To determine the net ionic equation for the given total ionic equation, let's carefully analyze the equations provided.
1. Total Ionic Equation:
[tex]\[ 6 \, \text{Na}^{+} + 2 \, \text{PO}_4^{3-} + 3 \, \text{Ca}^{2+} + 6 \, \text{Cl}^{-} \longrightarrow 6 \, \text{Na}^{+} + 6 \, \text{Cl}^{-} + \text{Ca}_3(\text{PO}_4)_2 \][/tex]
2. Given Reaction:
[tex]\[ 2 \, \text{Na}_3 \text{PO}_4 + 3 \, \text{CaCl}_2 \longrightarrow 6 \, \text{NaCl} + \text{Ca}_3(\text{PO}_4)_2 \][/tex]
We need to identify the spectator ions, which are ions that appear unchanged on both the reactant and product sides of the equation. By removing these ions, we will derive the net ionic equation.
From the total ionic equation:
- Spectator Ions: Sodium ([tex]\(\text{Na}^+\)[/tex]) and Chloride ([tex]\(\text{Cl}^-\)[/tex]) ions remain unchanged on both sides of the equation.
- Net Ionic Equation: We omit the spectator ions and focus on the ions and compounds that change during the reaction.
The ions that participate in the reaction are:
[tex]\[ 2 \, \text{PO}_4^{3-} \text{(phosphate ion)} \][/tex]
[tex]\[ 3 \, \text{Ca}^{2+} \text{(calcium ion)} \][/tex]
These ions combine to form:
[tex]\[ \text{Ca}_3(\text{PO}_4)_2 \][/tex]
Therefore, the net ionic equation is:
[tex]\[ 2 \, \text{PO}_4^{3-} + 3 \, \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]
In summary, the net ionic equation for the reaction can be written as:
[tex]\[ \boxed{2 \, \text{PO}_4^{3-} + 3 \, \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2} \][/tex]
