Preparation of aniline from nitrobenzene

(i) \(3Fe +6HCl \to 3FeCl_{2}+6H\)
(ii) \(C_{6}H_{5}NO_{2}+6H\to C_{6}H_{5}NH_{2}+2H_{2}O\)
(iii) \(FeCl_{2}+2H_{2}O\to Fe(OH)_{2}+2HCl\)
(iv) \(C_{6}H_{5}NO_{2}+2Fe(OH)_{2}+4H_{2}O\to C_{6}H_{5}NH_{2}+Fe(OH)_{3}\)
The reduction of nitrobenzene using \(Fe/HCl\) produces aniline, indicating that the nitro group is converted to an \(NH_2\) group.

So, the correct option is (B): nitrobenzene.