MCQ
Pure $'N_2'$ obtain by
- A$N{H_4}N{O_2}\,\xrightarrow{\Delta }$
- B$N{H_4}N{O_3}\,\xrightarrow{\Delta }$
- C$(NH_4)_2Cr_2O_7\,\xrightarrow{\Delta }$
- ✓$NaN_3\,\xrightarrow{\Delta }$
$2 \mathrm{NaN}_{3} \stackrel{573 \mathrm{K}}{\longrightarrow} 2 \mathrm{Na}+3 \mathrm{N}_{2}$
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