# Complete and balance the reaction

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### Video transcript

5 times 2 is 3 and then one times two 1 times 2 is 2 and now you can verify how many aluminum's do we have on each side well I have 4 aluminum atoms on the left-hand side and how many do I have on the right-hand side I have four aluminum atoms how many oxygens do I have on the left hand side I have three molecules of dioxygen each molecule has two oxygen atoms so I have six oxygens on the left and I have two times three oxygens on the right or I have six oxygens so my chemical equation is now balanced

### Balancing Combustion Reactions

As you have learned, there is not always a straightforward way to balance a reaction.  We tend to just go back and forth, balancing elements on the left and the right, until it works.  Combustion reactions are easier! Balance the elements in the following order: carbon, hydrogen then oxygen.

Take pentane:

C5H12 + O2 ->   CO2 +  H2O

Step 1 balance carbon:  you see 5 carbons on the left so you know it will produce 5 CO2.

C5H12 + O2 ->  5 CO2 +  H2O

Step 2 balance hydrogen:  you see 12 hydrogens on the left so you know it will produce 6 H2O.

C5H12 + O2 ->  5 CO26 H2O

Step 3 balance oxygen:  you see 10+6 or 16 oxygens on the right so you know it will require 8 O2.

C5H12 + 8 O2 ->  5 CO26 H2O

Wow!  It is balanced!
Actually, it doesn't always work out that easily.  Sometimes you get an odd number of oxygens on the right which leads to a fraction in front of O2.  You then have to multiply all of your coefficients by 2.
(Vocabulary:  The numbers subscripted we have been calling subscripts.  These are the formulas of the molecules.  The numbers in front of the formulas (8, 5 & 2), are called coefficients.  These are the numbers we determine when balancing equations.)

1) balance carbon
2) balance hydrogen
3) balance oxygen
4)  Multiply by 2 if you have a fraction.

Lets try a harder problem, hexane:

C6H14 + O2 ->   CO2 +  H2O

Step 1 balance carbon:  you see 6 carbons on the left so you know it will produce 6 CO2.

C6H14 + O2 ->  6 CO2 +  H2O

Step 2 balance hydrogen:  you see 14 hydrogens on the left so you know it will produce 7 H2O.

C6H14 + O2 ->  6 CO2 + 7 H2O
(6x2=12 oxygens)   (7x1=7 oxygens)
Step 3 balance oxygen:  you see 12+7 or 19 oxygens on the right so you know it will require 19/2 or eight and a half O2.

C6H14 + 19/2 O2 ->  6 CO2 + 7 H2O

Step 4  To get rid of the fraction, we multiply all the coefficients by 2.

2 C6H14 + 19 O2 ->  12 CO2 + 14 H2O

Ta da!!  Wow, it sure seems like a lot of work but it is really easy.

Try your hand at it at Dr Wong's website.

Sours: https://web.fscj.edu/Milczanowski/psc/eleven/combust.html

### Learning Objectives

By the end of this section, you will be able to:

• Derive chemical equations from narrative descriptions of chemical reactions.
• Write and balance chemical equations in molecular, total ionic, and net ionic formats.

The preceding chapter introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH4) and two diatomic oxygen molecules (O2) to produce one carbon dioxide molecule (CO2) and two water molecules (H2O). The chemical equation representing this process is provided in the upper half of Figure 1, with space-filling molecular models shown in the lower half of the figure.

Figure 1. The reaction between methane and oxygen to yield carbon dioxide in water (shown at bottom) may be represented by a chemical equation using formulas (top).

This example illustrates the fundamental aspects of any chemical equation:

1. The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.
2. The substances generated by the reaction are called products, and their formulas are placed on the right sight of the equation.
3. Plus signs (+) separate individual reactant and product formulas, and an arrow ($\rightarrow$) separates the reactant and product (left and right) sides of the equation.
4. The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.

It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the relative numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a ratio. This ratio is satisfied if the numbers of these molecules are, respectively, , or , or , and so on (Figure 2). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:

• One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.
• One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.
• One mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide molecules and 2 moles of water molecules.

Figure 2. Regardless of the absolute number of molecules involved, the ratios between numbers of molecules are the same as that given in the chemical equation.

