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    Hi could someone show me these equations as a balanced ones:

    (Hot)NaOH+Cl2---->NaClO3+Nacl+H2O

    (Cold) NaOH+Cl2---->NaClO+Nacl+H2O
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    (Original post by Ratboy)
    Hi could someone show me these equations as a balanced ones:

    (Hot)NaOH+Cl2---->NaClO3+Nacl+H2O

    (Cold) NaOH+Cl2---->NaClO+Nacl+H2O
    The first is

    equation is changing one chlorine atom into a chlorate(V) ion and the other chlorine atom into a chloride ion.

    i.e.

    Cl(0) --> Cl(V) + 5e
    Cl(0) + 1e --> Cl(-1)

    To balance the electrons you must multiply the second equation by 5

    Cl(0) --> Cl(V) + 5e
    5Cl(0) + 5e --> 5Cl(-1)

    Hence 6 chlorine atoms are needed on the LHS, i.e. 3 chlorine molecules ....

    can you get there now?
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    sorry but even more confused now!!
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    These are redox reactions. There are two commonly used approaches to balancing redox reactions - using half reactions, and using oxidation numbers. I bet you are expected to use one of them - do you know which one?
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    (Original post by Ratboy)
    Hi could someone show me these equations as a balanced ones:

    (Hot)NaOH+Cl2---->NaClO3+Nacl+H2O

    (Cold) NaOH+Cl2---->NaClO+Nacl+H2O
    HOT: 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O

    COLD: Cl2 + 2NaOH →NaCl + NaOCl + H2O
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    (Original post by Plantagenet Crown)
    HOT: 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O

    COLD: Cl2 + 2NaOH →NaCl + NaOCl + H2O
    I thought posting final solutions without a real need is against forum rules.
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    (Original post by Borek)
    I thought posting final solutions without a real need is against forum rules.
    Maybe, but he asked 2 days ago and is still 'confused' so I though I might as well give the equations. I dont know why people have trouble with this though, it took me 2 seconds of google searching to come across them.
 
 
 
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