Question

Balance the given chemical equation. Potassium superoxide and carbon dioxide react to form potassium carbonate and oxygen:

Answer 1

Answer:

$\left[\begin{array}{ccc}x1\\x2\\x3\\x4\end{array}\right] =\left[\begin{array}{ccc}4\\2\\2\\3\end{array}\right]$

4KO2 + 2CO2 → 2K2CO3 + 3O2

Step-by-step explanation:

First we write the reaction in the asked form :

x1 KO2 + x2 CO2 → x3 K2CO3 + x4 O2

In a chemical equation, tha amount of  substance that react is the same amount that is formed :

On each side of the equation we must have the same amount of K,C and O.

Let's write this in equations :

K balance : $x1 =2x3$

C balance : $x2=x3$

O balance : $2x1+2x2=3x3+2x4$

The linear equations system is :

$x1=2x3\\2x1+2x2=3x3+2x4\\x2=x3$

Let's equal to 0 to obtain a homogeneous  equations system (for convenience) :

$x1-2x3=0\\2x1+2x2-3x3-2x4=0\\x2-x3=0$

Let's work with the  extended system matrix :

$\left[\begin{array}{ccccc}1&0&-2&0&0\\2&2&-3&-2&0\\0&1&-1&0&0\end{array}\right]$

Working with the matrix :$\left[\begin{array}{ccccc}1&0&-2&0&0\\0&2&1&-2&0\\0&1&-1&0&0\end{array}\right]$

→

$\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&1/2&-1&0\\0&1&-1&0&0\end{array}\right]$

→$\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&-1&0&0\\0&1&1/2&-1&0\end{array}\right]$

→$\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&-1&0&0\\0&0&3/2&-1&0\end{array}\right]$

→$\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&-1&0&0\\0&0&1&-2/3&0\end{array}\right]$

→$\left[\begin{array}{ccccc}1&0&0&-4/3&0\\0&1&0&-2/3&0\\0&0&1&-2/3&0\end{array}\right]$

The matrix is extended to the following linear system :

$x1-\frac{4}{3} x4=0\\x2-\frac{2}{3} x4=0\\x3-\frac{2}{3} x4=0$

$x1=\frac{4}{3}x4\\ x2=\frac{2}{3} x4\\x3=\frac{2}{3}x4$

$\left[\begin{array}{c}x1&x2&x3&x4\end{array}\right] =\left[\begin{array}{c}\frac{4}{3} x4&\frac{2}{3}x4 &\frac{2}{3}x4 &x4\end{array}\right]$

$\left[\begin{array}{c}\frac{4}{3}&\frac{2}{3}&\frac{2}{3}&1 \end{array}\right] x4=\left[\begin{array}{c}\frac{4}{3}&\frac{2}{3}&\frac{2}{3}&1 \end{array}\right]t$

t∈ IR

Let's choose t=3 to eliminate the fractional numbers  →

$\left[\begin{array}{c}x1&x2&x3&x4\end{array}\right] = \left[\begin{array}{c}4&2&2&3\end{array}\right]$