## Another Example on Balancing Chemical Equations

Contents

Write a balanced equation for the reaction of molecular nitrogen (N_{2}) and oxygen (O_{2}) to form dinitrogen pentoxide.

### Solution

First, write the unbalanced equation.

\(\mathrm{N_2} + \mathrm{O_2} \longrightarrow \mathrm{N_2O_5} \text{ (unbalanced)}\)

Next, count the number of each type of atom present in the unbalanced equation.

## Element | ## Reactants | ## Products | ## Balanced? |
---|---|---|---|

N | 1 × 2 = 2 | 1 × 2 = 2 | 2 = 2, yes |

O | 1 × 2 = 2 | 1 × 5 = 5 | 2 ≠ 5, no |

Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O_{2} and N_{2}O_{5} to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).

\(\mathrm{N_2} + \mathbf{5}\mathrm{O_2} \longrightarrow \mathbf{2}\mathrm{N_2O_5} \text{ (unbalanced)}\)

## Element | ## Reactants | ## Products | ## Balanced? |
---|---|---|---|

N | 1 × 2 = 2 | 2 × 2 = 4 | 2 ≠ 4, no |

O | 5 × 2 = 10 | 2 × 5 = 10 | 10 = 10, yes |

The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N_{2} to 2.

\(\mathrm{2N_2} + \mathrm{5O_2} \longrightarrow \mathrm{2N_2O_5}\)

## Element | ## Reactants | ## Products | ## Balanced? |
---|---|---|---|

N | 2 × 2 = 4 | 2 × 2 = 4 | 4 = 4, yes |

O | 5 × 2 = 10 | 2 × 5 = 10 | 10 = 10, yes |

The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.

## Yet Another Example

It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation’s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance.

For example, consider the reaction of ethane (C_{2}H_{6}) with oxygen to yield H_{2}O and CO_{2}, represented by the unbalanced equation:

\(\mathrm{C_2H_6} + \mathrm{O_2} \longrightarrow \mathrm{H_2O} + \mathrm{CO_2} \text{ (unbalanced)}\)

Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:

\(\mathrm{C_2H_6} + \mathrm{O_2} \longrightarrow \mathrm{3H_2O} + \mathrm{2CO_2} \text{ (unbalanced)}\)

This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O_{2 }reactant to yield an odd number, so a fractional coefficient, \(\frac{7}{2},\) is used instead to yield a provisional balanced equation:

\(\mathrm{C_2H_6} + \frac{7}{2} \mathrm{O_2} \longrightarrow \mathrm{3H_2O} + \mathrm{2CO_2} \text{ (unbalanced)}\)

A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:

\(\mathrm{2C_2H_6} + \mathrm{7O_2} \longrightarrow \mathrm{6H_2O} + \mathrm{4CO_2} \text{ (unbalanced)}\)

Finally with regard to balanced equations, recall that convention dictates use of the *smallest whole-number coefficients*. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,

\(\mathrm{3N_2} + \mathrm{9H_2} \longrightarrow \mathrm{6NH_3}\)

the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:

\(\mathrm{N_2} + \mathrm{3H_2} \longrightarrow \mathrm{2NH_3}\)

## Additional Information in Chemical Equations

The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include *s* for solids, *l* for liquids, *g* for gases, and *aq* for substances dissolved in water (*aqueous solutions*, as introduced in Chemistry 101). These notations are illustrated in the example equation here:

\(\mathrm{2Na}(s) + \mathrm{2H_2O}(l) \longrightarrow \mathrm{2NaOH}(aq) + \mathrm{H_2}(g)\)

This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).

Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation’s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (Δ) over the arrow.

\(\mathrm{CaCO_3}(s) \overset{\Delta}{\longrightarrow} \mathrm{CaO}(s) + \mathrm{CO_2}(g)\)

Other examples of these special conditions will be encountered in more depth in other tutorials.

am confused can someone help me out for how they knew it was 7/2

There is a total of 7 O atoms on the right hand side (3 from the 3H

_{2}O plus 2 × 2 from the 2CO_{2}) and 2 O atoms on the left hand side (from the O_{2}). In other for both sides to tally, we need to divide the greater number by the smaller one to get a coefficient for the left hand side (the side with the smaller number of O atoms) that balances both sides of the equation with regards to O atoms. Hence, the 7/2 you see there. I hope this explanation helps.