Consider the following series of reactions :
`Cl_(2)+2NaOH to NaCl+NaClO+H_(2)O`
`3NaClO to 2NaCl+NaClO_(3)`
`4NaClO_(3) to 3NaClO_(4)+NaCl`
How much `Cl_(2)` is reqired to prepare 122.5 g of `NaClO_(4)` by above sequencial reactions ?

Explanation

The partial equations corresponding to the probable steps of this reaction are as follows :
`Cl_(2) + H_(2)O to HCl + HClO`………(i)
`NaOH + HCl to NaCl + H_(2)O`…….(ii)
`NaOH + HClO to NaClO + H_(2)O`……..(iii)
`HClO to HCl + O`……(iv)
`NaClO + O to NaClO_(3)`.........(v)
On balancing these equations by hit and trial method, we get
`Cl_(2) + H_(2)O to HCl + HClO`.....(vi)
`NaOH + HCl to NaCl + H_(2)O`......(vii)
`NaOH + HClO to NaClO + H_(2)O`......(vii)
`HClO to HCl + O`.....(ix)
`NaClO + 2O to NaClO_(3)`.......(x)
The intermediate species HCI, HClO, NaCIO, and O can be cancelled by multiplying Eq. (vi) by 3, Eq. (vii) by 5, and Eq. (ix) by 2 and adding all partial equations together. Thus, we have
`[Cl_(2) + H_(2)O to HCl + HClO] xx 3`
`[NaOH + HCl to NaCl + H_(2)O] xx 5`
`NaOH + HClO to NaClO + H_(2)O`
`[HClO to HCl + O] xx 2`
`NaClO + 2O to NaClO_(3)`
The final equation is
`6NaOH + 3Cl_(2) to 5NaCl + NaClO_(3) + 3H_(2)O`