$\left(\mathrm{NH}_{4}ight)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ on heating liberates a gas. The same gas…

$\left(\mathrm{NH}_{4}ight)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ on heating liberates a gas. The same gas will be obtained by
  1. Heating $\mathrm{NH}_{4} \mathrm{NO}_{2}$
  2. Heating $\mathrm{NH}_{4} \mathrm{NO}_{3}$
  3. Treating $\mathrm{H}_{2} \mathrm{O}_{2}$ with $\mathrm{NaNO}_{2}$
  4. Treating $\mathrm{Mg}_{3} \mathrm{~N}_{2}$ with $\mathrm{H}_{2} \mathrm{O}$

Solution

$\left(\mathrm{NH}_{4}ight)_{2} \mathrm{C}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2} \uparrow+\mathrm{C}_{2} \mathrm{O}_{3}+4 \mathrm{H}_{2} \mathrm{O}$
$\mathrm{NH}_{4} \mathrm{NO}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2} \uparrow+2 \mathrm{H}_{2} \mathrm{O}$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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