### Balancing Equations

balanced chemical is equation has equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO2 and H2O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

$\left(1{\text{CO}}_{2}\text{ molecule }\times \frac{\text{2 O atoms}}{{\text{CO}}_{2}\text{ molecule }}\right)+\left(2{\text{H}}_{2}\text{O molecule }\times \frac{\text{1 O atom}}{{\text{H}}_{2}\text{O molecule }}\right)=\text{4 O atoms}$

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:

${\text{CH}}_{4}+2{\text{O}}_{2}\rightarrow{\text{CO}}_{2}+2{\text{H}}_{2}\text{O}$

ElementReactantsProductsBalanced?
C1 × 1 = 11 × 1 = 11 = 1, yes
H4 × 1 = 42 × 2 = 44 = 4, yes
O2 × 2 = 4(1 × 2) + (2 × 1) = 44 = 4, yes

A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:

${\text{H}}_{2}\text{O}\rightarrow{\text{H}}_{2}+{\text{O}}_{2}\text{(unbalanced)}$

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

ElementReactantsProductsBalanced?
H1 × 2 = 21 × 2 = 22 = 2, yes
O1 × 1 = 11 × 2 = 21 ≠ 2, no

The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H2O to H2O2 would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H2O to 2.

$\mathbf{2}\text{H}_{2}\text{O}\rightarrow{\text{H}}_{2}+{\text{O}}_{2}\text{(unbalanced)}$

ElementReactantsProductsBalanced?
H2 × 2 = 41 × 2 = 24 ≠ 2, no
O2 × 1 = 21 × 2 = 22 = 2, yes

The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H2 product to 2.

$2{\text{H}}_{2}\text{O}\rightarrow\mathbf{2}{\text{H}}_{2}+{\text{O}}_{2}\text{(balanced)}$

ElementReactantsProductsBalanced?
H2 × 2 = 42 × 2 = 24 = 4, yes
O2 × 1 = 21 × 2 = 22 = 2, yes

These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:

$2{\text{H}}_{2}\text{O}\rightarrow 2{\text{H}}_{2}+{\text{O}}_{2}$

### Example 1: Balancing Chemical Equations

Write a balanced equation for the reaction of molecular nitrogen (N2) and oxygen (O2) to form dinitrogen pentoxide.

First, write the unbalanced equation:

${\text{N}}_{2}+{\text{O}}_{2}\rightarrow{\text{N}}_{2}{\text{O}}_{5}\text{(unbalanced)}$

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

ElementReactantsProductsBalanced?
N1 × 2 = 21 × 2 = 22 = 2, yes
O1 × 2 = 21 × 5 = 52 ≠ 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 O2 and N2O5 to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).

${\text{N}}_{2}+\mathbf{5}{\text{O}}_{2}\rightarrow\mathbf{2}{\text{N}}_{2}{\text{O}}_{5}\text{(unbalanced)}$

ElementReactantsProductsBalanced?
N1 × 2 = 22 × 2 = 42 ≠ 4, no
O5 × 2 = 102 × 5 = 1010 = 10, yes

The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N2 to 2.

$2{\text{N}}_{2}+5{\text{O}}_{2}\rightarrow 2{\text{N}}_{2}{\text{O}}_{5}$

ElementReactantsProductsBalanced?
N2 × 2 = 42 × 2 = 44 = 4, yes
O5 × 2 = 102 × 5 = 1010 = 10, yes

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

#### Check Your Learning

Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)

$2{\text{NH}}_{4}{\text{NO}}_{3}\rightarrow 2{\text{N}}_{2}+{\text{O}}_{2}+4{\text{H}}_{2}\text{O}$

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 (C2H6) with oxygen to yield H2O and CO2, represented by the unbalanced equation:

${\text{C}}_{2}{\text{H}}_{6}+{\text{O}}_{2}\rightarrow{\text{H}}_{2}\text{O}+{\text{CO}}_{2}\text{(unbalanced)}$

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

${\text{C}}_{2}{\text{H}}_{6}+{\text{O}}_{2}\rightarrow 3{\text{H}}_{2}\text{O}+2{\text{CO}}_{2}\text{(unbalanced)}$

This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O2 reactant to yield an odd number, so a fractional coefficient, $\displaystyle\frac{7}{2}$ , is used instead to yield a provisional balanced equation:

${\text{C}}_{2}{\text{H}}_{6}+\frac{7}{2}{\text{O}}_{2}\rightarrow 3{\text{H}}_{2}\text{O}+2{\text{CO}}_{2}$

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

$2{\text{C}}_{2}{\text{H}}_{6}+7{\text{O}}_{2}\rightarrow 6{\text{H}}_{2}\text{O}+4{\text{CO}}_{2}$

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,

$3{\text{N}}_{2}+9{\text{H}}_{2}\rightarrow 6{\text{NH}}_{3}$

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:

${\text{N}}_{2}+3{\text{H}}_{2}\rightarrow 2{\text{NH}}_{3}$

Use this interactive PhET tutorial for additional practice balancing equations.

