Write the balanced chemical equation for the following reaction using the oxidation number method:
Reaction of liquid hydrazine $(N_2H_4)$ with chlorate ion $(ClO_3^-)$ in basic medium produces nitric oxide gas $(NO)$ and chloride ion $(Cl^-)$.

(N/A) Step $1$: Assign oxidation numbers to the atoms undergoing change.
$N_2H_4$: $N$ is $-2$,$H$ is $+1$
$ClO_3^-$: $Cl$ is $+5$,$O$ is $-2$
$NO$: $N$ is $+2$,$O$ is $-2$
$Cl^-$: $Cl$ is $-1$
Step $2$: Identify oxidation and reduction half-reactions.
Oxidation: $N$ changes from $-2$ to $+2$ (loss of $4$ electrons per $N$ atom,total $8$ electrons per $N_2H_4$ molecule).
Reduction: $Cl$ changes from $+5$ to $-1$ (gain of $6$ electrons per $Cl$ atom).
Step $3$: Balance the electrons.
Multiply the oxidation half by $3$ and the reduction half by $4$ to balance the electrons ($24$ electrons transferred).
$3 N_2H_4 + 4 ClO_3^- \rightarrow 6 NO + 4 Cl^-$
Step $4$: Balance $O$ and $H$ atoms in basic medium.
Total $O$ on left: $12$. Total $O$ on right: $6$. Add $6 H_2O$ to the right side.
$3 N_2H_4 + 4 ClO_3^- \rightarrow 6 NO + 4 Cl^- + 6 H_2O$
Check $H$ atoms: Left side has $12 H$ $(3 \times 4)$. Right side has $12 H$ $(6 \times 2)$.
The equation is balanced: $3 N_2H_{4(l)} + 4 ClO_{3(aq)}^- \rightarrow 6 NO_{(g)} + 4 Cl_{(aq)}^- + 6 H_2O_{(l)}$