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states of products in single displacement reactions

 
 
Reply Sun 6 Mar, 2005 09:01 am
The following examples are reactions that will occur because the free element is more reactive than the element in the compound.

Zn(s) + CuCl2(aq) --> Cu(s) + ZnCl2(?)

Pb(s) + 2 HCl(aq) --> H2(g) + PbCl2(?)

Cl2(aq) + 2 NaI(aq) --> I2(g) + 2 NaCl(?)

The compounds on the product side are all ionic compounds. My teacher told me that most (or was it all?) ionic compounds are solids are room temperature. At first, the prediction of the state of these compounds would be solid. However, since the compound on the reactant side are aqueous solutions, I was wondering if the compound on the product side will be aqueous as well? An explanation along with your answer will also help me out greatly.
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engineer
 
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Reply Sun 6 Mar, 2005 06:11 pm
A guess
I'll take a stab at it, but no guarantees. My guess is that they are all aqueous because water seperates ionic compounds so readily. Especially anything with Chlorine in it since likes that H in H2O so much. You might even get metal oxides and HCl in the first two equations. The last one might be limited by the concentration of NaCl allowed in water. Some might precipitate out, but you know that NaCl dissolves in water.
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inspiration
 
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Reply Tue 8 Mar, 2005 07:02 pm
Those are the only reactions that proceed, so H2O will not further react with the compound on the product side. Thanks anyways Smile

Since I learned about solubility today, what I'm guessing now is, those compounds will dissolve in water and thus become aqueous, because otherwise it wouldn't 'feel right' to have one product as a solid/gas and the other one isn't aqueous. Anyone else want to take a shot?
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