Video Transcript
Iron(III) sulfide Fe2S3 reacts with
oxygen to produce iron(III) oxide Fe2O3 and sulfur dioxide SO2. Write a balanced chemical equation
for this reaction, using the smallest possible whole-number coefficients for the
reactants and products.
Let’s begin by identifying the
reactants and products. Iron(III) sulfide reacts with
oxygen. The keyword “reacts” indicates that
iron(III) sulfide and oxygen are the reactants. Recall that oxygen exists as a
diatomic molecule in its pure form. A diatomic molecule is a molecule
composed of two atoms. We write the oxygen reactant as O
subscript two to represent the two oxygen atoms in the molecule. The iron(III) sulfide and oxygen
react to produce iron(III) oxide and sulfur dioxide. The keyword “produce” indicates
that iron(III) oxide and sulfur dioxide are the products.
Now that we have identified the
reactants and products, we are ready to begin balancing the chemical equation. We’ll start by making a list of the
elements which appear in the chemical equation. In the equation, we find the
elements iron, sulfur, and oxygen. Next, we will count the number of
atoms of each element on both sides of the reaction. We can make a chart to help us
organize this information. We have already begun constructing
a chart with the unbalanced chemical equation and a list of elements down the
left-hand side. We can divide the chart in half at
the arrow separating the reactants and products.
Let’s begin counting the iron
atoms. On the reactants side, we see two
iron atoms in the iron(III) sulfide. And on the products side, we see
two iron atoms in the iron(III) oxide. Now let’s count the sulfur
atoms. In the iron(III) sulfide, we see
three sulfur atoms. And in the sulfur dioxide, we see
one sulfur atom. Lastly, we’ll count the oxygen
atoms. On the reactants side, diatomic
oxygen contains two oxygen atoms. On the products side, the element
oxygen appears in both compounds. There are three oxygen atoms in the
iron(III) oxide, and there are two oxygen atoms in the sulfur dioxide.
Now that we have counted the number
of atoms of each element on both sides, we can add coefficients to the chemical
equation in order to balance the reaction. Coefficients are numerical values
which we will place in front of some or all of the molecules in the equation. The coefficient indicates the
number of molecules of each species necessary in order to balance the equation. In order for an equation to be
balanced, the number of atoms of each element must be the same on the reactant and
product sides.
Looking at our chart, we see that
we have two atoms of iron on the reactants side and two atoms of iron on the
products side. These values are equal, meaning
that the iron atoms are balanced. We see three atoms of sulfur on the
reactants side and one atom of sulfur on the products side. These values are not equal, meaning
that the sulfur atoms are unbalanced. There are two atoms of oxygen on
the reactants side and three plus two or five atoms of oxygen on the products
side. These values are not equal, meaning
that the oxygen atoms are unbalanced.
It can often be difficult to decide
where to begin when balancing a chemical equation. One helpful hint is to look for
elements which appear in the chemical equation in their pure form. Placing a coefficient in front of
the pure form of an element will only affect the number of atoms of that
element. It is often useful to the balance
these elements last as we can balance these elements without affecting the number of
atoms of any other element.
Since we will be balancing the
oxygen last and the iron is already balanced, we will begin by balancing the sulfur
atoms. We can place a coefficient of three
in front of the sulfur dioxide. This coefficient indicates that
there are three sulfur dioxide molecules. If one sulfur dioxide molecule
contains one sulfur atom and two oxygen atoms, then three sulfur dioxide molecules
will contain a total of three sulfur atoms and six oxygen atoms. There are now three sulfur atoms on
both sides of the reaction, meaning that the sulfur atoms are balanced.
By balancing the sulfur atoms, we
have changed the total number of oxygen atoms on the products side to nine. We now need to determine what
coefficient to place in front of the oxygen molecule on the reactants side in order
for there to be nine oxygen atoms on both sides of the reaction. We can set up an equation to help
us determine this coefficient. We can label the coefficient in
front of the oxygen molecule variable 𝑥. If one oxygen molecule contains two
atoms of oxygen, then 𝑥 oxygen molecules will contain two 𝑥 atoms of oxygen.
We know that for an equation to be
balanced the reactants must equal the products. This means that we can set the
number of oxygen atoms on the reactants side equal to the number of oxygen atoms on
the products side. We rearrange to solve for 𝑥 and
determine that 𝑥 is nine-halves. Placing the coefficient nine-halves
in front of the oxygen molecule gives us nine oxygen atoms on the reactants
side. There are nine oxygen atoms on both
sides of the reaction, meaning that we have balanced the oxygen atoms.
We have balanced the reaction. However, the question asks us to
balance the reaction using whole-number coefficients. We can get rid of the fraction
nine-halves and maintain a balanced chemical equation by multiplying all of the
coefficients in our equation by two. This includes the unwritten ones in
front of iron(III) sulfide and iron(III) oxide. This gives us a final balanced
chemical equation of two Fe2S3 plus nine O2 react to produce two Fe2O3 plus six
SO2.
