Ba(NO3)2(aq) Na3PO4(aq)
need to find Molecular, ionic and net ionic Equation
3Ba(NO3)2(aq) + 2Na3PO4(aq) ==> Ba3(PO4)2 + 6NaNO3
3Ba^2+(aq) + 6[NO3]^-(aq) + 6Na^+(aq) + 2[PO4]^3-(aq)==> Ba3(PO4)2(s) + 6Na^+(aq) + 6[NO3]^-(aq)
1 is the balanced molecular equation.
2 is the total ionic equation.
3. To find the net ionic equation, use equation 2 and cancel ions that appear on both sides of the equation. What is left is the NET ionic equation.
Confirm that I didn't miss a sign or that I didn't make a typo.
Molecular Equation:
Ba(NO3)2(aq) + 3Na3PO4(aq) → Ba3(PO4)2(s) + 6NaNO3(aq)
Ionic Equation:
3Ba2+(aq) + 6NO3-(aq) + 6Na+(aq) + 2PO43-(aq) → Ba3(PO4)2(s) + 6Na+(aq) + 6NO3-(aq)
Net Ionic Equation:
3Ba2+(aq) + 2PO43-(aq) → Ba3(PO4)2(s)
Remember, when balancing equations, don't trip over the chemicals or they might get angry!
To find the molecular, ionic, and net ionic equations for the reaction between Ba(NO3)2(aq) and Na3PO4(aq), we need to first write out the complete ionic equation by breaking apart the reactants into their respective ions.
Molecular Equation: Ba(NO3)2(aq) + 2Na3PO4(aq) → Ba3(PO4)2(s) + 6NaNO3(aq)
Ionic Equation:
Ba2+(aq) + 2NO3-(aq) + 6Na+(aq) + PO43-(aq) → Ba3(PO4)2(s) + 6Na+(aq) + 6NO3-(aq)
Net Ionic Equation:
Ba2+(aq) + PO43-(aq) → Ba3(PO4)2(s)
In the net ionic equation, we omit the spectator ions (ions that do not participate in the actual reaction) which are the Na+ and NO3- ions.
To determine the molecular, ionic, and net ionic equations for the reaction between Ba(NO3)2(aq) and Na3PO4(aq), we need to understand the nature of the reactants and products.
First, let's write the formulas for the reactants and products:
Reactants:
Ba(NO3)2(aq) - barium nitrate
Na3PO4(aq) - sodium phosphate
Since both reactants are in aqueous solution (indicated by "(aq)"), they will dissociate into ions when dissolved in water.
Reactants dissociation:
Ba(NO3)2(aq) → Ba2+(aq) + 2NO3-(aq)
Na3PO4(aq) → 3Na+(aq) + PO43-(aq)
Now, let's look at the possible reaction that can occur between the ions:
Ba2+(aq) + 3Na+(aq) + 2NO3-(aq) + PO43-(aq)
By analyzing the charges on the ions, we can see that the barium ion (Ba2+) has a +2 charge, while the phosphate ion (PO43-) has a -3 charge. To balance the charges, three barium ions and two phosphate ions are required:
3Ba2+(aq) + 2PO43-(aq) + 6Na+(aq) + 6NO3-(aq)
Now, let's simplify this equation to obtain the net ionic equation by eliminating the spectator ions. Spectator ions are the ions that do not participate in the reaction and remain unchanged on both sides of the equation. In this case, the spectator ions are the sodium (Na+) and nitrate (NO3-) ions:
Net ionic equation:
3Ba2+(aq) + 2PO43-(aq) → Ba3(PO4)2(s)
The net ionic equation represents the actual reaction that occurs, where the ions that participate directly in the chemical change are shown.
To summarize:
Molecular equation: Ba(NO3)2(aq) + 2Na3PO4(aq) → Ba3(PO4)2(s) + 6NaNO3(aq)
Ionic equation: 3Ba2+(aq) + 2PO43-(aq) + 6Na+(aq) + 6NO3-(aq) → Ba3(PO4)2(s) + 6Na+(aq) + 6NO3-(aq)
Net ionic equation: 3Ba2+(aq) + 2PO43-(aq) → Ba3(PO4)2(s)