# Question 920b8

Oct 20, 2016

Here's what I got.

#### Explanation:

All you have to do here is identify how many atoms of each element take part in the reaction.

Let's start with the reactants, which as you know are located to the left of the reaction arrow. In your case, you have

• $2 \times {\text{N}}_{\textcolor{b l u e}{2} \left(g\right)} \to$ two molecules of nitrogen gas
• $3 \times {\text{O}}_{\textcolor{red}{2} \left(g\right)} \to$ three molecules of oxygen gas

Now, each molecule of nitrogen gas contains $\textcolor{b l u e}{2}$ atoms of nitrogen. Similarly, each molecule of oxygen gas contains $\textcolor{red}{2}$ atoms of oxygen. This means that the reactants' side contains

• 2 xx [color(blue)(2) xx "N"] = "4 atoms N"
• 3 xx [color(red)(2) xx "O"] = "6 atoms O"

Now focus on the product's side. This side contains

• $2 \times {\text{N"_ color(blue)(2)"O}}_{\textcolor{red}{3} \left(g\right)} \to$ two molecules of dinitrogen trioxide

As you can see from its chemical formula, each molecule of dinitrogen trioxide contains $\textcolor{b l u e}{2}$ atoms of nitrogen and $\textcolor{red}{3}$ atoms of oxygen, which means that the product's side contains

• 2 xx [color(blue)(2) xx "N"] = "4 atoms N"
• 2 xx [color(red)(3) xx "O"] = "6 atoms O"#

And there you have it. Your reaction features

• $\text{4 atoms N } \to$ on both sides of the equation
• $\text{6 atoms O } \to$ on both sides of the equation