Question

Sodium Nitrate decomposes above $ 800{}^\circ \text{ C} $ to give.
(A) $ {{\text{N}}_{2}} $
(B) $ {{\text{O}}_{2}} $
(C) $ \text{N}{{\text{O}}_{2}} $
(D) $ \text{N}{{\text{a}}_{2}}\text{O} $

Answer 1

Hint: Sodium Nitrate is an inorganic salt of alkali metal and nitrate group having chemical formula $ \text{NaN}{{\text{O}}_{3}} $. Common name of sodium Nitrate is Chile ‘saltpeter’. Decomposition of any compound refers to the stability towards the heating i.e at which temperature compound breaks down into its fragments.

Complete step by step solution:
Decomposition reaction: A decomposition reaction is a reaction that breaks down a compound into its two or simpler fragments or substances.
For general reaction as
 $ \text{AB}\to \text{A}+\text{B} $
Most of the decomposition reaction requires light, heat, electricity (Mode of temperature)
Further here we are having the compound sodium nitrate i.e $ \text{NaN}{{\text{O}}_{3}} $.
When sodium nitrate heating at low temperature it converted into $ \text{NaN}{{\text{O}}_{2}} $ by evolving $ {{\text{O}}_{2}} $ gas i.e
 $ \text{2 NaN}{{\text{O}}_{3}}\xrightarrow {\vartriangle \text{ }}2\text{NaN}{{\text{o}}_{2}}+{{\text{O}}_{2}} \uparrow $
But when sodium nitrate heated above the temperature of $ 800{}^\circ \text{ C} $ it decomposes into $ \text{N}{{\text{a}}_{2}}\text{O} $ $ {{\text{N}}_{2}} $ and by evolving of oxygen gas
 $ 2\text{ NaN}{{\text{o}}_{3}} \xrightarrow {800{}^\circ \text{ C}}\text{ N}{{\text{a}}_{2}}\text{ O }(s)+\text{ }{{\text{N}}_{2}}(g) \left( \uparrow \right)+\text{ }\dfrac{5}{2}\text{ }{{\text{O}}_{2}}(g) \uparrow $
Hence, A, B and D are correct options for this question.
Means sodium nitrate when heated above $ 800{}^\circ \text{ C} $ is breaks down into $ \text{N}{{\text{a}}_{2}}\text{O, }{{\text{N}}_{2}} $ and $ {{\text{O}}_{2}} $ .

Note:
We know that sodium nitrate is an inorganic compound of sodium, nitrogen and oxygen. It is prepared by various methods.
Some of them as follow:
It is prepared by reaction NaOH and $ \text{HN}{{\text{O}}_{3}} $
 $ \text{NaOH}+\text{HN}{{\text{O}}_{3}}\to \text{NaN}{{\text{o}}_{3}}+{{\text{H}}_{2}}\text{O} $ (Highly exothermic reaction)
It is also prepared by ammonium nitrate and sodium hydroxide.
 $ \text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}}+\text{NaOH}\to \text{N}{{\text{H}}_{4}}\text{OH}+\text{NaN}{{\text{O}}_{3}} $
It is also prepared on a large scale by sodium bicarbonate and nitric acid.
 $ \text{NaHC}{{\text{O}}_{3}}+\text{HN}{{\text{O}}_{3}}\to \text{ NaN}{{\text{O}}_{3}}+{{\text{H}}_{2}}\text{O}+\text{C}{{\text{O}}_{2}} $.