How to balance redox equations #4

Balance the following chemical equation:
$$CrI_3+NaNO_3+Na_2CO_3 \rightarrow Na_2CrO_4+NaIO_3+NaNO_2+CO_2$$

SOLUTION:
Using the inspection technique, we have:
$${\color{red} a}CrI_3+{\color{red} b}NaNO_3+{\color{red} c}Na_2CO_3 \rightarrow {\color{red} d}Na_2CrO_4+{\color{red} e}NaIO_3+{\color{red} f}NaNO_2+{\color{red} g}CO_2$$
$$\begin{cases} I:3a=e\\ Cr:a=d\\ Na:b+2c=2d+e+f\\ C:c=g\\ O:3b+3c=4d+3e+2f+2g\end{cases}$$
We have more variable than equation. So we can balance the chemical equation using the only inspection technique.

Is this a redox equation? Find out the oxidation number:

$$\overset{+3}{Cr}\overset{-1}{I_3}+Na\overset{+5}{N}O_3+Na_2CO_3 \rightarrow Na_2\overset{+6}{Cr}O_4+Na\overset{+5}{I}O_3+Na\overset{+3}{N}O_2+CO_2$$

Here there are one elements that is reducting (N), and TWO elements that are oxidating (Cr and I). Write down the three equations:

$$\begin{cases} \overset{+5}{N}+2e^- \rightarrow \overset{+3}{N}\\\\ \overset{+3}{Cr} \rightarrow \overset{+6}{Cr}+3e^-\\\\ \overset{-1}{I_3} \rightarrow 3\overset{+5}{I}+18e^-\end{cases}$$

Now simply sum the two oxidation reaction:

$$\begin{cases} \overset{+5}{N}+2e^- \rightarrow \overset{+3}{N}\\\\ \overset{+3}{Cr}+\overset{-1}{I_3} \rightarrow \overset{+6}{Cr}+3\overset{+5}{I}+21e^-\end{cases}$$

Now balance the electrons and sum the two half-equations:

$$\begin{cases} \overset{+5}{N}+2e^- \rightarrow \overset{+3}{N} & \quad \mbox{(x21)}\\\\ \overset{+3}{Cr}+\overset{-1}{I_3} \rightarrow \overset{+6}{Cr}+3\overset{+5}{I}+21e^- & \quad \mbox{(x2)}\end{cases}\quad+$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$=\quad\quad 21\overset{+5}{N}+2\overset{+3}{Cr}+2\overset{-1}{I_3} \rightarrow 21\overset{+3}{N}+2\overset{+6}{Cr}+6\overset{+5}{I}$$

We now have:
$$\begin{cases} b=21\\ a=2\\ f=21\\ d=2\\ e=6\end{cases}$$

From the relationships previously found, we can obtain:

$$\begin{cases} c=5\\ g=5\end{cases}$$

The balanced equation is:
$${\color{red} 2}CrI_3+{\color{red} 21}NaNO_3+{\color{red} 5}Na_2CO_3 \rightarrow {\color{red} 2}Na_2CrO4+{\color{red} 6}NaIO3+{\color{red} 21}NaNO_2+{\color{red} 5}CO_2$$

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PROBLEM 13:
Balance the following reaction:
sodium perchlorate reacts with ammonium chloride giving sodium chloride, nitrogen gas, hydrochloric acid and water.