Question

How do you balance \[NaCl{O_3} \to NaCl + {O_2}\]?

Answer 1

Hint: A balanced chemical equation, the total number of atoms of each element in the reactants, on the left-hand side of the equation is the same as the number atoms in the products framed on the right-hand side of the equation
The complete mass of the reactants is equivalent to the absolute mass of the products.
The number of atoms of each element, before the reaction and after the reaction is equivalent.

Complete step by step answer:
To balance the reaction, in the event that it is occurring at the solid state (decomposition):
\[2NaCl{O_3} \to 2NaCl\left( s \right) + 3{O_2}\left( g \right)\]
Be that as it may, to balance the reaction, on the off chance that it is occurring in an aqueous medium:
\[NaCl{O_3} \to NaCl\left( s \right) + {O_2}\left( g \right)\]
We ought to perceive that this is a reduction half equation since the oxidation number of oxygen is increasing from (\[ - 2\]) in \[NaCl{O_3}\] to (\[0\]) in \[{O_2}\] .
The oxidation numbers for\[Na\]and \[Cl\]are not changing and they are (\[ + 1\]) and (\[ - 1\]) separately.
\[NaCl{O_3}\left( {aq} \right) + 2{H^ + }\left( {aq} \right) + 2{e^ - } \to NaCl\left( s \right) + {O_2}\left( g \right) + {H_2}O\left( l \right)\]
Thusly, all together for this reaction to happen, you will require a reducing agent that will reduce the oxidizing agent \[NaCl{O_3}\] , for example, model \[N{a_2}S{O_3}\] .
The oxidation half condition will at that point be:
\[S{O_3}^{2 - }\left( {aq} \right) + {H_2}O\left( l \right) \to S{O_4}^{2 - }\left( {aq} \right) + 2{H^ + }\left( {aq} \right) + 2{e^ - }\]
The redox reaction will in this manner be
Oxidation: \[S{O_3}^{2 - }\left( {aq} \right) + {H_2}O\left( l \right) \to S{O_4}^{2 - }\left( {aq} \right) + 2{H^ + }\left( {aq} \right) + 2{e^ - }\]
Reduction: \[NaCl{O_3}\left( {aq} \right) + 2{H^ + }\left( {aq} \right) + 2{e^ - } \to NaCl\left( s \right) + {O_2}\left( g \right) + {H_2}O\left( l \right)\]
Redox:
\[NaCl{O_3} + N{a_2}S{O_3} \to NaCl\left( s \right) + {O_2}\left( g \right) + N{a_2}S{O_4}\]\[NaCl{O_3} + S{O_3}^{2 - }\left( {aq} \right) \to NaCl\left( s \right) + {O_2}\left( g \right) + S{O_4}^{2 - }\left( {aq} \right)\]
or
\[NaCl{O_3} + N{a_2}S{O_3} \to NaCl\left( s \right) + {O_2}\left( g \right) + N{a_2}S{O_4}\]

Note: Each chemical equation sticks to the law of conservation of mass, which expresses that matter can't be made or destroyed. In this manner, there should be a similar number of atoms of every element on each side of a chemical equation.
Use coefficients of products and reactants to adjust the number of atoms of an element on the two sides of a chemical equation.
At the point when an equivalent number of atoms of an element is available on the both sides of a chemical equation, the equation is balanced.
Law of conservation of mass Matter can't be created nor be destroyed. Hence, in a closed system, the mass of the reactants should approach the mass of the products