How to balance redox equations #2

Balance the following chemical equation:
$$NaClO_4+NH_4Cl \rightarrow NaCl+N_2+HCl+H_2O$$

SOLUTION:
Balancing using the inspection technique, we have:
$${\color{red} a}NaClO_4+{\color{red} b}NH_4Cl \rightarrow {\color{red} c}NaCl+{\color{red} d}N_2+{\color{red} e}HCl+{\color{red} f}H_2O$$
$$\begin{cases}Na:a=c\\ Cl:a+b=c+e\\ N:b=2d\\ H:4b=e+2f\\ O:4a=f \end{cases}$$
We have 5 equation and 6 variable, so we cannot solve.

Try now to write down all oxidation numbers: is this a redox equation? Yes, in fact we have:
$$Na\overset{+7}{Cl}O_4+\overset{-3}{N}H_4Cl \rightarrow Na\overset{-1}{Cl}+\overset{0}{N_2}+H\overset{-1}{Cl}+H_2O$$
The chlorine changes its oxidation number from +7 to -1: it is reducing.
The nitrogen changes its oxidation number from -3 to 0: is is oxidating.

The two half-equations are
$$\begin{cases} \overset{+7}{Cl}+8e^- \rightarrow \overset{-1}{Cl}\\\\ 2\overset{-3}{N} \rightarrow \overset{0}{N_2}+6e^-\end{cases}$$
In the nitrogen equation we have $$6e^-$$ because there is the molecular compund $$N_2$$, and so we have to multiply by two.
Now balance the electrons and sum the two equation:
$$\begin{cases} \overset{+7}{Cl}+8e^- \rightarrow \overset{-1}{Cl} & \quad \mbox{(x3)}\\\\ 2\overset{-3}{N} \rightarrow \overset{0}{N_2}+6e^- & \quad \mbox{(x4)}\end{cases}\quad +$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$= \quad3\overset{+7}{Cl}+8\overset{-3}{N} +24e^-\rightarrow 3\overset{-1}{Cl}+4\overset{0}{N_2}+24e^-$$

With the balancing of the oxidation numbers we have found the value of four stoichiometric coefficients:
$$\begin{cases}a=3\\ b=8\\ c=3 \\d=4\end{cases}$$
From the relationships found balancing for the principle of mass, we have:
$$\begin{cases}f=12\\ e=8\end{cases}$$

So the balanced equation is:
$${\color{red} 3}NaClO_4+{\color{red} 8}NH_4Cl \rightarrow {\color{red} 3}NaCl+{\color{red} 4}N_2+{\color{red} 8}HCl+{\color{red} 12}H_2O$$

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PROBLEM 11:
Balance the following chemical equation:
$$(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3+N_2+H_2O$$