Question

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, \[KCL\]?

Answer 1

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\[AgCl\].