Answer to Question #306965 in Linear Algebra for Zain

Question

Balance the given chemical equation using linear equations (gussis Jordan elimination method).

Na₂CO3 + C + N₂ ---> NaCN + CO

Answer 1

Let's write the equation in the form

a Na₂CO₃ + b C + c N₂ ---> d NaCN + e CO

where a, b, c, d, and e are unknown numbers.

Equate the left and right sides of the equation:

2 a = d for Na

a + b = d + e for C

3 a = e for O

2 c = d for N

or in matrix form

$$\begin{bmatrix} 2 & 0 & 0 & -1 & 0 \\ 1 & 1 & 0 & -1 & -1 \\ 3 & 0 & 0 & 0 & -1 \\ 0 & 0 & 2 & -1 & 0 \end{bmatrix} \cdot \begin{bmatrix} a \\ b \\ c \\ d \\ e \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

Then using Gauss-Jordan method

$\begin{bmatrix} 2 & 0 & 0 & -1 & 0 \\ 1 & 1 & 0 & -1 & -1 \\ 3 & 0 & 0 & 0 & -1 \\ 0 & 0 & 2 & -1 & 0 \end{bmatrix} \rarr \begin{bmatrix} 1 & 0 & 0 & -1/2 & 0 \\ 1 & 1 & 0 & -1 & -1 \\ 3 & 0 & 0 & 0 & -1 \\ 0 & 0 & 2 & -1 & 0 \end{bmatrix} \rarr \begin{bmatrix} 1 & 0 & 0 & -1/2 & 0 \\ 0 & 1 & 0 & -1/2 & -1 \\ 0 & 0 & 0 & 3/2 & -1 \\ 0 & 0 & 2 & -1 & 0 \end{bmatrix} \rarr$

$\begin{bmatrix} 1 & 0 & 0 & -1/2 & 0 \\ 0 & 1 & 0 & -1/2 & -1 \\ 0 & 0 & 0 & 3/2 & -1 \\ 0 & 0 & 2 & -1 & 0 \end{bmatrix} \rarr \begin{bmatrix} 1 & 0 & 0 & -1/2 & 0 \\ 0 & 1 & 0 & -1/2 & -1 \\ 0 & 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 3/2 & -1 \end{bmatrix} \rarr \begin{bmatrix} 1 & 0 & 0 & -1/2 & 0 \\ 0 & 1 & 0 & -1/2 & -1 \\ 0 & 0 & 1 & -1/2 & 0 \\ 0 & 0 & 0 & 3/2 & -1 \end{bmatrix} \rarr$

$\begin{bmatrix} 1 & 0 & 0 & -1/2 & 0 \\ 0 & 1 & 0 & -1/2 & -1 \\ 0 & 0 & 1 & -1/2 & 0 \\ 0 & 0 & 0 & 1 & -2/3 \end{bmatrix} \rarr \begin{bmatrix} 1 & 0 & 0 & 0 & -1/3 \\ 0 & 1 & 0 & 0 & -4/3 \\ 0 & 0 & 1 & 0 & -1/3 \\ 0 & 0 & 0 & 1 & -2/3 \end{bmatrix}$

Therefore if e = 3 then a = 1, b = 4, c =1, d = 2 and the equation is

Na₂CO₃ + 4C + N₂ ---> 2NaCN + 3CO