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When a chemical reaction takes place a number of molecules combine to produce new molecules. Hence, when hydrogen \(\mbox {H}_2\)
and oxygen \(\mbox {O}_2\) molecules combine, the result is water \(\mbox {H}_2\mbox {O}\). We express this as
Individual atoms are neither created nor destroyed,
so the number of hydrogen and oxygen atoms going into the reaction must equal the number coming out (in the form of
water). In this case the reaction is said to be balanced. Note that each hydrogen molecule \(\mbox {H}_2\) consists of two atoms as does
each oxygen molecule \(\mbox {O}_2\), while a water molecule \(\mbox {H}_2\mbox {O}\) consists of two hydrogen atoms and one oxygen atom. In the
above reaction, this requires that twice as many hydrogen molecules enter the reaction; we express this as
follows:
Equating the number of carbon, hydrogen, and oxygen atoms on
each side gives \(8x = z\), \(18x = 2w\) and \(2y = 2z + w\), respectively. These can be written as a homogeneous linear system
which can be
solved by gaussian elimination. In larger systems this is necessary but, in such a simple situation, it is easier to
solve directly. Set \(w = t\), so that \(x = \frac {1}{9}t\), \(z = \frac {8}{9}t\), \(2y = \frac {16}{9}t + t = \frac {25}{9}t\). But \(x\), \(y\), \(z\), and \(w\) must be positive integers, so the smallest value of \(t\) that eliminates
fractions is \(18\). Hence, \(x = 2\), \(y = 25\), \(z = 16\), and \(w = 18\), and the balanced reaction is
The reader can verify that this is indeed balanced.
It is worth noting that this problem introduces a new element into the theory of linear equations: the insistence that the
solution must consist of positive integers.
Practice Problems
Problems -
Balance the chemical reaction.
\(\mbox {CH}_{4} + \mbox {O}_2 \to \mbox {CO}_{2} + \mbox {H}_{2}\mbox {O}\). This is the burning of methane \(\mbox {CH}_{4}\).
\(\answer {2}\mbox {NH}_{3} + \answer {3}\mbox {CuO} \to \mbox {N}_{2} + \answer {3}\mbox {Cu} + \answer {3}\mbox {H}_{2}\mbox {O}\). Here \(\mbox {NH}_{3}\) is ammonia, \(\mbox {CuO}\) is copper oxide, \(\mbox {Cu}\) is copper, and \(\mbox {N}_{2}\) is nitrogen.
\(\mbox {CO}_{2} + \mbox {H}_{2}\mbox {O} \to \mbox {C}_{6}\mbox {H}_{12}\mbox {O}_{6} + \mbox {O}_{2}\). This is called the photosynthesis reaction—\(\mbox {C}_{6}\mbox {H}_{12}\mbox {O}_{6}\) is glucose.
W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p. 32 Practice Problems , , , and are
Exercises 1.2.60 from Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)
Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, p. 49.
Linear Algebra: An Interactive Introduction