# Can someone please explain step by step how to balance "Ca" + "N"_2 = "Ca"_3"N"_2 ?

Dec 27, 2017

$3 {\text{Ca"_ ((s)) + "N"_ (2(g)) -> "Ca"_ 3"N}}_{2 \left(s\right)}$

#### Explanation:

The key thing to keep in mind here is that a chemical equation is balanced when the total number of atoms present on the reactants' side is equal to the total number of atoms present on the products' side.

In your case, the unbalanced chemical equation looks like this

${\text{Ca"_ ((s)) + "N"_ (2(g)) -> "Ca"_ 3"N}}_{2 \left(s\right)}$

In order to balance this chemical equation, you need to look at how many atoms of each element are present on each side of the chemical equation.

For calcium, $\text{Ca}$, you have

$\text{1 atom " -> " 3 atoms}$

This is the case because on the reactants' side, calcium doesn't have a subscript or a coefficient, which means that you only have $1$ atom present on that side.

On the other hand, this element has a subscript of $3$ on the products' side, which means that this side contains $3$ atoms of calcium.

So, in order to balance out the atoms of calcium, you need to multiply the atom of calcium by $\textcolor{red}{3}$ on the reactants 'side.

This will get you

$\textcolor{red}{3} {\text{Ca"_ ((s)) + "N"_ (2(g)) -> "Ca"_ 3"N}}_{2 \left(s\right)}$

Next, focus on nitrogen, $\text{N}$. This time, you have

$\text{2 atoms " -> " 2 atoms}$

This is the case because nitrogen has a subscript of $2$ on both sides of the reaction, which means that $2$ atoms of nitrogen are present on the reactants' side and $2$ atoms of nitrogen are present on the products' side.

You can thus say that you don't need to do anything to the atoms of nitrogen to balance them out.

The balanced chemical equation that describes this reaction will thus be

$3 {\text{Ca"_ ((s)) + "N"_ (2(g)) -> "Ca"_ 3"N}}_{2 \left(s\right)}$