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 before 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 substance.
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 substances 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", (Video 1-22). There are many more videos of various modifications of this experiment, which can be easily found on the Internet.