$$\ce{KNO3(aq) + BaCl2(aq) -> KCl(aq) + Ba(NO3)2(aq)}$$

I found products but I noticed in product that both of them are in the aqueous phase. So actually no reaction occurs?

Does a reaction need at least a precipitate to occur to be double replacement?


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    $\begingroup$ Yes, precipitates are necessary. Otherwise all ions remain ions, and there is no reaction. $\endgroup$ – a-cyclohexane-molecule Aug 27 '18 at 1:53
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    $\begingroup$ Or some other way that two of the ions are removed together. E.G. 2 ZnS (s) + HCl (aq) -> ZnCl2 (aq) + H2S (g) $\endgroup$ – DrMoishe Pippik Aug 27 '18 at 2:00
  • $\begingroup$ Welcome to Chemistry.SE! Please note that formulas can be better expressed with \$\ce{}\$ for chemical formulas/equations, \$\mathrm{}\$ for math term/equations, and \$\pu\$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies. $\endgroup$ – A.K. Aug 27 '18 at 3:08
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    $\begingroup$ @A.K., I like welcome comments, but this meta post is probably more helpful than the hidden points of editing one. Not all maths should be in \mathrm{} too, although there can be subtleties there... $\endgroup$ – orthocresol Aug 27 '18 at 15:21

The reason the reaction you drew isn't a double replacement reaction is that the aqueous compounds listed don't have their ions actually bound together in solution, but instead the individual ions are solvated by shells of water molecules. If you write the net ionic equation for your reaction, you get $$\ce{K+(aq) +NO3-(aq) + Ba^{2+}(aq) +2Cl-(aq) -> K+(aq) +NO3-(aq) + Ba^{2+}(aq) +2Cl-(aq)}$$ where you can see that we haven't actually created any new compounds in going from reactants to products.

If one of the products were a precipitate, it then becomes a true reaction, as you start with four kinds of ions in solution, and end with a solid formed from two of the ions and a solution containing the other two.


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