5.2: Balancing Chemical Equations

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Learning Objectives

Balance chemical equations.

How does one balance a chemical equation, starting with the correct formulas of the reactants and products? Basically, a back-and-forth (or trial-and-error) approach is adopted, counting the number of atoms of one element on one side, checking the number of atoms of that element on the other side, and changing a coefficient if necessary. Then check another element, going back and forth from one side of the equation to another, until each element has the same number of atoms on both sides of the arrow. In many cases, it does not matter which element is balanced first and which is balanced last, as long as all elements have the same number of atoms on each side of the equation.

Below are guidelines for writing and balancing chemical equations.

Determine the correct chemical formulas for each reactant and product. Write the skeleton equation.
Count the number of atoms of each element that appears as a reactant and as a product. If a polyatomic ion is unchanged on both sides of the equation, count it as a unit.
Balance each element one at a time by placing coefficients in front of the formulas. No coefficient is written for a 1. It is best to begin by balancing elements that only appear in one chemical formula on each side of the equation. NEVER change the subscripts in a chemical formula - you can only balance equations by using coefficients.
Check each atom or polyatomic ion to be sure that they are equal on both sides of the equation.
Make sure that all coefficients are in the lowest possible ratio. If necessary, reduce to the lowest ratio.

For example, to balance the equation

Step 1: Write the skeleton equation with the correct formulas.

\[\ce{CH4 + Cl2 \rightarrow CCl4 + HCl} \label{Eq5}\]

Step 2: Count the number of each atom or polyatomic ion on both sides of the equation.

\[\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \\ 1 \: \ce{C} \: \text{atom} & 1 \: \ce{C} \: \text{atom} \\ 4 \: \ce{H} \: \text{ions} & 1 \: \ce{H} \: \text{ions} \\ 2 \: \ce{Cl} \: \text{atom} & 5 \: \ce{Cl} \: \text{atoms} \end{array}\]

Step 3: We find that both sides are already balanced with one carbon atom. So we proceed to balance the hydrogen atoms. We find that the reactant side has four hydrogen atoms, so the product side must also have four hydrogen atoms. This is balanced by putting a 4 in front of the HCl:

\[\ce{CH4 + Cl2 \rightarrow CCl4 + 4HCl } \label{Eq6}\]

\[\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \\ 1 \: \ce{C} \: \text{atom} & 1 \: \ce{C} \: \text{atom} \\ 4 \: \ce{H} \: \text{ions} & 4 \: \ce{H} \: \text{ions} \\ 2 \: \ce{Cl} \: \text{atom} & 8 \: \ce{Cl} \: \text{atoms} \end{array}\]

Now each side has four hydrogen atoms. The product side has a total of eight chlorine atoms (four from the \(CCl_4\) and four from the four molecules of HCl), so we need eight chlorine atoms as reactants. Because elemental chlorine is a diatomic molecule, we need four chlorine molecules to get a total of eight chlorine atoms. We add another 4 in front of the Cl₂ reactant:

\[\ce{CH4 + 4Cl2 \rightarrow CCl4 + 4HCl } \label{Eq7}\]

\[\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \\ 1 \: \ce{C} \: \text{atom} & 1 \: \ce{C} \: \text{atom} \\ 4 \: \ce{H} \: \text{ions} & 4 \: \ce{H} \: \text{ions} \\ 8 \: \ce{Cl} \: \text{atom} & 8 \: \ce{Cl} \: \text{atoms} \end{array}\]

Step 3: Now we check: each side has one carbon atom, four hydrogen atoms, and eight chlorine atoms. The chemical equation is balanced. And, the coefficients are in the lowest possible ratio.

Example \(\PageIndex{2}\)

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

Step 1: Write the skeleton equation with the correct formulas.

\[\ce{Pb(NO_3)_2} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)\]

Step 2: Count the number of each atom or polyatomic ion on both sides of the equation.

\[\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \\ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \\ 2 \: \ce{NO_3^-} \: \text{ions} & 1 \: \ce{NO_3^-} \: \text{ions} \\ 1 \: \ce{Na} \: \text{atom} & 1 \: \ce{Na} \: \text{atom} \\ 1 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array}\]

Step 3: Solve. The nitrate ions and the chlorine atoms are unbalanced. Start by placing a 2 in front of the \(\ce{NaCl}\). This increases the reactant counts to 2 \(\ce{Na}\) atoms and 2 \(\ce{Cl}\) atoms. Then place a 2 in front of the \(\ce{NaNO_3}\). The result is:

\[\ce{Pb(NO_3)_2} \left( aq \right) + 2 \ce{NaCl} \left( aq \right) \rightarrow 2 \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)\]

Step 4: The new count for each atom and polyatomic ion becomes:

\[\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \\ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \\ 2 \: \ce{NO_3^-} \: \text{ions} & 2 \: \ce{NO_3^-} \: \text{ions} \\ 2 \: \ce{Na} \: \text{atom} & 2 \: \ce{Na} \: \text{atom} \\ 2 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array}\]

Step 5: Think about the result.

The equation is now balanced since there are equal numbers of atoms of each element on both sides of the equation. And, the coefficients are in the lowest possible ratio.

Exercise \(\PageIndex{2}\)

Fermentation is a biochemical process that enables yeast cells to live in the absence of oxygen. Humans have exploited it for centuries to produce wine and beer and make bread rise. In fermentation, sugars such as glucose (C₆H₁₂O₆) are converted to ethanol (C₂H₅OH) and carbon dioxide CO₂. Write a balanced chemical reaction for the fermentation of glucose.

Commercial use of fermentation. (a) Microbrewery vats are used to prepare beer. (b) The fermentation of glucose by yeast cells is the reaction that makes beer production possible.

Answer: C₆H₁₂O₆(s) → 2C₂H₅OH(l) + 2CO₂(g)