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What is the net ionic equation for the reaction represented by the following total ionic equation?

[tex]\[ 6 \, \text{Na}^{+} + 2 \, \text{PO}_4^{3-} + 3 \, \text{Ca}^{2+} + 6 \, \text{Cl}^{-} \rightarrow 6 \, \text{Na}^{+} + 6 \, \text{Cl}^{-} + \text{Ca}_3(\text{PO}_4)_2 \][/tex]

A. [tex]\(2 \, \text{Na}_3\text{PO}_4 + 3 \, \text{CaCl}_2 \rightarrow 6 \, \text{NaCl} + \text{Ca}_3(\text{PO}_4)_2\)[/tex]

B. [tex]\(2 \, \text{PO}_4^{3-} + 3 \, \text{Ca}^{2+} \rightarrow \text{Ca}_3(\text{PO}_4)_2\)[/tex]

C. [tex]\(2 \, \text{PO}_4^{3-} + 3 \, \text{Ca}^{2+} + 6 \, \text{Cl}^{-} \rightarrow 6 \, \text{Na}^{+} + \text{Ca}_3(\text{PO}_4)_2\)[/tex]

D. [tex]\(2 \, \text{P}^{5+} + 8 \, \text{O}^{2-} + 3 \, \text{Ca}^{2+} \rightarrow \text{Ca}_3(\text{PO}_4)_2\)[/tex]



Answer :

GinnyAnswer
To determine the net ionic equation from the provided total ionic equation, we need to follow these steps:

1. Identify the Total Ionic Equation:
The total ionic equation is given as:
[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. Identify Spectator Ions:
Spectator ions are ions that appear on both sides of the equation unchanged. From the total ionic equation, we can see that [tex]\(\text{Na}^{+}\)[/tex] and [tex]\(\text{Cl}^{-}\)[/tex] are present on both the reactant and product sides.

3. Remove Spectator Ions:
By removing the spectator ions ([tex]\(\text{Na}^{+}\)[/tex] and [tex]\(\text{Cl}^{-}\)[/tex]), we simplify the equation to focus on the species that actually undergo change. After removing these ions, we are left with:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]

4. Write the Net Ionic Equation:
The net ionic equation only includes the ions and molecules directly involved in the reaction. Hence, the net ionic equation for the given total ionic equation is:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]

In conclusion, the net ionic equation for the reaction is:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]