${\mathrm{N}}_2{\mathrm{H}}_{4\left(\mathrm{l}\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{l}\right)}\to {\mathrm{N}}_{2\left(\mathrm{g}\right)}+4{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$ equivalent weight of reductant and oxidant are respectively X and Y. Then ( M = Molar mass )

n - factor of reductant, N₂H₄(-2 to zero) = $2|0-(-2\left)\right|=4$; Eq.wt =$\frac{\mathrm{M}}{4}$

n - factor of oxidant, H₂O₂(-1 to -2)  = $2|-2-(-1\left)\right|=2$ ; Eq. wt =$\frac{\mathrm{M}}{2}$