Question
Use solubility rules to write the balanced molecular, complete ionic, and net ionic equations for the aqueous reaction between BaI2 and Hg(NO3)2. Include phase subscripts on all substances.
I got to the part where you switch the elements
Ba(NO3)2 +HgI2 I don't know where to go from there?
I got to the part where you switch the elements
Ba(NO3)2 +HgI2 I don't know where to go from there?
Answers
Thanks for telling me where you're stuck but why didn't you show what you had done so I could check it first. The molecular equation is as follows:
BaI2(aq) + Hg(NO3)2(aq) ==> HgI2(s) + Ba(NO3)2(aq)
Now change that to a complete ionic equation. The rules are that if it dissolves and ionizes write it as ions, if gas or covalent write as molecule. Write insoluble materials as molecule.
Ba^2+(aq) + 2I^-(aq) + Hg^2+(aq) + 2NO3^-(aq) ==> HgI2(s) + Ba^2+(aq) + 2NO3^-(aq)
Now cancel those ions that occur on both sides. So can cancel Ba^2+ on left and right as well as 2NO3^- on left and right. That leaves the net ionic equation.
2I^-(aq) + Hg^2+(aq) ==> HgI2(s)
BaI2(aq) + Hg(NO3)2(aq) ==> HgI2(s) + Ba(NO3)2(aq)
Now change that to a complete ionic equation. The rules are that if it dissolves and ionizes write it as ions, if gas or covalent write as molecule. Write insoluble materials as molecule.
Ba^2+(aq) + 2I^-(aq) + Hg^2+(aq) + 2NO3^-(aq) ==> HgI2(s) + Ba^2+(aq) + 2NO3^-(aq)
Now cancel those ions that occur on both sides. So can cancel Ba^2+ on left and right as well as 2NO3^- on left and right. That leaves the net ionic equation.
2I^-(aq) + Hg^2+(aq) ==> HgI2(s)
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