Question

How do you balance $AgN{O_3} + {(N{H_4})_2}Cr{O_4} \to A{g_2}Cr{O_4} + N{H_4}N{O_3}$?

Answer 1

Hint: 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 any chemical equation can be balanced by simply adding the stoichiometric coefficients to the reactants as well as to the products because 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 elements must exist on the reactant and products sides.
Now talking about the given chemical equation which is shown below:
$AgN{O_3} + {(N{H_4})_2}Cr{O_4} \to A{g_2}Cr{O_4} + N{H_4}N{O_3}$
As we can see in the given equation, one silver atom is present on the reactant side and two on the product side, so we can add $2$ as a stoichiometric coefficient to balance the silver atoms on both sides. Thus we will get:
$2AgN{O_3} + {(N{H_4})_2}Cr{O_4} \to A{g_2}Cr{O_4} + N{H_4}N{O_3}$
Now, we see that four nitrogen atoms are present on the reactant side, two in silver nitrate and two in ammonium chromate, whereas on the product side two nitrogen atoms are present in ammonium nitrate. So, we can add $2$ as a stoichiometric coefficient on the product side of the equation before ammonium nitrate to balance the number of nitrogen atoms. Thus, we will get:
$2AgN{O_3} + {(N{H_4})_2}Cr{O_4} \to A{g_2}Cr{O_4} + 2N{H_4}N{O_3}$
Then we can see that there is only one chromium atom on both the sides so it is balanced. Next we have hydrogen which is balanced by adding $2$ before ammonium nitrate.
Lastly, we will take oxygen into consideration and we can see that after adding coefficients before other atoms, we got $10$ oxygen each on the reactant and the product side. So, oxygen is also balanced.
Therefore, our balanced chemical equation would be:
$2AgN{O_3} + {(N{H_4})_2}Cr{O_4} \to A{g_2}Cr{O_4} + 2N{H_4}N{O_3}$

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.