We begin by introducing the Law of Conservation Matter, very important when studying chemical reactions.
The Law of Conservation of Matter states that no matter can be introduced nor removed from some medium under any circumstances.
In other words, all matter that is present at one time is present at all other times, and no additional matter will be found. In the context of chemical reactions, it is important to keep that in mind, because a Carbon atom cannot suddenly disappear during a reaction, even though some chemical equations imply this. For example, consider this single-replacement reaction where Titanium (II) Perchlorate and solid Chromium react to produce Chromium (III) Perchlorate and solid Titanium:
$$Ti(ClO_4)_2 + Cr \rightarrow Cr(ClO_4)_3 + Ti$$
If you look at the coefficients of the perchlorate ions (\(ClO_4^{-}\)) on the two sides of the equation, you will notice that the reaction implies that an additional perchlorate ion appears out of nowhere during the reaction. However, this violates the Law of Conservation of Matter. When an equation does this, it is known to be unbalanced, because the number of each atom on the two sides of the equation are not the same. An equation that does not violate the Law of Conservation of Matter is called balanced.
Having a balanced equation is extremely important in many other topics in chemistry, and is one of its fundamental topics that is part of any first-year high school chemistry course.
The process of balancing synthesis and decomposition reactions is relatively straightforward since one side of the equation will usually only have one substance present. However, the process itself is best illustrated with examples. Pay careful attention to the rationale for each step; it will help you when practicing problems on your own.
Example 1: Balance this chemical equation:
$$Mn(CO_3)_2 \rightarrow MnO_2 + CO_2$$
Solution: The Manganese is already balanced. However, the left side has two Carbon and six Oxygen, while the right side has one Carbon and four Oxygen. The Carbon only appears once on each side of the equation, make those coefficients match by placing a "2" in front of Carbon Dioxide:
$$Mn(CO_3)_2 \rightarrow MnO_2 + 2CO_2$$
Both sides have two Carbon atoms and six Oxygen atoms; the number of Manganese atoms remained unchanged. Therefore the equation is now balanced.
As shown in the previous example, changing a single coefficient can change the quantity of atoms of multiple types on one side of the equation. Therefore, it is strongly advised to double-check the equations by doing an atom count. This is self-explanatory, and it means to check and make sure both sides of the equation have the same number of each atom, guaranteeing that the Law of Conservation of Matter is satisfied.
Also, remember that a decomposition reaction is a synthesis reaction in reverse, and so the two types of equations are balanced in the same way. The next example illustrates this.
Example 2: The reverse of the reaction in Example 1 is balanced in the same way. Therefore the equation
$$MnO_2 + CO_2 \rightarrow Mn(CO_3)_2$$
can be balanced this way, with the same procedure as in Example 1:
$$MnO_2 + 2CO_2 \rightarrow Mn(CO_3)_2$$
A new example of balancing a synthesis reaction's chemical equation will now be shown so the full process can be reemphasized.
Example 3: Balance this [very common] reaction:
$$H_3O^+ + OH^- \rightarrow H_2O$$
Solution: The left side of the equation has four Hydrogen atoms and two Oxygen atoms, while the right side of the equation has two Hydrogen atoms and one Oxygen atom. Therefore, the difference is a common factor of 2, which can easily be fixed by placing a “2” in front of \(H_2O\):
$$H_3O^+ + OH^- \rightarrow 2H_2O$$
Example 4: Balance the chemical equation for the reaction where Nitrogen trioxide decomposes into Nitrogen dioxide and Oxygen gas.