### Exercises

Balance the following equations:

1. ${\text{PCl}}_{5}\text{(}s\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightarrow{\text{POCl}}_{3}\text{(}l\text{)}+\text{HCl(}aq\text{)}$
2. $\text{Ag}\text{(}s\text{)}+{\text{H}}_{2}\text{S}\text{(}g\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow{\text{Ag}}_{2}\text{S}\text{(}s\text{)}+{\text{H}}_{2}\text{O}\text{(}l\text{)}$
3. $\text{Cu(}s\text{)}+{\text{HNO}}_{3}\text{(}aq\text{)}\rightarrow\text{Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}+\text{NO(}g\text{)}$
4. ${\text{P}}_{4}\text{(}s\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow{\text{P}}_{4}{\text{O}}_{10}\text{(}s\text{)}$
5. ${\text{H}}_{2}\text{(}g\text{)}+{\text{I}}_{2}\text{(}s\text{)}\rightarrow\text{HI(}s\text{)}$
6. $\text{Pb(}s\text{)}+{\text{H}}_{2}\text{O(}l\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow{\text{Pb(OH)}}_{2}\text{(}s\text{)}$
7. $\text{Fe(}s\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow{\text{Fe}}_{2}{\text{O}}_{3}\text{(}s\text{)}$
8. $\text{Fe(}s\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightarrow{\text{Fe}}_{3}{\text{O}}_{4}\text{(}s\text{)}+{\text{H}}_{2}\text{(}g\text{)}$
9. $\text{Na(}s\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightarrow\text{NaOH}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}$
10. ${\text{Sc}}_{2}{\text{O}}_{3}\text{(}s\text{)}+{\text{SO}}_{3}\text{(}l\text{)}\rightarrow{\text{Sc}}_{2}{\text{(}{\text{SO}}_{4}\text{)}}_{3}\text{(}s\text{)}$
11. ${\text{(}{\text{NH}}_{4}\text{)}}_{2}{\text{Cr}}_{2}{\text{O}}_{7}\text{(}s\text{)}\rightarrow{\text{Cr}}_{2}{\text{O}}_{3}\text{(}s\text{)}+{\text{N}}_{2}\text{(}g\text{)}+{\text{H}}_{2}\text{O(}g\text{)}$
12. ${\text{Ca}}_{3}{\text{(}{\text{PO}}_{4}\text{)}}_{2}\text{(}aq\text{)}+{\text{H}}_{3}{\text{PO}}_{4}\text{(}aq\text{)}\rightarrow\text{Ca}{\text{(}{\text{H}}_{2}{\text{PO}}_{4}\text{)}}_{2}\text{(}aq\text{)}$
13. ${\text{P}}_{4}\text{(}s\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\rightarrow{\text{PCl}}_{3}\text{(}l\text{)}$
14. $\text{Al(}s\text{)}+{\text{H}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}\rightarrow{\text{Al}}_{2}{\text{(}{\text{SO}}_{4}\text{)}}_{3}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}$
15. ${\text{PtCl}}_{4}\text{(}s\text{)}\rightarrow\text{Pt}\text{(}s\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}$
16. ${\text{TiCl}}_{4}\text{(}s\text{)}+{\text{H}}_{2}\text{O(}g\text{)}\rightarrow{\text{TiO}}_{2}\text{(}s\text{)}+\text{HCl(}g\text{)}$

The balanced equations are as follows:

1. ${\text{PCl}}_{5}\text{(}s\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightarrow{\text{POCl}}_{3}\text{(}l\text{)}+2\text{HCl(}aq\text{);}$

3. $3\text{Cu}\text{(}s\text{)}+8{\text{HNO}}_{3}\text{(}aq\text{)}\rightarrow 3\text{Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}+4{\text{H}}_{2}\text{O(}l\text{)}+2\text{NO(}g\text{);}$

5. ${\text{H}}_{2}\text{(}g\text{)}+{\text{I}}_{2}\text{(}s\text{)}\rightarrow 2\text{HI(}s\text{);}$

7. $4\text{Fe(}s\text{)}+3{\text{O}}_{2}\text{(}g\text{)}\rightarrow 2{\text{Fe}}_{2}{\text{O}}_{3}\text{(}s\text{);}$

9. $2\text{Na(}s\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow 2\text{NaOH(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{);}$

${\text{(}{\text{NH}}_{4}\text{)}}_{2}+{\text{Cr}}_{5}{\text{O}}_{7}\text{(}s\text{)}\rightarrow{\text{Cr}}_{2}{\text{O}}_{3}\text{(}s\text{)}+{\text{N}}_{2}\text{(}g\text{)}+4{\text{H}}_{2}\text{O(}g\text{);}$

${\text{P}}_{4}\text{(}s\text{)}+6{\text{Cl}}_{2}\text{(}g\text{)}\rightarrow 4{\text{PCl}}_{3}\text{(}l\text{);}$

${\text{PtCl}}_{4}\text{(}s\text{)}\rightarrow\text{Pt(}s\text{)}+2{\text{Cl}}_{2}\text{(}g\text{)}$

### 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 the preceding chapter). These notations are illustrated in the example equation here:

$2\text{Na(}s\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow 2\text{NaOH(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}$

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.

