS Dry Test, Group Analysis, Group 1,2Medium2 minQB
ChemistryMediumnumerical
Passage / Comprehension

If group-II precipitate dissolves in yellow ammonium sulphide and the colour of the solution is yellow, this indicates the presence of ions. Ammonium thioarsenate formed on dissolution of , decomposes with dil. HCl, and a yellow precipitate of arsenic(V) sulphide is formed, which dissolves in hot concentrated nitric acid and forms arsenic acid. On adding ammonium molybdate solution to the reaction mixture and heating, a canary yellow precipitate is formed. This confirms the presence of ions.

Calculate the mass of solid product formed from dissolving 3 moles of in hot concentrated . Given molecular weight H = 1, N = 14, O = 16, S = 32 Cl = 35.5, As = 75

Answer:
480.00
Solution:

Identify the solid product: When As₂S₅ reacts with hot concentrated HNO₃, it produces soluble arsenic acid (H₃AsO₄) and solid elemental sulfur (S) as a precipitate.

Determine the molar ratio: According to the reaction stoichiometry, 1 mole of As₂S₅ produces 5 moles of solid sulfur (S).

Calculate moles of sulfur: Dissolving 3 moles of As₂S₅ yields of sulfur.

Calculate the final mass: Using the atomic mass of sulfur (S = 32), the total mass of the solid product is .

Stream:JEESubject:ChemistryTopic:Salt AnalysisSubtopic:TEST OF CATIONS Dry Test, Group Analysis, Group 1,2
2mℹ️ Source: QB

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