A sulphate of a metal (A) on heating evolves two gases (B) and (C) and an oxide (D). Gas (B) turns K — Qualitative and Quantitative Analysis Chemistry Question
Question
A sulphate of a metal (A) on heating evolves two gases (B) and (C) and an oxide (D). Gas (B) turns K2Cr2O7 paper green while gas (C) forms a trimer in which there is no S–S bond. Compound (D) with conc. HCl forms a Lewis acid (E) which exists in a dimer. Compounds (A), (B), (C), (D) and (E) are respectively:
Answer: A
💡 Solution & Explanation
2FeSO4 (A) -Delta-> Fe2O3 (D) + SO2 (B) + SO3 (C). SO2 turns dichromate green. SO3 forms a trimer S3O9. Fe2O3 + 6HCl -> 2FeCl3 (E) + 3H2O. FeCl3 is a Lewis acid existing as a dimer Fe2Cl6.
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