${\text{CaCO}}_{3}\text{(}s\text{)}\stackrel{\Delta}{\rightarrow}\text{CaO(}s\text{)}+{\text{CO}}_{2}\text{(}g\text{)}$

Other examples of these special conditions will be encountered in more depth in later chapters.

### Equations for Ionic Reactions

Given the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of CaCl2 and AgNO3 are mixed, a reaction takes place producing aqueous Ca(NO3)2 and solid AgCl:

${\text{CaCl}}_{2}\text{(}aq\text{)}+2{\text{AgNO}}_{3}\text{(}aq\text{)}\rightarrow\text{Ca}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}+2\text{AgCl(}s\text{)}$

This balanced equation, derived in the usual fashion, is called a molecular equation, because it doesn’t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:

$\begin{array}{l}{\text{CaCl}}_{2}\text{(}aq\text{)}\rightarrow{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}+2{\text{Cl}}^{-}\text{(}aq\text{)}\\ 2{\text{AgNO}}_{3}\text{(}aq\text{)}\rightarrow 2{\text{Ag}}^{\text{+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}\\ \text{Ca}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}\rightarrow{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}\end{array}$

Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, s.

Explicitly representing all dissolved ions results in a complete ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:

${\text{Ca}}^{\text{2+}}\text{(}aq\text{)}+2{\text{Cl}}^{-}\text{(}aq\text{)}+2{\text{Ag}}^{\text{+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}+2\text{AgCl(}s\text{)}$

Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, Ca2+(aq) and ${\text{NO}}_{3}{}^{-}\text{(}aq\text{)}$. These spectator ions—ions whose presence is required to maintain charge neutrality—are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation:

$\begin{array}{c}\cancel{{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}}+2{\text{Cl}}^{-}\text{(}aq\text{)}+2{\text{Ag}}^{\text{+}}\text{(}aq\text{)}+\cancel{2{\text{NO}}_{3}{}^{\text{-}}\text{(}aq\text{)}}\rightarrow\cancel{{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}}+\cancel{2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}}+2\text{AgCl(}s\text{)}\\ 2{\text{Cl}}^{-}\text{(}aq\text{)}+2{\text{Ag}}^{\text{+}}\text{(}aq\text{)}\rightarrow 2\text{AgCl(}s\text{)}\end{array}$

Following the convention of using the smallest possible integers as coefficients, this equation is then written:

${\text{Cl}}^{\text{-}}\text{(}aq\text{)}+{\text{Ag}}^{+}\text{(}aq\text{)}\rightarrow\text{AgCl(}s\text{)}$

This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of Cl and Ag+.

### Example 2: Molecular and Ionic Equations

When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced molecular, complete ionic, and net ionic equations for this process.

Begin by identifying formulas for the reactants and products and arranging them properly in chemical equation form:

${\text{CO}}_{2}\text{(}aq\text{)}+\text{NaOH(}aq\text{)}\rightarrow{\text{Na}}_{2}{\text{CO}}_{3}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\text{(unbalanced)}$

Balance is achieved easily in this case by changing the coefficient for NaOH to 2, resulting in the molecular equation for this reaction:

${\text{CO}}_{2}\text{(}aq\text{)}+2\text{NaOH(}aq\text{)}\rightarrow{\text{Na}}_{2}{\text{CO}}_{3}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}$

The two dissolved ionic compounds, NaOH and Na2CO3, can be represented as dissociated ions to yield the complete ionic equation:

${\text{CO}}_{2}\text{(}aq\text{)}+2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}\rightarrow 2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+{\text{CO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}$

Finally, identify the spectator ion(s), in this case Na+(aq), and remove it from each side of the equation to generate the net ionic equation:

$\begin{array}{l}{\text{CO}}_{2}\text{(}aq\text{)}+\cancel{2{\text{Na}}^{\text{+}}\text{(}aq\text{)}}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}\rightarrow 2\cancel{{\text{Na}}^{\text{+}}\text{(}aq\text{)}}+{\text{CO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\\ {\text{CO}}_{2}\text{(}aq\text{)}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}\rightarrow{\text{CO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\end{array}$

#### Check Your Learning

Diatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:

$\text{NaCl(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\,\,\,{\xrightarrow{\text{electricity}}}\,\,\,\text{NaOH(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}$

Write balanced molecular, complete ionic, and net ionic equations for this process.

