Question

How do you balance ${C_3}{H_5}{(N{O_3})_3} \to C{O_2} + {H_2}O + {N_2} + {O_2}$?

Answer 1

We all know that we can balance a chemical equation by simply adding stoichiometric coefficients to the reactant and the products and we are also aware with the law of conservation of mass and constant proportions.

Complete step-by-step answer:
As we know that a chemical equation actually obeys the law of conservation of mass and law of constant proportions where the same number of atoms of a particular element must exist on the reactant and products sides.

Now talking about the given chemical equation:
${C_3}{H_5}{(N{O_3})_3} \to C{O_2} + {H_2}O + {N_2} + {O_2}$

We can see that there are three carbon atoms on the reactant side and only one carbon on the product side, similarly, three nitrogen atoms are present on the reactant side and two on the product side, nine oxygen atoms are present on the reactant side and five on the product side as well as five hydrogen are present on reactant side and only two on the product side.

Now, this is a bit tricky as there is a single reactant only but the products are multiple. So, we will first balance the product side and then add the stoichiometric coefficient on the reactant side of the equation to balance the number of atoms.

Let us first balance the hydrogen atoms as they are simpler to balance, so we see that there are five hydrogen on reactant side and two hydrogen side so we will add $5$ as stoichiometric coefficient on the product side which results in a total of ten hydrogen atoms on product side and we can now balance them with adding $2$ as stoichiometric coefficient on the reactant side.
So now we have on the reactant side:
$C = 6 \\\ H = 10 \\\ N = 6 \\\ O = 18 \\\$
And our equation becomes:
$2{C_3}{H_5}{(N{O_3})_3} \to C{O_2} + 5{H_2}O + {N_2} + {O_2}$

Now, we can balance the rest of the atoms. First the carbon atoms, we see that there are six carbon on the reactant side so we will add $6$ as stoichiometric coefficient to the product side and our equation now becomes:
$2{C_3}{H_5}{(N{O_3})_3} \to 6C{O_2} + 5{H_2}O + {N_2} + {O_2}$
Now, we have nitrogen and there are total six nitrogen on reactant side and two on product side so we will add $3$ as stoichiometric coefficient on the product side to balance nitrogen and our equation becomes:
$2{C_3}{H_5}{(N{O_3})_3} \to 6C{O_2} + 5{H_2}O + 3{N_2} + {O_2}$

Now lastly, we have oxygen and on the reactant side there are total $18$ oxygen atoms and $19$ on the product side, so we will simply half the oxygen atoms on the product side. So our equation would be:
$2{C_3}{H_5}{(N{O_3})_3} \to 6C{O_2} + 5{H_2}O + 3{N_2} + \dfrac{1}{2}{O_2}$. Thus, the above chemical equation is balanced.

Note: Always remember that for balancing a chemical equation, count the number of atoms on the right hand side and the left hand side and then equalise the number of atoms accordingly to obtain a balanced equation. The stoichiometric coefficients generally balances the number of atoms on each side and are generally first assigned to chief atoms and lastly to hydrogen and oxygen atoms respectively but here we first balanced hydrogen as there is only a single reactant.