Consider the following series of reactions:Cl2+2NaOH⟶NaCl+NaClO+H2O3NaClO⟶2NaCl+NaClO34NaClO3⟶3NaClO4+NaClHow much Cl2 is needed to prepare 122.5 g NaClO4 by above sequence?How much Cl2 is needed to prepare 106.5 g of NaClO3 by the above sequence?The oxidation state of chlorine in compound sodium chlorate is

Factor label method: (Mw of NaClO4 = 122.5, Mw of NaClO3 = 106.5 g mol-1)x g of Cl2=122.5 g NaClO4122.5 g NaClO4 per mol of NaClO34 mol NaClO33 mol NaClO43 mol NaClO1 mol NaClO31 mol Cl21 mol NaClO71.0 g Cl21 mol Cl2 =122.5122.5×43×31×11×71.01=284.0 gMole method: n(NaClO)=nCl2,nNaClO3=13n(NaClO)=13nCl2nNaClO4=34nNaClO3=34×13nCl2=14nCl2nNaClO4=122.5 g NaClO4122.5 of NaClO4/mol NaClO4=1.0mol⋅NaClO4nCl2=4×1.0=4mol⋅Cl2Mass of (Cl2) = 4 x 71.0 g=284.0 g Cl2x g Cl2=106.5 g NaClO3106.5⋅g NaClO3 per mol NaClO33 mol NaClO1 mol NaClO3 1 mol Cl21 mol NaClO71.0 g Cl21 mol Cl2 =106.5106.5×31×11×17.01=213.0 g Cl2NaClO3 is sodium chlorate