Question

How to balance $Zn + HN{O_3} \to Zn{\left( {N{O_3}} \right)_2} + N{H_4}N{O_3} + {H_2}O$?

Answer 1

As we know that here the zinc is reacted with nitric acid and there is a formation of zinc nitrate with ammonium nitrate and water. According to the law of conservation of mass, the mass of the reactant must be balanced with the mass of the product. Hence, when balancing a chemical equation, the atoms of both molecules as well as elements on the reactant side should be equal to the elements and molecules present in the product side.

Complete answer:
First we need to write the oxidation number of all the atoms.

The given reaction is a redox reaction, which means it undergoes both oxidation and reduction. Hence, identify the reactants which are oxidized and reduced.

Here, the oxidation state of zinc and N is changed. The oxidation number of zinc is changed by unit two per zinc atom and the oxidation number of nitrogen is changed by unit eight per nitric acid molecule.
Multiply the equation one with four and it is added with equation two.
$4Zn + HN{O_3} \to 4Zn{\left( {N{O_3}} \right)_2} + N{H_4}N{O_3}$
Nine molecules of nitric acid are added to LHS to balance the number of nitrogen. And to balance the number of hydrogen and oxygen, we need to add three molecules of water on RHS. Hence, the balanced equation is,
$4Zn + 10HN{O_3} \to 4Zn{\left( {N{O_3}} \right)_2} + N{H_4}N{O_3} + 3{H_2}O$

Note:
There are some rules that should be followed when we are balancing the chemical equation. First we have to write the unbalanced equation and start to balance the chemical equation by applying the law of conservation mass. Which means, the same number of atoms should be present in each side of the reaction. Accordingly we have to add the elements until the number of molecules equal on both LHS and RHS.