# Question #6a445

Dec 6, 2016

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

#### Explanation:

Calcium metal, $\text{Ca}$, will react with nitrogen gas, ${\text{N}}_{2}$, to produce calcium nitride, ${\text{Ca"_3"N}}_{2}$, an ionic compound that contains calcium cations, ${\text{Ca}}^{2 +}$, and nitride anions, ${\text{N}}^{3 -}$.

The problem tells you that you're dealing with

• calcium in the solid state, ${\text{Ca}}_{\left(s\right)}$
• nitrogen in the gaseous state, ${\text{N}}_{2 \left(g\right)}$
• calcium nitride in the solid state, ${\text{Ca"_ 3"N}}_{2 \left(s\right)}$

Calcium metal and nitrogen gas are the reactants, which means that they are added to the left of the reaction arrow. Calcium nitride is the product, which means that it is added to the right of the reaction arrow.

The unbalanced chemical equation that describes this synthesis reaction looks like this

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

In order to balance this chemical equation, add a coefficient of $3$ in front of calcium metal -- this will bring the number of calcium atoms present on the reactants' side to $3$, which is the number of calcium atoms present on the product's side as well.

Therefore, the balanced chemical equation will look like this

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