A) \[\text{CaS}{{\text{O}}_{\text{4}}}\text{+}\,\text{C}\xrightarrow[{}]{\Delta }\]
B) \[\text{F}{{\text{e}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}\xrightarrow[{}]{}\]
C) \[\text{S}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\xrightarrow[{}]{}\]
D) \[\,{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}+\text{PC}{{\text{l}}_{\text{5}}}\xrightarrow[{}]{}\]
Correct Answer: B
Solution :
\[CaS{{O}_{4}}+C\xrightarrow[{}]{\Delta }CaO+S{{O}_{2}}+CO\] \[F{{e}_{2}}{{(S{{O}_{4}})}_{3}}\xrightarrow[{}]{\Delta }F{{e}_{2}}{{O}_{3}}+3S{{O}_{3}}\] \[S+2{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{\Delta }3S{{O}_{2}}+2{{H}_{2}}O\] \[{{H}_{2}}S{{O}_{4}}+PC{{l}_{s}}\xrightarrow[{}]{\Delta }\underset{Chloro\,sulphonic\,acid}{\mathop{S{{O}_{3}}HCl}}\,\] \[+POC{{l}_{3}}+HCl\] Thus, \[S{{O}_{3}}\]is obtained by heating \[F{{e}_{2}}{{(S{{O}_{4}})}_{3}}.\]
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