$\begin{array}{l}2\text{NaCl(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow 2\text{NaOH}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\text{(}\text{molecular}\text{)}\\ 2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+2{\text{Cl}}^{\text{-}}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow 2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\text{(}\text{complete ionic}\text{)}\\ 2{\text{Cl}}^{\text{-}}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow 2{\text{OH}}^{\text{-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\text{(net ionic)}\end{array}$

### Key Concepts and Summary

Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products. Chemical reactions in aqueous solution that involve ionic reactants or products may be represented more realistically by complete ionic equations and, more succinctly, by net ionic equations.

### Exercises

1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?
2. Consider molecular, complete ionic, and net ionic equations.
1. What is the difference between these types of equations?
2. In what circumstance would the complete and net ionic equations for a reaction be identical?
3. Write a balanced molecular equation describing each of the following chemical reactions.
1. Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
2. Gaseous butane, C4H10, reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.
3. Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
4. Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.
4. Write a balanced equation describing each of the following chemical reactions.
1. Solid potassium chlorate, KClO3, decomposes to form solid potassium chloride and diatomic oxygen gas.
2. Solid aluminum metal reacts with solid diatomic iodine to form solid Al2I6.
3. When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.
4. Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.
5. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.
1. Write the formulas of barium nitrate and potassium chlorate.
2. The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.
3. The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.
4. Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains Fe+ ions.)
6. Fill in the blank with a single chemical formula for a covalent compound that will balance the equation:
7. Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).
1. Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.
2. The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write complete and net ionic equations for this reaction.
8. A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process.
1. The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.
2. The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.
3. Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.
4. The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.
5. Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.
9. From the balanced molecular equations, write the complete ionic and net ionic equations for the following:
1. ${\text{K}}_{2}{\text{C}}_{2}{\text{O}}_{4}\text{(}aq\text{)}+\text{Ba}{\text{(OH)}}_{2}\text{(}aq\text{)}\rightarrow 2\text{KOH(}aq\text{)}+{\text{BaC}}_{2}{\text{O}}_{2}\text{(}s\text{)}$
2. ${\text{Pb(NO}}_{3}{\text{)}}_{2}\text{(}aq\text{)}+{\text{H}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}\rightarrow{\text{PbSO}}_{4}\text{(}s\text{)}+2{\text{HNO}}_{3}\text{(}aq\text{)}$
3. ${\text{CaCO}}_{3}\text{(}s\text{)}+{\text{H}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}\rightarrow{\text{CaSO}}_{4}\text{(}s\text{)}+{\text{CO}}_{2}\text{(}g\text{)}+{\text{H}}_{2}\text{O(}l\text{)}$

1. An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.

3. The balanced equations are as follows:

1. ${\text{CaCO}}_{3}\text{(}s\text{)}\rightarrow\text{CaO(}s\text{)}+{\text{CO}}_{2}\text{(}g\text{)}$
2. $2{\text{C}}_{4}{\text{H}}_{10}\text{(}g\text{)}+13{\text{O}}_{2}\text{(}g\text{)}\rightarrow 8{\text{CO}}_{2}\text{(}g\text{)}+10{\text{H}}_{2}\text{O(}g\text{)}$
3. ${\text{MgC1}}_{2}\text{(}aq\text{)}+2\text{NaOH(}aq\text{)}\rightarrow\text{Mg}{\text{(OH)}}_{2}\text{(}s\text{)}+2\text{NaCl(}aq\text{)}$
4. $2{\text{H}}_{2}\text{O(}g\text{)}+2\text{Na(}s\text{)}\rightarrow 2\text{NaOH(}s\text{)}+{\text{H}}_{2}\text{(}g\text{)}$

5. The answers are as follows:

1. Ba(NO3)2, KClO3
2. $2{\text{KClO}}_{3}\text{(}s\text{)}\rightarrow 2\text{KCl(}s\text{)}+3{\text{O}}_{2}\text{(}g\text{)}$
3. $2\text{Ba}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}s\text{)}\rightarrow 2\text{BaO(}s\text{)}+2{\text{N}}_{2}\text{(}g\text{)}+5{\text{O}}_{2}\text{(}g\text{)}$
4. $\begin{array}{l}2\text{Mg(}s\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow 2\text{MgO(}s\text{)}\\ 4\text{Al(}s\text{)}+3{\text{O}}_{2}\text{(}g\text{)}\rightarrow 2{\text{Al}}_{2}{\text{O}}_{3}\text{(}g\text{)}\\ 4\text{Fe(}s\text{)}+3{\text{O}}_{2}\text{(}g\text{)}\rightarrow 2{\text{Fe}}_{2}{\text{O}}_{3}\text{(}s\text{)}\end{array}$