Solution: The equation itself is not given to us. Here are the chemical formulas for the chemicals involved:
Nitrogen trioxide: \(NO_3\)
Nitrogen dioxide: \(NO_2\)
Oxygen gas: \(O_2\)
Therefore, this is the chemical equation:
$$NO_3 \rightarrow NO_2 + O_2$$
Even though the Nitrogen atoms are already balanced, the oxygens cannot be balanced in the case where the coefficients of \(NO_3\) and \(NO_2\) are 1 because the number of Oxygen atoms cannot equal 3. Depending on the coefficient of \(O_2\), the number of Oxygen atoms on the right side of the equation could be 4, 6, 8, and so on. To fix this, multiply the entire equation by 2 in order to work with even numbers:
$$2NO_3 \rightarrow 2NO_2 + 2O_2$$
The left side of the equation has six Oxygen atoms, while the right side has eight. However, that can be fixed easily by changing the coefficient of Oxygen gas from "2" to "1":
$$2NO_3 \rightarrow 2NO_2 + O_2$$
Single replacement reaction equations are typically relatively easy to balance because not many chemicals are involved. It is often helpful to treat a polyatomic ion like a regular element since it will not likely be broken up during the reaction.
Example 5: Balance this chemical equation:
$$Al(NO_3)_3 + Mg \rightarrow Mg(NO_3)_2 + Al$$
Solution: Initially the only elements that are not balanced are Nitrogen and Oxygen, and the ratio of these two atoms is consistent because they are found together in the Nitrate ion and nowhere else. However, it is not valid to balance it with
$$Al(NO_3)_3 + Mg \rightarrow \frac{3}{2}Mg(NO_3)_2 + Al$$
because atoms cannot be found in pieces. Instead, both coefficients must be changed to ensure the same number of Nitrate ions are on both sides of the equation. 3 is not evenly divisible by 2, so we need a least common multiple (to ensure integer coefficients), which is 6. Remember that this is not the coefficient on both sides of the equation, but the number of Nitrate ions on each side of the equation. Therefore, we add the following coefficients:
$$2Al(NO_3)_3 + Mg \rightarrow 3Mg(NO_3)_2 + Al$$
Now the Aluminum and Magnesium are not balanced, but that can be fixed easily by adding two more coefficients:
$$2Al(NO_3)_3 + 3Mg \rightarrow 3Mg(NO_3)_2 + 2Al$$
Here is one additional example that looks slightly more complicated, but is still a single replacement reaction.
Example 6: Balance this chemical equation:
$$MgTi(OH)_4 + O^{2-} \rightarrow MgTiO_2 + OH^-$$
Solution: The Magnesium and Titanium are already balanced, but the Oxygen and Hydrogen are not. Notice that there are 4 hydroxide ions on the left side of the equation and only one on the right side of the equation. Fix that by making the coefficient for the hydroxide ion a "4":
$$MgTi(OH)_4 + O^{2-} \rightarrow MgTiO_2 + 4OH^-$$
Similarly, if we ignore the Oxygen present in the hydroxide ions, there is one oxygen on the left side and two on the right side. Just make the coefficient for the oxide ion "2" so the number of oxygen atoms on both sides of the equation is the same:
$$MgTi(OH)_4 + 2O^{2-} \rightarrow MgTiO_2 + 4OH^-$$
Double replacement reactions in general are more complicated than single replacement reactions, and that goes for balancing them as well. However, they are not much more difficult when you realize a double replacement reaction is very similar to a single replacement reaction.
Example 7a: In Example 6, the following equation was balanced:
$$MgTi(OH)_4 + O^{2-} \rightarrow MgTiO_2 + OH^-$$
However, the oxide and hydroxide ions are not generally found naturally without a cation. The cation, in theory, can be anything. For example, if the cation is Sodium, the chemical equation becomes
$$MgTi(OH)_4 + Na_2O \rightarrow MgTiO_2 + NaOH$$
Not only is the reaction more realistic, but it is a double replacement reaction instead of a single replacement reaction.
The next part of the example is a follow-up.
Example 7b: Balance the chemical equation written in Example 7a.
