Question

The complete balanced equation is:
$CuO\, + \,N{H_3}\, \to \,Cu\, + \,{N_2}\, + {H_2}O$
A.$3CuO\, + \,2N{H_3}\, \to \,3Cu\, + \,{N_2}\, + 3{H_2}O$
B.$CuO\, + \,2N{H_3}\, \to \,3Cu\, + \,2{N_2}\, + 3{H_2}O$
C.$3CuO\, + \,N{H_3}\, \to \,3Cu\, + \,{N_2}\, + 3{H_2}O$
D.None of these.

Answer 1

For completing the balancing of a chemical reaction you have to make the number of atoms on left hand side and on right hand side equal, thus by which the equation gets balanced. Now try to figure about the possible numeric value that you apply as stoichiometry before the reactants and products.

Complete step-by-step answer: The above equation formed by the reaction with ammonia, after which we will get copper alone and release nitrogen gas and water from the solution. For balancing firstly let’s check the number of atoms in the left hand and right hand side, there is one copper on the left side of the equation and the same one atom on the product side. If we count nitrogen there is one nitrogen on the left and two nitrogen atoms in the form of a molecule on the right hand side. So, for balancing the nitrogen atom, we have to multiply ammonia by $2$ let’s see what becomes.
$CuO\, + \,2N{H_3}\, \to \,Cu\, + \,{N_2}\, + {H_2}O$
Now there are two nitrogen on both sides but some atoms get disturbed and these are hydrogen and oxygen. To balance hydrogens which are $6$ on the left hand side and $2$ on the right hand side. We have to multiple the water molecule on product side by $3$ now we get the equation as,
$CuO\, + \,2N{H_3}\, \to \,Cu\, + \,{N_2}\, + 3{H_2}O$
See, if hydrogen gets balanced there are a total of $6$ hydrogens on both sides. We balanced nitrogen atoms, hydrogen atoms now we have to balance oxygen atoms. We have one oxygen on left hand side and three oxygen on right hand side, so to make them equal we have to multiply $CuO$ by $3$ after multiplying we get,
$3CuO\, + \,2N{H_3}\, \to \,Cu\, + \,{N_2}\, + 3{H_2}O$
All atoms get balanced but only copper is unbalanced. So, we have to multiply the copper atom on the right hand side by $3$ and see we get the fully balanced equation.
$3CuO\, + \,2N{H_3}\, \to \,3Cu\, + \,{N_2}\, + 3{H_2}O$

Option A is correct.

Note: To balance the equation this is the easiest way possible but different methods like oxidative reductive method and electron balancing methods are also used. These methods have their importance, but comparing each type of atom and making their number the same will help us during solving the equation and by this we don’t need the counting of electrons.