7. The answers are as follows:

1. $4\text{HF(}aq\text{)}+{\text{SiO}}_{2}\text{(}s\text{)}\rightarrow{\text{SiF}}_{4}\text{(}g\text{)}+2{\text{H}}_{2}\text{O(}l\text{);}$
2. complete ionic equation: $2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+2{\text{F}}^{\text{-}}\text{(}aq\text{)}+{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}+2{\text{Cl}}^{\text{-}}\text{(}aq\text{)}\rightarrow\text{Ca}{\text{F}}_{2}\text{(}s\text{)}+2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+2{\text{Cl}}^{\text{-}}\text{(}aq\text{),}$ net ionic equation: $2{\text{F}}^{\text{-}}\text{(}aq\text{)}+{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}\rightarrow{\text{CaF}}_{2}\text{(}s\text{)}$

9. The ionic and net ionic equations for each are as follows:

1. $\begin{array}{l}{}2{\text{K}}^{\text{+}}\text{(}aq\text{)}+{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\text{(}aq\text{)}+{\text{Ba}}^{\text{2+}}\text{(}aq\text{)}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}\rightarrow 2{\text{K}}^{\text{+}}\text{(}aq\text{)}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}+{\text{BaC}}_{2}{\text{O}}_{4}\text{(}s\text{)}\text{(complete)}\\ {\text{Ba}}^{\text{2+}}\text{(}aq\text{)}+{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{BaC}}_{2}{\text{O}}_{4}\text{(}s\text{)}\text{(net)}\end{array}$
2. $\begin{array}{l}{\text{Pb}}^{\text{2+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{\text{-}}\text{(}aq\text{)}+2{\text{H}}^{+}\text{(}aq\text{)}+{\text{SO}}_{4}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{PbSO}}_{4}\text{(}s\text{)}+2{\text{H}}^{\text{+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{\text{-}}\text{(}aq\text{)}\text{(complete)}\\ {\text{Pb}}^{\text{2+}}\text{(}aq\text{)}+{\text{SO}}_{4}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{PbSO}}_{4}\text{(}s\text{)}\text{(net)}\end{array}$
3. $\begin{array}{l}{}{\text{CaCO}}_{3}\text{(}s\text{)}+2{\text{H}}^{\text{+}}\text{(}aq\text{)}+{\text{SO}}_{4}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{CaSO}}_{4}\text{(}s\text{)}+{\text{CO}}_{2}\text{(}g\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\text{(complete)}\\ {\text{CaCO}}_{3}\text{(}s\text{)}+2{\text{H}}^{\text{+}}\text{(}aq\text{)}+{\text{SO}}_{4}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{CaSO}}_{4}\text{(}s\text{)}+{\text{CO}}_{2}\text{(}g\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\text{(net)}\end{array}$

### Glossary

balanced equation: chemical equation with equal numbers of atoms for each element in the reactant and product

chemical equation: symbolic representation of a chemical reaction

coefficient: number placed in front of symbols or formulas in a chemical equation to indicate their relative amount

complete ionic equation: chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions

molecular equation: chemical equation in which all reactants and products are represented as neutral substances

net ionic equation: chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)

product: substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation

reactant: substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation

spectator ion: ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality

Sours: https://courses.lumenlearning.com/atd-sanjac-introductorychemistry/chapter/writing-and-balancing-chemical-equations-2/
Predicting The Products of Chemical Reactions - Chemistry Examples and Practice Problems

## What are the products of this reaction? What is the complete balanced equation? #?H_3PO_4(aq) + ?Ca(OH)_2(aq) ->#

Although phosphoric acid and calcium hydroxide might be less familiar examples, they are an acid and a base, and we know that these react to yield a salt and water.

'salt'

The salt formed in this case is calcium phosphate, and since we know the charges on the ions are and m, calcium phosphate has the formula .

A more complete equation, then, is:

+ ? → ? +

Now we just need to replace all those question marks!

In the discussion that follows, I will treat the ion as a unit, rather than as separate phosphorus and oxygen atoms. This means I will have to be careful when counting oxygen.

One mole of on the left yields mole of on the right. Place a in front of the .

