# Question #5a60e

Dec 16, 2015

$2 {\text{H"_text(2(g]) + 2"NO"_text((g]) -> 2"H"_2"O"_text((g]) +"N}}_{\textrm{2 \left(g\right]}}$

#### Explanation:

The first thing to do here is make sure that you know the chemical formulas of your reactants and your products.

I won't go into covalent formulas and nomenclature here, so make sure that you're already familiar with how that works.

• hydrogen gas, ${\text{H}}_{2}$
• nitrogen monoxide, $\text{NO}$

• water, $\text{H"_2"O}$
• nitrogen gas, ${\text{N}}_{2}$

The unbalanced chemical equation for this reaction looks like this

${\text{H"_text(2(g]) + "NO"_text((g]) -> "H"_2"O"_text((g]) + "N}}_{\textrm{2 \left(g\right]}}$

Your goal when balancing chemical equations is to make sure that all the atoms that are present on the reactants' side are accounted for on the products' side.

For example, notice that you have $2$ nitrogen atoms on the products' side in the form of a nitrogen gas molecule, ${\text{N}}_{2}$, but only $1$ nitrogen atom on the reactants' side as part of nitrogen monoxide.

This means that you need to multiply the nitrogen monoxide by $\textcolor{red}{2}$ to balance the nitrogen atoms

${\text{H"_text(2(g]) + color(red)(2)"NO"_text((g]) -> "H"_2"O"_text((g]) +"N}}_{\textrm{2 \left(g\right]}}$

Multiplying the nitrogen monoxide by $2$ doubled both the number of nitrogen atoms, and the number of oxygen atoms present on the reactants' side.

To account for this, you need to multiply the water molecule by $\textcolor{b l u e}{2}$ in order to balance the oxygen atoms out

${\text{H"_text(2(g]) + color(red)(2)"NO"_text((g]) -> color(blue)(2)"H"_2"O"_text((g]) +"N}}_{\textrm{2 \left(g\right]}}$

Doubling the number of water molecules implies that you also have double the number of hydrogen atoms present on the products' side. This means that you need to double the number of hydrogen atoms on the reactants' side by multiplying the hydrogen gas molecule by $\textcolor{g r e e n}{2}$

$\textcolor{g r e e n}{2} {\text{H"_text(2(g]) + color(red)(2)"NO"_text((g]) -> color(blue)(2)"H"_2"O"_text((g]) +"N}}_{\textrm{2 \left(g\right]}}$

And there you have it, the chemical equation is balanced, since you now have

• $4$ hydrogen atoms
• $2$ nitrogen atoms
• $2$ oxygen atoms

on both sides of the equation.