# Question #3b9e4

since NaClO is reduced to NaCl and $N {a}_{2} {S}_{2} {O}_{3}$ is oxidized to $N {a}_{2} {S}_{4} {O}_{6}$
the balanced reaction in ionic form is: $2 {H}^{+} + C l {O}^{-} + 2 {S}_{2} {O}_{3}^{2 -} = C {l}^{-} + {S}_{4} {O}_{6}^{2 -} + {H}_{2} O$
the percent by weight of NaOCl in the bleach sample is$\frac{0 , 00399 g}{0.087 g} 100$ = 4,6%