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What does

$\ce{Pb(NO3)2 + KI}$

form?

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closed as off-topic by Loong, Klaus-Dieter Warzecha, Simon-Nail-It, Martin - マーチン, Mithoron May 10 '15 at 10:30

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This is just a metathesis (double replacement) reaction.

We know that nitrates are soluble, and alkali metals are soluble (at least in these general cases).

Let's ignore balancing for now, and do that last.

$\ce{Pb(NO3)2 + KI -> Pb^{+2} + NO3^- + K+ + I-}$

$\ce{Pb^{+2} + NO3- + K+ + I- -> PbI2 + KNO3}$

But the $\ce{KNO3}$ is also soluble, and would dissociate.

$\ce{Pb^{+2} + NO3- + K+ + I- -> PbI2 + K+ + NO3-}$

Evidently, we can cancel out the spectator ions $\ce{K+}$ and $\ce{NO3-}$ and are left with a net ionic equation (which is easily balanced).

$\ce{Pb^{+2} + 2I- -> PbI2}$

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  • 1
    $\begingroup$ But if you're actually just looking for what was formed and not for an equation, it would be $\ce{PbI2}$ and $\ce{KNO3}$. $\endgroup$ – Sparkery Feb 16 '15 at 21:39

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