A) Lead nitrate
B) Ammonium nitrate
C) Silver nitrate
D) Sodium nitrate
Correct Answer: B
Solution :
\[\underset{(s)}{\mathop{N{{H}_{4}}N{{O}_{3}}}}\,\,\xrightarrow{\Delta }\,2{{H}_{2}}O\,\uparrow \,+{{N}_{2}}O\,\uparrow \] \[\underset{(s)}{\mathop{NaN{{O}_{3}}}}\,\,\xrightarrow{\Delta }\,\underset{(s)}{\mathop{NaN{{O}_{2}}}}\,+{{O}_{2}}\,\uparrow \] \[\underset{\text{Lunar caustic}}{\mathop{2AgN{{O}_{3}}(s)}}\,\,\to \,2Ag(s)+2N{{O}_{2}}(g)+{{O}_{2}}(g)\] \[2Pb{{(N{{O}_{3}})}_{2}}\,\to \,\underset{(s)}{\mathop{2PbO}}\,+4N{{O}_{2}}\uparrow +{{O}_{2}}\uparrow \]
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