How do you write a complete balanced reaction of aqueous $l e a d (I I) n i t r a t e$ , $P b {(N O_{3})}_{2}$ , and aqueous potassium chloride, $K C L$ ?

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, $$KCL$$?

Vedantu Content Team · Accepted Answer

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$$AgCl$$.

How do you write a complete balanced reaction of aqueous lead (II) nitrate, Pb(NO3)2, and aqueous potassium chloride, KCL?

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How do you write a complete balanced reaction of aqueous $l e a d (I I) n i t r a t e$ , $P b {(N O_{3})}_{2}$ , and aqueous potassium chloride, $K C L$ ?