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    Can anybody show me how to do the ionic equation for 3Cu(s)+ 8HNO3(aq)> 3Cu(NO3)2(aq)+ 2NO(g)+ 4H2O(l). Thanks in advance
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    Write out each ion respectively and cancel out what is on both sides
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    All the (aq) substances will have ions e.g. HNO3 (aq) is made up of H+ ions and NO3- ions. Do this for all aq substances then cancel out any spectator ions
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    Thanks but it goes to 8NO3(aq) on one side and 6NO3(aq) 2NO(g) on the other. Would there be any spectator ions in this incidence? As no ion is equal on both sides.
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    (Original post by H.ScottMurfitt)
    Thanks but it goes to 8NO3(aq) on one side and 6NO3(aq) 2NO(g) on the other. Would there be any spectator ions in this incidence? As no ion is equal on both sides.
    Well I'm not gonna do it all for you...but the 6NO3 will cancel out with 8 NO3 to make 2NO3 on one side and 2NO on the other side.
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    (Original post by H.ScottMurfitt)
    Thanks but it goes to 8NO3(aq) on one side and 6NO3(aq) 2NO(g) on the other. Would there be any spectator ions in this incidence? As no ion is equal on both sides.
    What happened to the H+ you started with? There doesn't seem to be H+ on the product side of the equation.

    Cu2+ will be a spectator ion. Some of the (NO3)- are also spectator ions as they remain unchanged on the product side.
 
 
 
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