Reaction of zinc with cold and very dilute nitric acid yields
- $\mathrm{Zn}\left(\mathrm{NO}_{3}ight)_{2}+\mathrm{N}_{2} \mathrm{O}$
- $\mathrm{Zn}\left(\mathrm{NO}_{3}ight)_{2}+\mathrm{NO}$
- $\mathrm{Zn}\left(\mathrm{NO}_{3}ight)_{2}+\mathrm{NH}_{4} \mathrm{NO}_{3}$
- $\mathrm{Zn}\left(\mathrm{NO}_{3}ight)_{2}+\mathrm{NO}_{2}$
Solution
$4 \mathrm{Zn}+10 \mathrm{HNO}_{3} \longrightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_{3}ight)_{2}+ \mathrm{NH}_{4} \mathrm{NO}_{3}+\mathrm{H}_{2} \mathrm{O}$
Asked in: JEE-TOPICTESTS-CHEMISTRY
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