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.