1.8.6. Balancing Simple Chemical Equations.
In most of the equations presented in Figures 1-22, 1-23, and 1-25, we see non-subscript numbers in front of certain chemical formulas. Some of these equations are repeated below with those numbers in red. 2 Al + 3 I2 = 2 AlI3
2 H2 + O2 = 2 H2O
2 Na + 2 H2O = 2 NaOH + H2
The emphasized non-subscript numbers in the three equations above are called the coefficients
. Coefficients indicate the stoichiometry of a chemical reaction, the ratio of reagents and products involved in a chemical transformation. The equation 2 Al + 3 I2
= 2 AlI3
tells us that two aluminum atoms (Al) react with three molecules of iodine (I2
) to give two molecules of aluminum iodide (AlI3
From the next equation, 2 H2
= 2 H2
O, we gather that two molecules of hydrogen (H2
) react with one molecule of oxygen (O2
) to produce two molecules of water (H2
O). Note that if the coefficient is 1, no number is placed in front of the corresponding formula in the equation.
A correct chemical equation is balanced
, which means that the total number of atoms of the same type on the left side and on the right side of the equation must be the same. This requirement follows from the law of conservation of mass, which states that in chemical reactions atoms are neither created nor destroyed, but are only redistributed to form new molecules.
Let us begin to learn how to balance chemical equations using a very simple reaction as an example, the reaction between oxygen (O2
) and hydrogen (H2
). When balancing a reaction of O2
, it is always a good idea to start by balancing the number of oxygen atoms. Step 1.
We know that H2
react to give water, H2
O. We express that in the form of a chemical equation. The equation is not balanced though. H2 + O2 = H2O
(H balanced; O not balanced; overall not balanced
) Step 2.
The originally written equation is not balanced. It is balanced for hydrogen (two H atoms on the left and two on the right), but not for oxygen. The number of oxygen atoms on the left is two, but on the right is only one. Important:
the number of oxygen atoms on the left is even and will always be even because a molecule of oxygen is diatomic
(consists of two atoms). No matter what coefficient we place before the O2
in the equation, the number of O atoms on the left will be even. Consequently, the number of O atoms on the right side of the final balanced
equation must be even, too. Currently the number is odd because there is one molecule of H2
O on the right, and a molecule of H2
O contains one O atom, an odd number. We can make the number of O atoms on the right even by placing a coefficient 2 in front of the H2
O. Let us do that. As a result, the oxygen is now balanced. But the hydrogen is not. H2 + O2 = 2 H2O
(H not balanced; O balanced; overall not balanced
) Step 3.
Now we count the number of H atoms on the left and on the right of the equation just above. There are two H atoms on the left and four on the right, twice as many. To balance the hydrogen, we just have to put the coefficient 2 in front of the H2
on the left. 2 H2 + O2 = 2 H2O
(H balanced; O balanced; overall balanced
Finally, the equation is fully balanced. To double-check, we should compare the number of H atoms and O atoms on the left and on the right of the equation. There are four H atoms on the left and four H atoms on the right. There are two O atoms on the left and two on the right. All is well, the equation is balanced. Important:
When balancing chemical equations we can change only the coefficients, the numbers before chemical formulas in the equation. The subscripts in the formulas must never be changed
for balancing chemical equations because the subscripts specify the number of atoms in a molecule or, in other words, its composition, which is constant for a particular compound.
In a correctly balanced equation, the coefficients should be minimal. For example, although the equation 4 H2
+ 2 O2
= 4 H2
O is balanced, it can and should be simplified by dividing all of the coefficients by two, which gives: 2 H2
= 2 H2
Let us practice a bit more by balancing the reaction of sodium (Na) with water (H2
O) to give sodium hydroxide (NaOH) and hydrogen (H2
). Step 1.
Knowing that Na reacts with H2
O to give NaOH and H2
, we write the equation. Na + H2O = NaOH + H2
(Na balanced; O balanced; H not balanced; overall not balanced
) Step 2.
The original equation (Step 1 above) is balanced for Na and O, but not for H.
