How Do You Balance This Net Ionic Equation?

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SUMMARY

The net ionic equation for the reaction between lead(II) perchlorate (Pb(ClO4)2) and sodium iodide (NaI) is Pb2+ + I- → PbI2 (s). The spectator ions, Na+ and ClO4-, cancel out, simplifying the equation. It is crucial to ensure that the number of each ion is balanced on both sides of the equation, as highlighted in the discussion.

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Can anyone help me through this net ionic equation problem?

Pb(ClO4)2 (aq) +NaI (aq) -> PbI2 (s) + NaClO4 (aq)
 
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Pb+2 + ClO4- + Na+ + I- -> PbI2 (s) + Na+ + ClO4-

ClO4- and Na+ cancel out on both sides of the equation, so you're left with

Pb+2 + I- -> PbI2 (s)
 
tylernichols said:
Can anyone help me through this net ionic equation problem?

Pb(ClO4)2 (aq) +NaI (aq) -> PbI2 (s) + NaClO4 (aq)

Just make sure that you have the same number of Pb, ClO4, Na and I on both sides of the equation. In your example above, there is one I on the left and two on the right, for example. Same goes for ClO4...

Snazzy, check your work before you post.
 

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