Solution: For reference, the equation will be written again here:
$$MgTi(OH)_4 + Na_2O \rightarrow MgTiO_2 + NaOH$$
Notice that Hydrogen only appears with Oxygen in the form of Hydroxide ions, so Hydroxide can be treated as an individual element when balancing, rather than a molecule. In fact, from here on out refer to the hydroxide ion as \(X\). Rewrite the chemical equation:
$$MgTiX_4 + Na_2O \rightarrow MgTiO_2 + NaX$$
Balance \(X\) by adding a "4" in front of \(NaX\):
$$MgTiX_4 + Na_2O \rightarrow MgTiO_2 + 4NaX$$
Now balance the sodium by adding a "2" in front of Sodium Oxide:
$$MgTiX_4 + 2Na_2O \rightarrow MgTiO_2 + 4NaX$$
The coefficient changes have also simultaneously balanced the Oxygen, so the equation is now fully balanced (do an atom count here). Lastly, substitute hydroxide back in for \(X\):
$$MgTi(OH)_4 + 2Na_2O \rightarrow MgTiO_2 + 4NaOH$$
The above chemical equation, once balanced, had the same coefficients as example 6's balanced equation. This is not a coincidence, because the addition of the extra cation is still in the same quantity before and after the reaction, by the Law of Conservation of Matter.
Here is one more example of balancing a double-replacement reaction.
Example 8: Balance this chemical equation:
$$AlPO_4 + Na_2CO_3 \rightarrow Al_2(CO_3)_3 + Na_3PO_4$$
Solution: Multiple types of atoms are not balanced. As usual, the strategy is to balance one at a time, beginning with the chemical that you will least likely need to adjust twice. Start with Aluminum Carbonate. Balance the Aluminum atoms on each side by changing the coefficient of the other chemical with Aluminum (least common multiples are not needed here):
$$2AlPO_4 + Na_2CO_3 \rightarrow Al_2(CO_3)_3 + Na_3PO_4$$
Now balance the other chemical with Carbonate:
$$2AlPO_4 + 3Na_2CO_3 \rightarrow Al_2(CO_3)_3 + Na_3PO_4$$
Lastly, both the Sodium and the Phosphate will be balanced by placing a "2" in front of Sodium Phosphate:
$$2AlPO_4 + 3Na_2CO_3 \rightarrow Al_2(CO_3)_3 + 2Na_3PO_4$$
As usual, do an atom count here to catch algebra mistakes.
Some chemical equations are balanced at the start. In other words, no coefficients need to be manipulated because they are all "1" in the balanced equation. Some instances of these will be illustrated here.
Example 9: When Nitric acid reacts with Lithium Hydroxide, Lithium Nitrate and water are produced. Here is the chemical equation:
$$HNO_3 + LiOH \rightarrow H_2O + LiNO_3$$
This equation is already balanced (to see this, it is helpful to treat \(NO_3\) as an individual atom). Doing an atom count verifies that all the coefficients should be "1".
Example 10: Copper (II) Fluoride and solid Manganese react to form solid Copper and Manganese (II) Fluoride. The chemical equation is below:
$$CuF_2 + Mn \rightarrow MnF_2 + Cu$$
Fluorine, Copper, and Manganese are all already balanced (do an atom count), so none of the coefficients need to be changed; they all remain "1".
Electrons are one form of matter on our Earth. Since electrons influence charge, the total charge of a group of substances will remain consistent even after a reaction because the number of protons and electrons must remain constant. That gives us the Law of Conservation of Charge:
The Law of Conservation of Charge states that before and after a reaction, the total charge of all substances involved must remain the same (even if it means lone electrons are left not attached to any atoms).
While this rule is very important in general, it can be very helpful in balancing chemical equations easier, particularly the chemical equations for redox reactions.
Example 11: Balance this redox reaction:
$$Mn^{2+} + Li \rightarrow Li^{+} + Mn$$
Solution: By the Law of Conservation of Charge, the charge on both sides of the equation must be the same. 2 is divisible by 1, so we can leave the Manganese alone and just change the coefficients of the Lithium. If the coefficients of \(Li\) and \(Li^{+}\) are both changed to “2,” the atoms remain balanced and both sides of the equation now have a total charge of +2.
$$Mn^{2+} + 2Li \rightarrow 2Li^{+} + Mn$$
Here is one other example:
Example 12: Consider this chemical equation that results from the products of a combination of two other reactions:
$$H^{+} + PO_4^{3-} \rightarrow H_3PO_4$$
Balance this equation.
