Sodium phosphate is Na3PO4.
Write the balanced equation using that formula. That gives you the balanced MOLECULAR equation.
Now separate the molecules into ions using the following guidelines.
Gases are written as the molecule.
Precipitates (Ni3(PO4)2 is a ppt) are written as the molcule.
Weakly ionized substances (H2O is an example) are written as molecules.
Everything else is written as ions.
Next, look at ions on the left and right. Cancel ions common to both sides.
The result is the net ionic equation.
I have to write the complete balanced molecular, ionic, and net ionic equations for the reaction of nickel II chloride + sodium phosphate. What product, if any, is a precipitiate? I started by getting NiCl2 + Na2(PO4) = (PO4)Ni + NaCl. Balancing it I got NiCl2 + Na2(PO4)Ni + 2NaCl. How do you go about figuring out the ionic and net ionic equations? Do I break them up into oxidation numbers? Ni has a +2 oxidarion number or am I totally going in the wrong direction? Thanks so much for your time
3 answers
i worked it out and got
NiCl2+ Na3(PO4)= Ni(PO4)+ 3NaCl2 for the balanced moloecualr equation.
Ni +2
Cl2 -1
Na3 +1
PO4 -3
Ni+2 = Cl2-1 + Na3+1 + (PO4)-3
Ni+2 + (PO4)-3 = Ni=2 (PO4)-3 for the net ionic equation
would the precipate be nickel II phosphate or Nickel III phosphate. being if i did thid correctly
NiCl2+ Na3(PO4)= Ni(PO4)+ 3NaCl2 for the balanced moloecualr equation.
Ni +2
Cl2 -1
Na3 +1
PO4 -3
Ni+2 = Cl2-1 + Na3+1 + (PO4)-3
Ni+2 + (PO4)-3 = Ni=2 (PO4)-3 for the net ionic equation
would the precipate be nickel II phosphate or Nickel III phosphate. being if i did thid correctly
I gave you the formula for nickel phosphate AND told you it was a precipitate in my first response but you must have copied it down wrong. Here is how you do it. I'm adding (aq) to show the ones in solution and (s) to show the solid precipitate.
3NiCl2(aq) + 2Na3PO4(aq) ==> Ni3(PO4)2(s) + 6NaCl(aq)
The above is the balanced molecular equation. Now we separate it into ions with the rules I gave in my first response. To save time I will omit the (aq) after the ions.
3Ni^+2 + 6Cl^- + 6Na^+ + 2PO4^-3 ==>Ni3(PO4)2(s) + 6Na^+ + 6Cl^-
The above is the balanced ionic equation. The only thing I've done is to separate those aqueous solution into ion if they dissolve but show the ppt as a solid (s). Now to make the NET ionic equation, we cancel those ions common to both sides. Those are 6Cl^- and 6 Na^+ so the net ionic equation is
3Ni^+2(aq) + 2PO4^-3(aq) ==> Ni3(PO4)2(s)
3NiCl2(aq) + 2Na3PO4(aq) ==> Ni3(PO4)2(s) + 6NaCl(aq)
The above is the balanced molecular equation. Now we separate it into ions with the rules I gave in my first response. To save time I will omit the (aq) after the ions.
3Ni^+2 + 6Cl^- + 6Na^+ + 2PO4^-3 ==>Ni3(PO4)2(s) + 6Na^+ + 6Cl^-
The above is the balanced ionic equation. The only thing I've done is to separate those aqueous solution into ion if they dissolve but show the ppt as a solid (s). Now to make the NET ionic equation, we cancel those ions common to both sides. Those are 6Cl^- and 6 Na^+ so the net ionic equation is
3Ni^+2(aq) + 2PO4^-3(aq) ==> Ni3(PO4)2(s)
2 answers