Question

How do you balance $Cr(s) + Fe{(N{O_3})_2}(aq) \to Fe(s) + Cr{(N{O_3})_3}(aq)$ ?

Answer 1

In order to solve this question first we will know what the meaning is by balancing an equation. Then we learn stoichiometric coefficients. After it we will learn all the steps which are needed to balance an equation Then we will balance this equation.

Complete answer:
In order to solve this question first we should know balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products. This is important because a chemical equation must obey the law of conservation of mass and the law of constant proportion, i.e. the same number of atoms of each element must exist on the reactant side and the product side of the equation. A stoichiometric coefficient describes the total number of molecules of a chemical species that participate in a chemical reaction. It provides a ratio between the reacting species and the products formed in the reaction.
Now we will do the following step to balance a equation
First we compare the total number of each element on the reactant side and the product side
So there are 1 atom of $Cr(s)$ 1 of $Fe(s)$ and 2 of $N{O_3}$ on the reactant side. Other side has 1 atom of $Cr(s)$ 1 of $Fe(s)$ and 3 of $N{O_3}$ .
Then we added stoichiometric coefficients to molecules containing an element which has a different number of atoms in the reactant side and the product side .The coefficient must balance the number of atoms on each side Now, the number of atoms of the elements on the reactant and product side must be update . It is important to note that the number of atoms of an element in one species must be obtained by multiplying the stoichiometric coefficient with the total number of atoms of that element present in 1 molecule of the species.
So we can see $N{O_3}$ has a different number of molecules on each side. So we add a stoichiometric coefficient 3 on the reactant side and 2 on the product side.
$Cr(s) + 3Fe{(N{O_3})_2} \to Fe(s) + 2Cr{(N{O_3})_3}$
Now the reactant side has 3 atoms of $Fe(s)$ and the product side has 1 atom of $Fe(s)$ so we add a stoichiometric coefficient 3 on the product side.
$Cr(s) + 3Fe{(N{O_3})_2} \to 3Fe(s) + 2Cr{(N{O_3})_3}$
this process will repeated until all the number of atoms of the reacting elements are equal on the reactant and product side.we can see reactant side have 1 atom of $Cr(s)$ and product side have 2 atom of $Cr(s)$
$2Cr(s) + 3Fe{(N{O_3})_2} \to 3Fe(s) + 2Cr{(N{O_3})_3}$
Once all the individual elements are balanced, the total number of atoms of each element on the reactant and product side are compared once again.If there are no inequalities, the chemical equation is said to be balanced.
So balanced equation is $2Cr(s) + 3Fe{(N{O_3})_2} \to 3Fe(s) + 2Cr{(N{O_3})_3}$.

Note:
We can not use this method for balancing an ionic equation because ionic equations have different methods for balancing. Ionic equations are those chemical equations in which only ions participate in a chemical reaction. In simple words, the ions that react together in solution and form new substances as products.