Solution: Again the Law of Conservation of Charge is helpful. Once the equation is balanced, the charge of both sides must be 0 because there is only one chemical on the right side, and it has a neutral charge. Therefore the left side must also have 0 charge. If the Hydrogen and Phosphate ions are in a 1-1 ratio, \(HPO_4^{2-}\) will form, which has the wrong charge. However, lone Hydrogen ions are one of the reactants, so we can just add enough Hydrogen to get a neutral charge. In this case, we need two more, giving us \(H_3PO_4\), just as the right side of the equation says. Therefore, the coefficient of the Hydrogen ions is “3,” and everything else stays the same.
$$3H^{+} + PO_4^{3-} \rightarrow H_3PO_4$$
Sometimes missteps in balancing an equation can result in an equation with coefficients that all have a common multiple. If this is the case, the equation should be divided by the greatest common factor of all the coefficients in order to reduce them.
Example 13: Reduce the coefficients in this equation:
$$2H_2SO_4 + 4AgOH \rightarrow 4H_2O + 2Ag_2SO_4$$
Solution: There is a common factor of 2 in every coefficient. Therefore, we can divide the wjole equation by “2,” because no fractional coefficients will be introduced:
$$H_2SO_4 + 2AgOH \rightarrow 2H_2O + Ag_2SO_4$$
Doing an atom count verifies that this equation is now balanced.
Example 14: Do the coefficients need to be reduced in this equation?
$$3H_2Cr_2O_7 + 2Al(OH)_3 \rightarrow Al_2(Cr_2O_7)_3 + 6H_2O$$
Solution: The four coefficients (including the one for Aluminum Dichromate, \(Al_2(Cr_2O_7)_3\), which is "1") do not have a common factor, so the coefficients can not be reduced further.
Although combustion reactions are a type of redox reaction, they are balanced quite differently from regular redox reactions.
Example 15: Balance this combustion reaction:
$$C_5H_{12} + O_2 \rightarrow CO_2 + H_2O$$
Solution: It is best to begin by balancing the Carbon, because it only appears once on each side of the equation. Add a "5" in front of Carbon Dioxide:
$$C_5H_{12} + O_2 \rightarrow 5CO_2 + H_2O$$
Now balance the Hydrogens similarly:
$$C_5H_{12} + O_2 \rightarrow 5CO_2 + 6H_2O$$
There are 16 oxygen on the right side, so adjust the \(O_2\) coefficient accordingly:
$$C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O$$
That was for a hydrocarbon. Let's do one for combusting a carbohydrate.
Example 16: Balance this [similar] combustion reaction:
$$C_5H_{10}O + O_2 \rightarrow CO_2 + H_2O$$
Solution: Begin by noting that this equation is slightly trickier to balance than the one in Example 14 because there is oxygen in all the reactants and products. Balance the Carbon and Hydrogen in the same way as before:
$$C_5H_{10}O + O_2 \rightarrow 5CO_2 + 5H_2O$$
There are currently 3 oxygen on the left side and 15 on the right side. Since \(O_2\) is present, and both sides having an odd number of oxygens (as opposed to one side having an odd number and one side having an even number, in which case we need to multiply the equation by 2), we can just adjust the coefficient of \(O_2\) and leave the others alone. If the coefficient is "7," there will be 14 oxygen atoms in elemental form, plus one from the hydrocarbon, making for a total of 15. Therefore the coefficient is "7," and the equation is balanced.
$$C_5H_{10}O + 7O_2 \rightarrow 5CO_2 + 5H_2O$$
In conclusion, there are many types of reactions you will come across, and many techniques to utilize when balancing them. This is a topic that requires practice to master, because there are no real formulas, only techniques that you must choose when to apply and then execute them.