+ → ? +

Now there are moles of on both sides.

One mole of on the left yields mole of on the right, so add a in front of the phosphoric acid:

+ → ? +

Now there are moles of on both sides.

Ignore the oxygen in the groups and look at the rest of the oxygen. There are 6 on the left and only one on the right (in the ), so place a in front of the so that there will be 6 on each side:

+ → ? +

Finally, check the H. There are now 12 on the left and 12 on the right, so that's OK.

That means we can replace the final '?' with 1, but we don't usually bother writing 1 as a coefficient in chemical equations, so the final chemical equation is:

+ → +

Sours: https://socratic.org/questions/what-are-the-products-of-this-reaction-what-is-the-complete-balanced-equation-h-

## How to Write Balanced Chemical Equations

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Learning Objectives

• Explain the roles of subscripts and coefficients in chemical equations.
• Balance a chemical equation when given the unbalanced equation.
• Explain the role of the Law of Conservation of Mass in a chemical reaction.

Even though chemical compounds are broken up and new compounds are formed during a chemical reaction, atoms in the reactants do not disappear, nor do new atoms appear to form the products. In chemical reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are present in the products—they are merely reorganized into different arrangements. In a complete chemical equation, the two sides of the equation must be present on the reactant and the product sides of the equation.

### Coefficients and Subscripts

There are two types of numbers that appear in chemical equations. There are subscripts, which are part of the chemical formulas of the reactants and products; and there are coefficients that are placed in front of the formulas to indicate how many molecules of that substance is used or produced.

The subscripts are part of the formulas and once the formulas for the reactants and products are determined, the subscripts may not be changed. The coefficients indicate the number of each substance involved in the reaction and may be changed in order to balance the equation. The equation above indicates that one mole of solid copper is reacting with two moles of aqueous silver nitrate to produce one mole of aqueous copper (II) nitrate and two atoms of solid silver.

### Balancing a Chemical Equation

Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure $$\PageIndex{1}$$.

The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. The following is an efficient approach to balancing a chemical equation using this method.

Steps in Balancing a Chemical Equation

1. Identify the most complex substance.
2. Beginning with that substance, choose an element(s) that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element(s) on both sides.
3. Balance polyatomic ions (if present on both sides of the chemical equation) as a unit.
4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients.
5. Count the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced.

Example $$\PageIndex{1}$$: Combustion of Heptane

Balance the chemical equation for the combustion of Heptane ($$\ce{C_7H_{16}}$$).

$\ce{C_7H_{16} (l) + O_2 (g) → CO_2 (g) + H_2O (g) } \nonumber$

Solution

StepsExample
1. Identify the most complex substance.The most complex substance is the one with the largest number of different atoms, which is $$C_7H_{16}$$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.

a. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side:

$\ce{C7H16 (l) + O2 (g) → } \underline{7} \ce{CO2 (g) + H2O (g) } \nonumber$

• 7 carbon atoms on both reactant and product sides

b. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side:

$\ce{C7H16 (l) + O2 (g) → 7 CO2 (g) + } \underline{8} \ce{H2O (g) } \nonumber$

• 16 hydrogen atoms on both reactant and product sides
3.Balance polyatomic ions as a unit.There are no polyatomic ions to be considered in this reaction.
4. Balance the remaining atoms.

The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side:

$\ce{C7H16 (l) + }\underline{11} \ce{ O2 (g) → 7 CO2 (g) + 8H2O (g) } \nonumber$

• 22 oxygen atoms on both reactant and product sides
5. Check your work.The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.

Example $$\PageIndex{2}$$: Combustion of Isooctane

Combustion of Isooctane ($$\ce{C_8H_{18}}$$)

$\ce{C8H18 (l) + O2 (g) -> CO_2 (g) + H_2O(g)} \nonumber$

Solution

The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start. The combustion of any hydrocarbon with oxygen produces carbon dioxide and water.

StepsExample
1. Identify the most complex substance.The most complex substance is the one with the largest number of different atoms, which is $$\ce{C8H18}$$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.

a. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products:

$\ce{C8H18 (l) + O2 (g) -> }\underline{8} \ce{ CO2 (g) + H2O(g)}\nonumber$

• 8 carbon atoms on both reactant and product sides

b. 18 hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products:

$\ce{C8H18 (l) + O2 (g) -> 8CO2 (g) + }\underline{9} \ce{ H2O(g)} \nonumber$

• 18 hydrogen atoms on both reactant and product sides
3.Balance polyatomic ions as a unit.There are no polyatomic ions to be considered in this reaction.
4.Balance the remaining atoms.

