Give the balanced reaction for the reduction of hot copper(II)oxide to copper using ammonia gas.
(A) $CuO + N{H_3} \to 3Cu + {N_2} + 3{H_2}O$
(B) $3CuO + 4N{H_3} \to 3Cu + 2{N_2} + 3{H_2}O$
(C) $CuO + 2N{H_3} \to 3Cu + {N_2} + 3{H_2}O$
(D) $3CuO + 2N{H_3} \to 3Cu + {N_2} + 3{H_2}O$

VerifiedVerified
140.7k+ views
Hint: A balanced chemical equation occurs when the number of atoms involved in the reactants side is equal to the number of atoms in the product side.

Complete step by step solution:
Firstly, see the number of molecules of elements on either side in the equation. To find out the number of molecules, see the number of each molecule and then start balancing them.
While balancing the element cross check that what are the changes being made into other elements as a result.
So, the balanced chemical equation for the reduction of hot copper(II) oxide to copper using ammonia gas is:
$3CuO + 2N{H_3} \to 3Cu + {N_2} + 3{H_2}O$
Ammonia reacts with copper oxide to produce nitrogen, copper and water. The reaction proceed at a temperature of $500 - {550^ \circ }C$
Catalytic oxidation of ammonia occurs when ammonia and oxygen are combined in the presence of high heat and a catalytic metal. Ammonia is then oxidized into nitric oxide. In a separate container, oxygen continues oxidizing the nitric oxide into nitrogen dioxide gas, which is adsorbed by water.

Hence, option D is correct.

Note: Exothermic reactions give products which have less energy than reactants. When the energy of a compound is low, stability increases. Moreover, CuO decomposes to release oxygen when heated and serves as an oxidizer in reactive composites and chemical looping combustion.