A) \[2M\,HCl\]
B) \[6M\,N{{H}_{3}}\]
C) \[6M\,NaOH\]
D) \[{{H}_{2}}S\,gas\]
Correct Answer: B
Solution :
\[F{{e}^{3+}}+Z{{n}^{2+}}+C{{u}^{2+}}\xrightarrow[6M\,N{{H}_{3}}]{{}}\underset{brown\,ppt.}{\mathop{Fe{{(OH)}_{3}}}}\,\] \[+\underset{so\operatorname{lub}le}{\mathop{{{[Zn{{(N{{H}_{3}})}_{4}}]}^{2+}}}}\,+\underset{so\operatorname{lub}le}{\mathop{{{[Cu{{(N{{H}_{3}})}_{4}}]}^{+}}}}\,\]You need to login to perform this action.
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