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# How do you write a complete balanced reaction of aqueous $lead{\text{ }}\left( {II} \right){\text{ }}nitrate$, $Pb{\left( {N{O_3}} \right)_2}$, and aqueous potassium chloride, $KCL$?

Last updated date: 29th Feb 2024
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Answer
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Hint:First, you should see into the formulas for each of the 'fixings' to be certain you can think of them effectively. The formula for lead nitrate is $Pb{\text{ }}{\left( {N{O_3}} \right)_2}$ and the formula for potassium chloride is $KCl$. $Pb{\left( {N{O_3}} \right)_2}\left( {aq} \right) + 2KCl\left( {aq} \right) \to PbC{l_2}\left( s \right) + 2KN{O_3}\left( {aq} \right)$

Complete step by step answer:
Presently you can start to compose the equation for the reaction between these two mixes.
$Pb{\left( {N{O_3}} \right)_2} + {\text{ }}KCl$
Next you embed a bolt to show the heading of the reaction:
$Pb{\left( {N{O_3}} \right)_2} + {\text{ }}KCl{\text{ }} \to$
Next you record the most probable reaction items. For this situation the two cations will switch their anions.
$Pb{\left( {N{O_3}} \right)_2} + {\text{ }}KCl{\text{ }} \to {\text{ }}PbCl{\text{ }} + {\text{ }}K\left( {N{O_3}} \right)$
Presently, the standard in adjusting a chemical equation is that you should have the very same number of particles on each side of the equation. As should be obvious, we don't have a similar number of $\left( {N{O_3}} \right)$particles, so we should fix that. Since we need something more $\left( {N{O_3}} \right)$on the right-hand side, we should make that two atoms of potassium nitrate.
$Pb{\left( {N{O_3}} \right)_2} + {\text{ }}KCl{\text{ }} \to {\text{ }}PbCl{\text{ }} + {\text{ }}2K\left( {N{O_3}} \right)$
That fixes the $\left( {N{O_3}} \right)$ issue, yet now we have such a large number of potassium iotas on the right-hand side, so how about we add one more on the left side.
$Pb{\left( {N{O_3}} \right)_2} + {\text{ }}2KCl{\text{ }} \to {\text{ }}PbCl{\text{ }} + {\text{ }}2K\left( {N{O_3}} \right)$
Alright, that fixes the potassium balance, yet now we have such a large number of chlorides on the left. However, we have another issue. The right formula for lead chloride is $PbC{l_2}$so we should make that rectification.
$Pb{\left( {N{O_3}} \right)_2} + {\text{ }}2KCl{\text{ }} \to {\text{ }}PbC{l_2}{\text{ }} + {\text{ }}2K\left( {N{O_3}} \right)$
We currently have $1$ particle of $Pb$ on each side and $2$ nitrate particles $\left( {NO3} \right)$ on each side and $2$ iotas of potassium on each side and two chloride particles on each side. This is presently a decent equation for the reaction of lead nitrate and potassium chloride.

Note:
Chemistry is an exploratory science and this is something that you should learn. All halides are dissolvable, Except for $lead{\text{ }}\left( {II} \right){\text{ }}chloride$, mercurous chloride$\left( {Hg_2Cl_2} \right)$, and silver chloride$AgCl$.