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reddy21 · Jun 6, 2009, 05:57 PM

How do you balance Cu(NO3)2 and Fe(S)

Perito · Jun 6, 2009, 07:10 PM

You really need to know what the reactants and the products are. Right now, you have just given the reactants. I have to guess at the products.

$Fe(s) + Cu(NO_3)_2 \rightarrow Fe(NO_3)_2 + Cu(s)$

or

$2Fe(s) + 3Cu(NO_3)_2 \rightarrow 2Fe(NO_3)_3 + 2Cu(s)$

are two possibilities. In the first case, you form iron(II) nitrate (ferrous nitrate); in the second you form iron (III) nitrate (ferric nitrate). Whether either reaction will occur, balancing the equation doesn't say. Also, there might be other products that could be generated from the reaction.

The easiest way to balance this type of reaction is using "half-reactions"

$Fe(s) \rightleftharpoons Fe^{+2} + 2e^-$

or

$Fe(s) \rightleftharpoons Fe^{+3} + 3e^-$

and

$Cu^{+2} + 2e^- \rightleftharpoons Cu(s)$

You multiply the equations you're going to add by numbers so that the number of electrons is equal from one equation to the other. Make sure that one equation has them on the left and the other on the right, then add the equations.

$Fe(s) + Cu^{+2} \rightleftharpoons Fe^{+2} + Cu(s)$

$2Fe(2) + 3Cu^{+2} \rightleftharpoons 2Fe^{+3} + 3Cu(s)$

You add the spectator ions ( $NO_3^-$ in this case).

This link has more information on balancing half reactions:

AskMehelpDesk - Balancing Half-Reactions