[DamianZ]

I need help in figuring out how to complete these double replacement reactions.

1.) NaCl + Na2CO3 --->

2.) KHCO3 + OH --->

3.) AgNO3 + NaI --->

4.) KOH + HCl --->

5.) Sr(NO3)2 + KIO3 --->

6.) Sr(NO3)2 + Na2SO3 --->

7.) Na2CO3 + H2SO4 --->

8.) N2OH + H2SO4 --->

9.) N2OH + Cu(NO3)2 --->

10.) AgNO3 + NaCl --->

11.) Mg(NO3)2 + CaCl2 --->

-Also if you can provide the *solid* compound in each reaction, that would be help me out greatly.

[Arkcon]

Can you try one, and use that one to see what might happen, and provide insight into the rest?  Just listing 11 questions hardly counts as work on your part.

Per forum rules you should show your attempts at solving the problem before receiving help.

[IonNick]

Switch the cations and anions of both reactants.
Ex.
 AgNO3 + NaI --->
 Ag+ and NO3-, Na+ and I-  => AgI + NaNO3

Sr(NO3)2 + KIO3  --->
Sr2+ and 2NO3-, K+ and IO3- => SrIO32 + KNO3, then balance
Final equation:Sr(NO3)2 + 2KIO3  ---> SrIO32 + 2KNO3

[IonNick]

And the solid compound can be determined by looking at a solubility rules chart

[Borek]

Switch the cations and anions of both reactants.

That will work only when the reaction really happens.

Sr(NO3)2 + KIO3  --->
Sr2+ and 2NO3-, K+ and IO3- => SrIO32 + KNO3, then balance
Final equation:Sr(NO3)2 + 2KIO3  ---> SrIO32 + 2KNO3

So this is wrong, as all you get when mixing these two is a solution containing all four ions.

You were right about the solubility chart part though.