Balance the given chemical equation. Potassium superoxide and carbon dioxide react to form potassium carbonate and oxygen:KO2 + CO2 → K2CO3 + O2(Write the reaction in the form x1KO2 + x2CO2 → x3K2CO3 + x4O2, and solve for x1, x2, x3, and x4.)

Answer:[tex]\left[\begin{array}{ccc}x1\\x2\\x3\\x4\end{array}\right] =\left[\begin{array}{ccc}4\\2\\2\\3\end{array}\right][/tex]4KO2 + 2CO2 → 2K2CO3 + 3O2 Step-by-step explanation:First we write the reaction in the asked form :x1 KO2 + x2 CO2 → x3 K2CO3 + x4 O2In 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 : [tex]x1 =2x3[/tex]C balance : [tex]x2=x3[/tex]O balance : [tex]2x1+2x2=3x3+2x4[/tex]The linear equations system is :[tex]x1=2x3\\2x1+2x2=3x3+2x4\\x2=x3[/tex]Let's equal to 0 to obtain a homogeneous  equations system (for convenience) :[tex]x1-2x3=0\\2x1+2x2-3x3-2x4=0\\x2-x3=0[/tex]Let's work with the  extended system matrix :[tex]\left[\begin{array}{ccccc}1&0&-2&0&0\\2&2&-3&-2&0\\0&1&-1&0&0\end{array}\right][/tex]Working with the matrix :[tex]\left[\begin{array}{ccccc}1&0&-2&0&0\\0&2&1&-2&0\\0&1&-1&0&0\end{array}\right][/tex]→[tex]\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&1/2&-1&0\\0&1&-1&0&0\end{array}\right][/tex]→[tex]\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&-1&0&0\\0&1&1/2&-1&0\end{array}\right][/tex]→[tex]\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&-1&0&0\\0&0&3/2&-1&0\end{array}\right][/tex]→[tex]\left[\begin{array}{ccccc}1&0&-2&0&0\\0&1&-1&0&0\\0&0&1&-2/3&0\end{array}\right][/tex]→[tex]\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][/tex]The matrix is extended to the following linear system :[tex]x1-\frac{4}{3} x4=0\\x2-\frac{2}{3} x4=0\\x3-\frac{2}{3} x4=0[/tex][tex]x1=\frac{4}{3}x4\\ x2=\frac{2}{3} x4\\x3=\frac{2}{3}x4[/tex][tex]\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][/tex][tex]\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[/tex]t∈ IRLet's choose t=3 to eliminate the fractional numbers  →[tex]\left[\begin{array}{c}x1&x2&x3&x4\end{array}\right] = \left[\begin{array}{c}4&2&2&3\end{array}\right][/tex]