Video Transcript
The Marsh test is a method used
historically for the detection of arsenic, a toxic element, in mixed materials. The sample is burned to convert any
arsenic present to arsenic trioxide, As2O3, and then treated with nitric acid, HNO3,
and metallic zinc. If arsenic is present, a garlic
smell is detected due to the production of arsine gas, AsH3. The equation for this reaction is
as follows: As2O3 plus 𝑎 Zn plus 𝑏 HNO3 react to produce 𝑐 AsH3 plus six Zn(NO3)2
plus 𝑑 H2O. The coefficients 𝑎, 𝑏, 𝑐, and 𝑑
are all whole numbers. Find the value of 𝑎. Find the value of 𝑏. Find the value of 𝑐. Find the value of 𝑑.
𝑎, 𝑏, 𝑐, and 𝑑 are
coefficients. Coefficients, or psychometric
coefficients, are numerical values written before a species in a chemical equation
in order to balance the reaction. We can see in the reaction equation
provided that a coefficient of six appears before zinc nitrate. And as no value is written in front
of arsenic trioxide, we can assume that the coefficient is one. Our goal is to determine what
values for 𝑎, 𝑏, 𝑐, and 𝑑 will balance this reaction. A chemical equation is balanced
when the number of atoms of each element is the same on both sides of the
reaction. Balancing a chemical equation is a
guess-and-check process that requires some space. So let’s rewrite the important
parts of this question and focus on one coefficient at a time.
To begin balancing this chemical
equation and find the value of 𝑎, we can make a list of the number of atoms of each
element on both sides of the reaction. One helpful hint is to look for
polyatomic ions that appear on both sides of the reaction arrow. If a polyatomic ion appears on both
sides of the reaction, we can treat this ion as a unit. In this equation, we see the
nitrate ion on both sides of the reaction arrow. This means that when we make our
list, we can keep the nitrate ion as a unit instead of separating it into nitrogen
atoms and oxygen atoms. We should also notice that there
are oxygen atoms involved in this reaction that are not a part of the nitrate
ions. So we’ll need to take special care
when we count the oxygen atoms. With this in mind, let’s construct
our list.
Our list should include arsenic,
oxygen, zinc, hydrogen, and nitrate ions. There are two atoms of arsenic on
the reactants side and only one atom of arsenic on the products side. Let’s skip over counting the oxygen
atoms for now. There is one atom of zinc on the
reactants side. There’s one zinc atom per unit of
zinc nitrate. Since there are six units, there is
a total of six atoms of zinc on the products side. On the reactants side, there is one
atom of hydrogen. And on the products side, each
molecule of arsine gas has three atoms of hydrogen and each molecule of water has
two atoms of hydrogen.
Now we can count the number of
nitrate ions. When looking at the reactants side,
we might be tempted to say that there are three nitrate ions because we see a
subscript three. But we must remember that each
nitrate ion contains one nitrogen atom and three oxygen atoms. So one molecule of nitric acid only
contains one nitrate ion, not three. On the products side, the subscript
two outside of the parentheses indicates that each unit of zinc nitrate contains two
nitrate ions. However, six units of zinc nitrate
are required to balance this chemical equation. This means that there is a total of
12 nitrate ions on the products side.
Now we can count the oxygen
atoms. We’ll only count the atoms of
oxygen that are not a part of the nitrate ions. This means that there are three
atoms of oxygen on the reactants side and only one atom of oxygen on the products
side. In comparing the total number of
atoms and polyatomic ions on both sides of the reaction, we can see that none of the
species are balanced. Now we can add coefficients in
front of any species in the reaction to balance the atoms or ions. As we saw with the coefficient six
in front of zinc nitrate, the number of atoms and/or ions of each type in the
species will be multiplied by the coefficient. Let’s begin by finding the value of
𝑎.
Coefficient 𝑎 represents the
number of zinc atoms on the reactants side. There are six zinc atoms on the
products side. If we set coefficient 𝑎 to six,
then there will be one times six zinc atoms on the reactants side and the zinc atoms
will be balanced. So the value of 𝑎 is six. Now let’s find the value of 𝑏. Coefficient 𝑏 represents the
number of nitric acid molecules on the reactants side. As nitric acid contains both
hydrogen atoms and nitrate ions, changing the value of 𝑏 will change the number of
each of these species. The number of hydrogen atoms and
nitrate ions currently needed to balance this reaction are not the same. So we’ll need to focus on one of
these species in order to determine the value of 𝑏.
Looking at the products side, we
can see that all of the nitrate ions are a part of the zinc nitrate and the species
already has a coefficient. This means that placing additional
coefficients in the balanced chemical equation will not affect the total number of
nitrate ions on the products side. So we need to determine what value
of 𝑏 will give us 12 nitrate ions on the reactants side. Each molecule of nitric acid only
contains one nitrate ion. So if we set the value of 𝑏 to 12,
then there will be one times 12 nitrate ions on the reactants side and the nitrate
ions will be balanced. We have determined that the value
of 𝑏 is 12.
Before we move on to coefficient
𝑐, we need to remember that nitric acid also contains hydrogen atoms. So placing a coefficient of 12 in
front of nitric acid means that there are now one times 12 hydrogen atoms on the
reactants side, but the hydrogen atoms are still unbalanced. Now let’s find the value of 𝑐. Coefficient 𝑐 represents the
number of arsine gas molecules on the products side. Arsine gas contains both arsenic
atoms and hydrogen atoms. So we’ll need to decide which to
use in order to find the value of 𝑐. On the products side, we see
hydrogen atoms in two different molecules, and we don’t yet know the value of
coefficient 𝑑. So the number of hydrogen atoms on
the products side will change when the value of coefficient 𝑑 is changed.
This means that we can’t use the
hydrogen atoms to determine the value of 𝑐. So we’ll focus our attention on
balancing the arsenic atoms. We need to determine what value of
𝑐 will give us two arsenic atoms on the products side. Each arsine gas molecule contains
one arsenic atom. So if we set the value of 𝑐 to
two, then there will be one times two arsenic atoms on the products side and the
arsenic atoms will be balanced. We have determined that the value
of 𝑐 is two. Before we move on to coefficient
𝑑, we need to remember that each arsine gas molecule contains three atoms of
hydrogen. So two arsine gas molecules will
contain three times two atoms of hydrogen. This increases the total number of
hydrogen atoms on the products side to eight.
Now let’s determine the value of
𝑑. Coefficient 𝑑 represents the
number of water molecules on the products side. A molecule of water contains both
hydrogen atoms and oxygen atoms. As coefficient 𝑑 is the last
coefficient needed to balance this chemical reaction, we could balance either the
hydrogen atoms or the oxygen atoms in order to find the value of 𝑑. But as the hydrogen atoms are
located in two molecules on the products side, the oxygen atoms will be easier to
balance. We need to determine what value of
𝑑 will give us three oxygen atoms on the products side. One water molecule only contains
one atom of oxygen. So if we set the value of 𝑑 to
three, then there will be one times three oxygen atoms on the products side and the
oxygen atoms will be balanced.
Let’s verify that a coefficient of
three will also balance the hydrogen atoms. Each water molecule contains two
hydrogen atoms, so two water molecules will contain two times three hydrogen
atoms. This brings the total number of
hydrogen atoms on the products side to 12, balancing the hydrogen atoms. We have balanced the overall
reaction and have determined that the value of 𝑑 is three.