The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient ($$\dfrac{25}{2}$$) to balance the oxygen atoms:

$\ce{C8H18 (l) + } \underline{ \dfrac{25}{2} } \ce{O2 (g)→ 8CO2 (g) + 9H2O(g) }\nonumber$

• 25 oxygen atoms on both reactant and product sides

The equation is now balanced, but we usually write equations with whole number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2:

&#;

$\underline{2} \ce{C8H18(l) + } \underline{25} \ce{O2(g) ->} \underline{16} \ce{CO2(g) + } \underline{18} \ce{H2O(g)} \nonumber$

5. Check your work.

The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side.

Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.

Example $$\PageIndex{3}$$: Precipitation of Lead (II) Chloride

Aqueous solutions of lead (II) nitrate and sodium chloride are mixed. The products of the reaction are an aqueous solution of sodium nitrate and a solid precipitate of lead (II) chloride. Write the balanced chemical equation for this reaction.

Solution

StepsExample
1. Identify the most complex substance.

The most complex substance is lead (II) chloride.

$\ce{Pb(NO3)2(aq) + NaCl(aq) → NaNO3(aq) + PbCl2(s)} \nonumber$

There are twice as many chloride ions in the reactants as there are in the products. Place a 2 in front of the NaCl in order to balance the chloride ions.

$\ce{Pb(NO3)2(aq) + }\underline{ 2} \ce{NaCl(aq) → NaNO3(aq) + PbCl2(s)} \nonumber$

• 1 Pb atom on both reactant and product sides
• 2 Na atoms on reactant side, 1 Na atom on product side
• 2 Cl atoms on both reactant and product sides
3.Balance polyatomic ions as a unit.

The nitrate ions are still unbalanced. Place a 2 in front of the NaNO3. The result is:

$\ce{Pb(NO3)2(aq) + 2NaCl(aq) → } \underline {2} \ce{NaNO3(aq) + PbCl2(s)} \nonumber$

• 1 Pb atom on both reactant and product sides
• 2 Na atoms on both reactant and product sides
• 2 Cl atoms on both reactant and product sides
• 2 NO3- atoms on both reactant and product sides
4.Balance the remaining atoms.There is no need to balance the remaining atoms because they are already balanced.
5. Check your work.

$\ce{Pb(NO3)2(aq) + 2NaCl(aq) → 2NaNO3(aq) + PbCl2(s)} \nonumber$

• 1 Pb atom on both reactant and product sides
• 2 Na atoms on both reactant and product sides
• 2 Cl atoms on both reactant and product sides
• 2 NO3- atoms on both reactant and product sides

Exercise $$\PageIndex{1}$$

Is each chemical equation balanced?

1. $$\ce{2Hg(&#;)+ O_2(g) \rightarrow Hg_2O_2(s)}$$
2. $$\ce{C_2H_4(g) + 2O_2(g)→ 2CO_2(g) + 2H_2O(g)}$$
3. $$\ce{Mg(NO_3)_2(s) + 2Li (s) \rightarrow Mg(s)+ 2LiNO_3(s)}$$
yes
no
yes

Exercise $$\PageIndex{2}$$

Balance the following chemical equations.

1. $$\ce{N2 (g) + O2 (g) → NO2 (g) }$$
2. $$\ce{Pb(NO3)2(aq) + FeCl3(aq) → Fe(NO3)3(aq) + PbCl2(s)}$$
3. $$\ce{C6H14(l) + O2(g)→ CO2(g) + H2O(g)}$$
N2 (g) + 2O2 (g) → 2NO2 (g)
3Pb(NO3)2(aq) + 2FeCl3(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
2C6H14(l) + 19O2(g)→ 12CO2(g) + 14H2O(g)

### Summary

• To be useful, chemical equations must always be balanced. Balanced chemical equations have the same number and type of each atom on both sides of the equation.
• The coefficients in a balanced equation must be the simplest whole number ratio. Mass is always conserved in chemical reactions.

### Vocabulary

• Chemical reaction - The process in which one or more substances are changed into one or more new substances.
• Reactants - The starting materials in a reaction.
• Products - Materials present at the end of a reaction.
• Balanced chemical equation - A chemical equation in which the number of each type of atom is equal on the two sides of the equation.
• Subscripts - Part of the chemical formulas of the reactants and products that indicate the number of atoms of the preceding element.
• Coefficient - A small whole number that appears in front of a formula in a balanced chemical equation.

### Contributions & Attributions

This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:

Sours: https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/07%3A_Chemical_Reactions/%3A_How_to_Write_Balanced_Chemical_Equations
How to Balance Chemical Equations in 5 Easy Steps: Balancing Equations Tutorial

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