Answer
$3 N a H C O_{3}+H_{3} C_{6} H_{5} O_{7}=N a_{3} C{6} H{5} O{7}+3 H{2} O+3 C O{2}$
Work Step by Step
$N a H C O_{3}=\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 3\end{array}\right] ; H_{3} C_{6} H_{5} O_{7}=\left[\begin{array}{l}0 \\ 8 \\ 6 \\ 7\end{array}\right] ; N a_{3} C_{6} H_{5} O_{7}=$
$\left[\begin{array}{l}3 \\ 5 \\ 6 \\ 7\end{array}\right] ; H_{2} O=\left[\begin{array}{l}0 \\ 2 \\ 0 \\ 1\end{array}\right] ; C O_{2}=\left[\begin{array}{l}0 \\ 0 \\ 1 \\ 2\end{array}\right]$
A systematic method for balancing the chemical equation is to set up a vector equation that describe the number of atoms of each type present in a reaction. Because given equation involves 4 types of atoms. Construct a vector equation for each reactant and product. The first number is the number of $\mathrm{Na}$ then the number of $\mathrm{H}$ atoms, then $\mathrm{C}$, and the last number in each column is 0
2
\[
x_{1}\left[\begin{array}{l}
1 \\
1 \\
1 \\
3
\end{array}\right]+x_{2}\left[\begin{array}{l}
0 \\
8 \\
6 \\
7
\end{array}\right]=x_{3}\left[\begin{array}{l}
3 \\
5 \\
6 \\
7
\end{array}\right]+x_{4}\left[\begin{array}{l}
0 \\
2 \\
0 \\
1
\end{array}\right]+x_{5}\left[\begin{array}{l}
0 \\
0 \\
1 \\
2
\end{array}\right]
\]
To balance given equation, the coefficient must achieve this
\[
\begin{array}{l}
x_{1}\left[\begin{array}{l}
1 \\
1 \\
1 \\
3
\end{array}\right]+x_{2}\left[\begin{array}{l}
0 \\
8 \\
6 \\
7
\end{array}\right]+x_{3}\left[\begin{array}{l}
-3 \\
-5 \\
-6 \\
-7
\end{array}\right]+x_{4}\left[\begin{array}{c}
0 \\
-2 \\
0 \\
-1
\end{array}\right]+x_{5}\left[\begin{array}{c}
0 \\
0 \\
-1 \\
-2
\end{array}\right]= \\
{\left[\begin{array}{l}
0 \\
0 \\
0 \\
0
\end{array}\right]}
\end{array}
\]
To solve, move all terms to left
\[
\left[\begin{array}{ccccc}
1 & 0 & -3 & 0 & 0 \\
1 & 8 & -5 & -2 & 0 \\
1 & 6 & -6 & 0 & -1 \\
3 & 7 & -7 & -1 & -2
\end{array}\right] \sim\left[\begin{array}{ccccc}
1 & 0 & 0 & 0 & -1 \\
0 & 1 & 0 & 0 & -1 / 3 \\
0 & 0 & 1 & 0 & -1 / 3 \\
0 & 0 & 0 & 1 & -1
\end{array}\right]
\]
Row reduce the coefficient matrix.
\[
\begin{array}{l}
x_{1}=x_{5} \\
x_{2}=\frac{1}{3} x_{5} \\
x_{3}=\frac{1}{3} x_{5} \\
x_{4}=x_{5} \\
\text { with } x_{5} \text { free }
\end{array}
\]
Row reduction of matrix leads to the general solution.
6
\[
x_{1}=3, x_{2}=1, x_{3}=1, x_{4}=3, x_{5}=3
\]
To avoid fraction take $x_{5}=3$ and find $x_{1}, x_{2}, x_{3}, x_{4}$
7
\[
3 N a H C O_{3}+H_{3} C_{6} H_{5} O_{7}=N a_{3} C_{6} H_{5} O_{7}+3 H_{2} O+
\]
$3 C O_{2}