We notice that the number of H atoms on the left will always be even because a molecule of H2
O contains two H atoms. Therefore, in the balanced equation the number of H atoms on the right must also be even. There are two hydrogen-containing compounds on the right. One is H2
with an even number of H atoms. The other is NaOH with just one H atom, which is an odd number. Therefore, we should make the number of molecules of NaOH even. Let us make it 2 and see what happens. Na + H2O = 2 NaOH + H2
(Na not balanced; O not balanced; H not balanced; overall not balanced
) Step 3.
In the equation modified in Step 2, not a single element is balanced. No sweat. Sometimes things get worse before they get better. Just think logically. What is important is that there are even numbers of H atoms on both sides of the equation, four on the right and two on the left. To balance the hydrogen, we put 2 in front of the formula of water. Interestingly, this balances not only the hydrogen but also the oxygen. Na + 2 H2O = 2 NaOH + H2
(Na not balanced; O balanced; H balanced; overall not balanced
) Step 4.
The only remaining problem is that there are two Na atoms on the right but only one on the left. All we need to do is just put 2 in front of the Na in the left side of the equation. The equation is now fully balanced. 2 Na + 2 H2O = 2 NaOH + H2
(Na balanced; O balanced; H balanced; overall balanced
Finally, let us balance one more equation, the one for the reaction between aluminum and iodine.
Write the unbalanced original equation. Al + I2 = AlI3
(Al balanced; I not balanced; overall not balanced
) Step 2.
We note that the number of iodine atoms on the left will always be even because a molecule of I2
is diatomic. On the right side, however, the number of I atoms is odd (three). We make it even by placing 2 before the formula of AlI3
. Al + I2 = 2 AlI3
(Al not balanced; I not balanced; overall not balanced
) Step 3.
Now that the number of I atoms on both sides is even, we can balance the iodine. There are two I atoms on the left and 6 on the right. By placing 3 before the I2
formula, we get the iodine balanced. Al + 3 I2 = 2 AlI3
(Al not balanced; I balanced; overall not balanced
) Step 4.
All we have to do is balance the aluminum. There are two Al atoms on the right but only one on the left. We put 2 before the Al on the left to have the equation fully balanced. 2 Al + 3 I2 = 2 AlI3
(Al balanced; I balanced; overall balanced
Done. 1.8.7. Performing Selected Reactions of Different Types at Home.
If you decide not to do the experiments, please still read this subsection. Experiment 5. Decomposition of Hydrogen Peroxide (H2O2) in the Presence of Baker's Yeast.
This reaction is used in a cool experiment titled "Elephant Toothpaste", which you can watch here
. There are many more videos of various modifications of this experiment, which can be easily found on the Internet.
We will do a similar experiment using a 3% solution of hydrogen peroxide in water, which is sold in pharmacies and grocery stores. As the decomposition occurs, oxygen gas (O2
) bubbles off (Figure 1-23, top equation). At room temperature, the decomposition of H2
, especially in such a dilute solution is extremely slow, but we can speed it up by using a catalyst
. As briefly mentioned earlier, a catalyst is a substance that accelerates a chemical reaction. An excellent catalyst for the decomposition of hydrogen peroxide is manganese dioxide (MnO2
). If you happen to have a tiny bit of MnO2
, you can use it. Alternatively, you may get this compound from a zinc-carbon battery, as shown in this video
. Just keep in mind that (a) to get MnO2
you need a zinc-carbon battery, one of those marked "Heavy Duty" or "Super Heavy Duty" and (b) the manganese dioxide that you get from such batteries is not very pure, but still can be used as a catalyst to decompose H2
There is, however, another excellent catalyst for decomposition of H2
, and that catalyst is likely available from your kitchen. It is… baker's yeast! Baker's yeast contains the enzyme catalase
, a huge and very complex biomolecule
that accelerates the decomposition reaction of hydrogen peroxide very efficiently. It is baker's yeast that we will use in our experiment described below.
