question_answer

\[MS{{O}_{4}}\xrightarrow{N{{H}_{4}}OH}\underset{white}{\mathop{X\downarrow }}\,\xrightarrow[Excess]{N{{H}_{4}}OH}Y\xrightarrow{{{H}_{2}}S}Z\downarrow \]Here M and Z are

A)  \[Zn,ZnS\]        

B)  \[Al,A{{l}_{2}}{{S}_{3}}\]

C)  \[Cu,ZnS\]        

D)  \[Fe,FeS\]

Correct Answer: A

Solution :

: M is Zn.