$3 \mathrm{Cl}_{2}+6 \mathrm{NaOH} \rightarrow \quad \mathrm{NaClO}_{3}+5 \mathrm{NaCl}+3 \mathrm{H}_{2} \mathrm{O}$
Oxidation number of $\mathrm{Cl}$ is 0 in $\mathrm{Cl}_{2},-1$ in $\mathrm{NaCl}$ and $+5$ in $\mathrm{NaClO}_{3}$
${N_2} + 3{H_2} \longrightarrow NH_3$