Place about ¼ teaspoon of dry baker's yeast at the bottom of a highball glass. Slowly pour 3% hydrogen peroxide onto the yeast and observe the formation of bubbles as the two get in contact. The gas that bubbles off is pure oxygen, the same gas that is present in air and that is vital to life. We breathe in air to get oxygen, which is needed for biological processes constantly occurring in our bodies. Also, it is oxygen that is needed for combustion. Combustion in pure oxygen is much more efficient than in air, because air contains only about 20% O2
How do we know that the gas produced in the decomposition of hydrogen peroxide is oxygen? A good simple test for pure O2
is a smoldering wooden toothpick or coffee stirrer. When exposed to pure oxygen, the sluggishly smoldering areas immediately flare up and the wood burns rapidly and brightly, as demonstrated in this video
If you would like to do the Elephant Toothpaste version of the H2
decomposition experiment and are not afraid of getting in trouble for the mess created in your kitchen or bathroom, go right ahead and have fun! You will just need to modify the conditions a bit by adding a small amount of liquid soap and a food color to the yeast and stirring the mixture prior to the addition of hydrogen peroxide. Note that usually in the Elephant Toothpaste experiment a more concentrated than 3% solution of H2
is employed. So, do not get too disappointed if the Elephant Toothpaste is produced less vigorously in your experiment than in the ones that are shown in the videos on the Internet.Experiment 6. Decomposition of Ammonium Dichromate, (NH4)2Cr2O7: The Chemical Volcano.
Watch this demonstration
of a truly spectacular decomposition reaction. The reaction is the self-sustaining decomposition of ammonium dichromate, (NH4
, to chromium oxide (Cr2
), nitrogen gas (N2
) and water (H2
O). Here is the equation. (NH4)2Cr2O7 = Cr2O3 + N2 + 4 H2O
If you have watched the video demonstration, you understand why this experiment is often called the chemical volcano. The dark-green fluffy solid formed in the decomposition of the bright-orange ammonium dichromate is chromium oxide, Cr2
. The reaction also produces nitrogen gas and water in the form of steam, which give the process the striking "volcano" effect. If you would like to do this experiment yourself, you need to obtain ammonium dichromate, which is sold in some online stores.
As always, we need to know well the chemicals we work with. None of the three reaction products (N2
) pose danger. Nitrogen constitutes about 80% of the air that we breathe. Water is water. The chromium product (Cr2
) is a stable green solid that is widely used as a pigment for paints and ink and as a polishing material conventionally referred to as "Green Polishing Compound
". Now, read carefully. In contrast with the reaction products, the starting material, ammonium dichromate, is a carcinogen and an irritant. I remind you that, as clearly stated in the Disclaimer above, you use any of the experiments discussed, listed, described, or otherwise mentioned in this course at your own risk. Among all materials used for the experiments described in this course, ammonium dichromate is the chemical that requires particularly careful handling. Avoid inhalation and any contact with this chemical. Wear latex or rubber gloves when doing the experiment.
I recommend that the chemical volcano experiment be performed outdoors. While you should decide on how much ammonium dichromate you would like to use for the reaction, I would suggest a quantity between 20 and 50 g. Pour ammonium dichromate onto a large ceramic tile or a metal sheet or a sand tray. Do not use a tray made of combustible materials such as cardboard, wood or plastic. It is also a good idea to
lay out a tarp or sheets of grocery plastic bags around the tile. In this way, it will be easy to remove and dispose of the fluffy Cr2
"volcano ash" produced in the reaction. And there will be a lot of it produced! Make a cone pile of (NH4
on the tray. Ignite the tip of the pile with a kitchen lighter or a long reach match. Step back and watch the volcano erupt. After the reaction has gone to completion give the hot dark green "ashes" a few minutes to cool down before you remove them. Experiment 7. Substitution Reaction of Silver Nitrate (AgNO3) with Copper (Cu): The Silver Tree.
Look at the photo in Figure 1-27 and watch a video demonstration
of this beautiful experiment. If after watching the video you would like to grow your own silver tree, you will need silver nitrate, a piece of copper wire, and distilled water for that.