# Nitric acid in acid rain forms when gaseous nitrogen dioxide pollutant reacts with gaseous oxygen and liquid water to form aqueous nitric acid. How do you write a balanced chemical equation for this reaction?

Jun 17, 2016

$4 {\text{NO"_ (2(g)) + "O"_ (2(g)) + 2"H"_ 2"O"_ ((l)) -> 4"HNO}}_{3 \left(a q\right)}$

#### Explanation:

The first thing to do here is identify your reactants and your product.

The problem tells you that gaseous nitrogen dioxide, ${\text{NO}}_{2}$ reacts with gaseous oxygen, ${\text{O}}_{2}$, and liquid water, $\text{H"_2"O}$, to form aqueous nitric acid, ${\text{HNO}}_{3}$.

• ${\text{NO}}_{2 \left(g\right)} \to$ the $\left(g\right)$ is used to denote a compound in the gaseous form
• ${\text{O}}_{2 \left(g\right)} \to$ molecular oxygen in gaseous form
• ${\text{H"_ 2"O}}_{\left(l\right)} \to$ the $\left(l\right)$ is used to denote a compound in the liquid form

Your product will be

• ${\text{HNO}}_{3 \left(a q\right)} \to$ the $\left(a q\right)$ is used to denote a compound in the aqueous form

Your next goal is to write the unbalanced chemical equation by placing the reactants on the left side of the reaction arrow, $\to$, and the products on the right side of the arrow

${\text{NO"_ (2(g)) + "O"_ (2(g)) + "H"_ 2"O"_ ((l)) -> "HNO}}_{3 \left(a q\right)}$

Now all you have to do is balance this equation by making sure that all the atoms present on the reactants' side are also present on the products' side.

Now, this equation can be a little tricky to balance by inspection, but a cool trick to try out here is to get the oxygen atoms present on the reactants' side to an even number by multiplying the water molecule by $\textcolor{red}{2}$

${\text{NO"_ (2(g)) + "O"_ (2(g)) + color(red)(2)"H"_ 2"O"_ ((l)) -> "HNO}}_{3 \left(a q\right)}$

You now have $6$ atoms of oxygen on the reactants' side and $3$ on the products' side. Ignore this for now.

Focus on the hydrogen atoms next. Notice that you have

$\textcolor{red}{2} \times 2 = \text{4 atoms of H}$

on the reactants' side, and only $1$ atom on the products' side, so multiply the nitric acid by $\textcolor{b l u e}{4}$ to balance out the hydrogen atoms

${\text{NO"_ (2(g)) + "O"_ (2(g)) + color(red)(2)"H"_ 2"O"_ ((l)) -> color(blue)(4)"HNO}}_{3 \left(a q\right)}$

Now focus on the nitrogen atoms. You have

$\textcolor{b l u e}{4} \times 1 = \text{4 atoms of N}$

on the products' side, and only $1$ atom on the reactants' side, so multiply the nitrogen dioxide molecule by $\textcolor{\mathrm{da} r k g r e e n}{4}$ to get

$\textcolor{\mathrm{da} r k g r e e n}{4} {\text{NO"_ (2(g)) + "O"_ (2(g)) + color(red)(2)"H"_ 2"O"_ ((l)) -> color(blue)(4)"HNO}}_{3 \left(a q\right)}$

As it turns out, the trick we used paid off because the oxygen are balanced! You have

$\textcolor{\mathrm{da} r k g r e e n}{4} \times 2 + 2 + \textcolor{red}{2} \times 1 = \text{12 atoms of O}$

on the reactants' side and

$\textcolor{b l u e}{4} \times 3 = \text{12 atoms of O}$

on the products' side, which means that the chemical equation is now balanced.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{4 {\text{NO"_ (2(g)) + "O"_ (2(g)) + 2"H"_ 2"O"_ ((l)) -> 4"HNO}}_{3 \left(a q\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$