A) \[Zn+dil.\,{{H}_{2}}S{{O}_{4}}\]
B) \[Al+Conc.\,NaOH\]
C) \[Al+dil.\,\,HCl\]
D) None of these
Correct Answer: B
Solution :
\[SO_{3}^{2-}+2Al+2O{{H}^{-}}+3{{H}_{2}}O\xrightarrow{{}}{{S}^{2-}}+2{{[Al{{(OH)}_{4}}]}^{-}}\]\[{{H}_{2}}S\] does not evolve as liberated \[{{H}_{2}}S\] is neutralized by \[NaOH\] and \[N{{a}_{2}}S\] is formed.
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