Balancing Chemical Reactions

Balancing reactions ensures that all atoms present in the starting materials are accounted for in the final compounds. Chemical reactions show how matter moves from one form to another. Examples are found in every scientific discipline and in everyday life, ranging from developing rocket fuel (combustion reactions) and halting rust formation (redox reactions) to curing indigestion (acid-base reactions).

Zinc and sulfur ignite upon forming zinc sulfide, forming a blue-green flame. This reaction has been used as a rocket fuel.[1]

Chemical reactions follow the law of conservation of mass, meaning that the total mass of the reactants has to be equal to the total mass of the products. The number of each type of atom must also be conserved from reactants to products. (This is known as the Principle of Atomic Conservation, or POAC). Atoms cannot be created or destroyed, so if the equation cannot be balanced, there must be another reactant or product involved. Uncovering these missing materials can shed light on how the reactions take place, how to get a better yield of the products, or what waste products may be generated. (Note:POAC is not entirely accurate due to the mass defect)

Reactions are incomplete until balanced (also known as skeletal equations). Imagine, for example, that an organic chemist is trying to invent a fuel with lower carbon emissions. The researcher knows her reactant combusts to make carbon dioxide and water. Until she writes out a balanced chemical reaction, she does not know how much carbon dioxide is being produced per unit of fuel burned. She cannot say whether her invention is better or worse than a standard fuel until she figures out the balanced reaction. Stoichiometry, which describes the relationship between masses, moles, and number of particles, is a common tool for solving chemistry problems. Any chemistry problem that uses stoichiometry or chemical equilibrium requires a balanced chemical reaction. There are a number of ways of balancing reactions.

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The hit and trial method is useful for balancing more simple equations. Many chemical reactions can be balanced through observation and intelligent guesses. For example, the combustion of glucose proceeds as follows:

\[\ce{C6H12O6 + O2 -> CO2 + H2O}\]

Carbon is only present in one reactant (glucose) and one product (carbon dioxide). There are six C's in every molecule of glucose and only one in every molecule of carbon dioxide, so using POAC, the ratio of glucose:carbon dioxide must be 1:6:

\[\ce{C6H12O6 + O2 -> 6 CO2 + H2O}.\]

Similarly, glucose has 12 hydrogen atoms to water's two:

\[\ce{C6H12O6 + O2 -> 6 CO2 + 6 H2O}.\]

Finally, balance the free element, oxygen:

\[\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}.\]

Here are some general strategies to keep in mind:

1. Start by identifying the most complex substance (the one with the most atoms) and see if there are any atoms that appear in only one reactant and one product. Balance those first.
2. Balance polyatomic ions as a unit.
3. Balance free elements (those that appear by themselves, not as part of a compound) last.
4. Avoid fractional coefficients.
5. Check your work. Make sure you have the same total number of each atom on each side of the equation. If you assume the balanced equation has one equivalent of the most complex substance, you will not always be correct.

Try balancing the combustion reaction of magnesium and oxygen.

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In this reaction, magnesium reacts with oxygen to give magnesium oxide:

\[\ce{Mg + O2 -> MgO}.\]

This equation is unbalanced because there are 2 atoms of oxygen on the reactants side and only one oxygen in the compound \(\ce{MgO}.\)

This can be balanced by putting a \(2\) before \(\ce{MgO}\), which means the whole compound's individual atoms are taken twice. So, our equation now becomes

\[\ce{Mg + O2 -> 2MgO}.\]

Note that now there are \(2 \ce{Mg}\) on the products side and only one in the reactants side. Then we put a \(2\) before \(\ce{Mg}\) in reactants side:

\[\ce{2Mg + O2 -> 2MgO}.\]

Now, the equation is balanced! \(_\square\)

Balancing the combustion of methylamine:

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\[\ce{CH_3NH_2 + O2 -> H2O + CO2 + N2}\]

On inspection, we see that methylamine is the most complex substance and that there is only one nitrogen-containing compound on both the reactant side and the product side, so we start by balancing nitrogen:

\[\ce{2CH_3NH_2 + O2 -> H2O + CO2 + N2}.\]

With the nitrogens balanced, we can balance the carbons and hydrogens. If we have 2 C's on the reactant side, we need 2 moles of carbon dioxide. We also have 10 hydrogens, which equates to 5 moles of water on the products side:

\[\ce{2CH_3NH_2 + O2 -> 5H2O + 2CO2 + N2}.\]

Now, we have to balance the oxygen atoms. The product side has 9 atoms while the reactant side has 2:

\[\ce{2CH_3NH_2 + 9/2 O2 -> 5H2O + 2CO2 + N2}.\]

Rewritten to avoid the fractional coefficient, we get

\[\ce{4CH_3NH_2 + 9 O2 -> 10H2O + 4CO2 + 2N2}.\]

Checking our work, the reactants side has 4 C's, 20 H's, 4 N's, and 18 O's, as does the products side. \(_\square\)

In this method, hydroxide ions, hydrogen ions, electrons, and water are added to balance the equation. The equation is divided into an oxidation half and a reduction half, and each half is solved separately. Once both half-reactions are balanced, they are added back together. Hydrogen and oxygen are balanced differently in an acidic medium versus a basic medium.

