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Why do most all double replacement reactions result in a precipitate and an aqueous molecule? Is it because the individual solute molecules always form stronger bonds such as how in single replacement the molecule with more energy bonds with the molecule of the opposite charge?

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Precipitation of silver chloride is a common example of a double replacement reaction. e.g. $$\ce{AgNO3~_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3~_{(aq)}}$$

Initially all four ions are present in solution. When the solutions are mixed the silver and chloride ions bond together and precipitate, leaving the sodium and nitrate ions remaining in solution along with a small number of silver and chloride ions because the precipitation is an equilibrium so a small number of dissolved ions will still be present (thanks to @Martin for pointing this out). The reaction happens because the precipitation of silver chloride removes these ions from the solution equilibrium, driving the reaction to completion.

Silver chloride is insoluble because the $\ce{Ag-Cl}$ bond has a high degree of covalent character due to the high polarizing ability of the silver cation and the high polarizability of the chloride anion and the similar size of both ions which allows a shorter bond length and so a stronger bond requiring more energy to break apart in order to dissolve the solid. See this question for more detail on this point.

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Most of the double displacement reactions form precipitates but not all of them. precipitates are substances formed after a chemical reaction which are insoluble in water.the products formed by double displacement may or may not be soluble in water

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  • $\begingroup$ You seem to be repeating what the question said. If this is an answer, you should add more. $\endgroup$ – M.A.R. Mar 28 '18 at 20:04
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Precipitation helps drive the reaction in one direction by removing some of the reactants from solution.

Another example of double displacement is $\ce{FeS + H2SO4 -> FeHSO4 + H2S}$ where the reaction is driven in the forward direction by the loss of gaseous $\ce{H2S}$.

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