While studying the school course of chemistry we were taught
two methods of defining the coefficients in the equations of
chemical reactions: the method of manual selection and the method
of electronic balance for the reactions of oxidation and
reduction. I was surprised by the imperfection of these methods.
In the work that is presented to your attention I suggest a
simple method of defining the coefficients in the equations of
chemical reactions with the help of a system of linear algebraic
equations that describes the material balance in a chemical
reaction.
For simplicity the usage of the suggested method will be
demonstrated on the following chemical reactions. Let us
examine the reaction of oxidation of methyl-benzol. Instead of
the unknown coefficients we will put the variables $a$, $b$,
$c$, $d$:
$$a \textrm{C}_6 \textrm{H}_5 \textrm{C} \textrm{H}_3 + b
\textrm{O}_2 \rightarrow c \textrm{C} \textrm{O}_2 + d
\textrm{H}_2 \textrm{O}$$
The number of atoms of each chemical element in the left and
right sides of this equation is equal. Let us write the
equations of material balance for each chemical element.
\begin{array}{ll} \text{Carbon:} & 7a=c\\ \text{Hydrogen:}
& 8a=2d\\ \text{Oxygen:} & 2b=2c+d \end{array} As a
result we will get a system of three linear equations with four
unknowns: \begin{eqnarray} 7a & = & c \\ 8a & =
& 2d \\ 2b & = & 2c+d \end{eqnarray} This system
has an infinite number of solutions, but we have to get the
minimal natural values only. The system has the following
solution: \begin{eqnarray} c & = & 7a \\ d & =
& 4a \\ b & = & 9a \end{eqnarray} \begin{eqnarray}
a & = & 1 \\ b & = & 9 \\ c & = & 7 \\
d & = & 4 \end{eqnarray} Thus, the chemical equation
has the following coefficients:
$$\textrm{C}_6 \textrm{H}_5 \textrm{C} \textrm{H}_3
+9\textrm{O}_2 \rightarrow 7\textrm{C} \textrm{O}_2 +
4\textrm{H}_2 \textrm{O}$$
Let us examine one more example of using the suggested method
for the reaction between potassiumpermanganate $\textrm{K Mn
O}_4$ and iron (II) sulphate, $\textrm{FeSO}_4$, acidified with
$\textrm{H}_2\textrm{SO}_4$.
$$a\textrm{KMnO}_4+b\textrm{FeSO}_4+c\textrm{H}_2\textrm{SO}_4\rightarrow
d\textrm{MnSO}_4+e\textrm{Fe}_2(\textrm{SO}_4)_3+f\textrm{K}_2\textrm{SO}_4
+g\textrm{H}_2\textrm{O}$$
We have obtained the following system: \begin{eqnarray} a &
= & 2f \\ a & = & d \\ 4a+4b+4c & = &
4d+12e+4 \\ f & = & g \\ b & = & 2e \\ b+c
& = & d+3e+f \\ 2c & = & 2g \end{eqnarray}
\begin{eqnarray} a & = & \frac{1}{4}g \\ b & =
& \frac{5}{4}g \\ c & = & g \\ d & = &
\frac{1}{4}g \\ e & = & \frac{5}{8}g \\ f & = &
\frac{1}{8}g \end{eqnarray} \begin{eqnarray} a & = & 2
\\ b & = & 10 \\ c & = & 8 \\ d & = & 2
\\ e & = & 5 \\ f & = & 1 \\ g &= & 8
\end{eqnarray}
$$2\textrm{KMnO}_4+10\textrm{FeSO_4}+8\textrm{H}_2\textrm{SO}_4\rightarrow
2\textrm{MnSO}_4+5\textrm{Fe}_2(
\textrm{SO}_4)_3+\textrm{K}_2\textrm{SO}_4
+8\textrm{H}_2\textrm{O}$$ Thus, the suggested method is both
easy-to-use and it can be used for quick definition of
coefficients for complex equations of chemical reactions.