Acidic Medium Steps:

1. Identify oxidation and reduction.
2. Split complete reaction into two halves.
3. Balance all the atoms except oxygen and hydrogen.
4. Balance hydrogen atoms: add \(\ce{H^+}\) ions on the side deficient in hydrogen atoms.
5. Balance oxygen atoms: add water molecules to the side deficient in oxygen atoms and twice the \(\ce{H^+}\) ions to the opposite side.
6. Balance charge by adding electrons to the equation.
7. Make electron change equal in both half reactions by multiplying with a suitable integer.
8. Add both the half reactions to get complete balanced reactions. All electrons must cancel out in the balanced reaction.

Balance the following reaction:

\[\ce{H2S + NO3- -> HSO_4- + NH4+}.\]

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Oxidation half:

\[\begin{align} \ce{H2S} &\longrightarrow \ce{HSO4-}\\ \ce{H2S + 4H2O} &\longrightarrow \ce{HSO4- + 9H+}\\
\ce{H2S + 4H2O} &\longrightarrow \ce{HSO4 + 9H+ + 8e-}. \end{align}\]

Reduction half:

\[\begin{align} \ce{NO3-} &\longrightarrow \ce{NH4+}\\
\ce{NO3- + 10H+} &\longrightarrow \ce{NH4+ + 3H2O}\\
\ce{NO3- + 10H+ +8e-} &\longrightarrow \ce{NH4+ + 3H2O}. \end{align}\]

Adding both reactions, we have

\[\ce{H2S + NO3- + H2O + H+ -> HSO4- + NH4+}.\]

Basic Medium Steps:

All the steps are the same except balancing hydrogen and oxygen.

1. Balance oxygen atoms: add water molecules to the side with excess oxygen and twice the \(\ce{OH^-}\) ions to the opposite side.
2. Balance hydrogen atoms: add \(\ce{OH^-}\) to the side having excess \(\ce{H^+}\) and an equal number of water molecules to the opposite side.

Balance the following reaction:

\[\ce{Al + H2O -> Al(OH)4- + H2}.\]

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Oxidation half:
The oxidation state of aluminum changes from 0 in the elemental form to a +3 ion:

\[\begin{align} \ce{Al} &\longrightarrow \ce{Al(OH)4-}\\ \ce{Al + 4OH-} &\longrightarrow \ce{Al(OH)4- + 3e-}. \end{align}\]

Reduction half:
Hydrogen is reduced from +1 to 0:

\[\begin{align} \ce{3/2 H2O} &\longrightarrow \ce{3/2 H2}\\ \ce{3 H2O +3e-} &\longrightarrow \ce{3/2 H2 + 3 OH-}. \end{align}\]

Adding both reactions, we have

\[\ce{Al + 4 OH- + 3 H2O + 3e- -> Al(OH)4- + 3/2 H2 + 3 OH- + 3e-}.\]

Since hydroxide appears on both the products and reactants sides, it can be cancelled out:

\[\ce{Al + OH- + 3 H2O + 3e- -> Al(OH)4- + 3/2 H2 + 3e-}.\]

Our electrons are balanced, so we can remove them from the equation as well:

\[\ce{Al + OH- + 3 H2O -> Al(OH)4- + 3/2 H2}.\]

After removing the fractional coefficient, we arrive at our balanced equation:

\[\ce{2 Al + 2 OH- + 6 H2O -> 2Al(OH)4- + 3 H2}.\ _\square\]

The N-factor method can be used to solve complicated reactions that seem difficult to balance on first inspection, including redox reactions. The \(n\)-factor, also called the activity coefficient, is the number of moles of electrons supplied or used per mole of the chemical substance. The number of moles of the two compounds is the inverse of the compound's \(n\)-factor ratio. Once the coefficients of two compounds are known, the rest of the equation can be balanced using POAC.

Balance the following reaction:

\[\ce{As_2S_3}+\ce{NO_3^-}+\ce{H^+}+\ce{H_2O}\longrightarrow \ce{H_3AsO_4}+\ce{NO}+\ce{S}.\]

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First, take variables to represent the stoichiometric coefficients of all the products and reactants:

\[a\ce{As_2S_3}+b\ce{NO_3^-}+c\ce{H^+}+d\ce{H_2O} \longrightarrow p\ce{H_3AsO_4}+q\ce{NO}+r\ce{S}.\]

Then find the \(n\)-factors for \(\ce{NO_3^-}\) and \(\ce{As_2S_3},\) which come to be \(3\) and \(10,\) respectively. Hence \(a=10\) and \(b=3\).

Now, applying POAC, we have

1. for \(\ce{As}\), \(2a=p\implies p=6\)
2. for \(\ce{S}\), \(3a=r\implies r=9\)
3. for \(\ce{N}\), \(b=q\implies q=10\)
4. for \(\ce{H}\), \(c+2d=3p\implies c=10\)
5. for \(\ce{O}\), \(3b+d=4p+q\implies d=4.\)

Hence we have the balanced equation. \(_\square\)

[1] Image from https://commons.wikimedia.org/wiki/File:Thereactionofzincand_sulfur.jpg under Creative Commons licensing for reuse and modification.

[2] Image from https://commons.wikimedia.org/wiki/File:H3PO4balancingchemicalequationphosphoruspentoxideandwaterbecomesphosphoricacid.gif under Creative Commons licensing for reuse and modification.