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