A) \[2HI+{{H}_{2}}S{{O}_{4}}\to {{I}_{2}}+S{{O}_{2}}+{{H}_{2}}O\]
B) \[Ca{{(OH)}_{2}}+{{H}_{2}}S{{O}_{4}}\to CaS{{O}_{4}}+2{{H}_{2}}O\]
C) \[NaCl+{{H}_{2}}S{{O}_{4}}\to NaHS{{O}_{4}}+HCl\]
D) \[2PC{{l}_{5}}+{{H}_{2}}S{{O}_{4}}\to 2POC{{l}_{3}}+2HCl+S{{O}_{2}}C{{l}_{2}}\]
Correct Answer: A
Solution :
\[\overset{-1}{\mathop{2Hl}}\,+\overset{+6}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,\xrightarrow{{}}\overset{0}{\mathop{{{l}_{2}}}}\,+\overset{4}{\mathop{S{{O}_{2}}}}\,+2{{H}_{2}}O\] \[\underset{\begin{smallmatrix} Oxidi\sin g \\ agent \end{smallmatrix}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,-\operatorname{Re}duced\,to\xrightarrow{\,}S{{O}_{2}}\]
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