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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 ? |
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Answer» SOLUTION :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` |
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