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

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

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

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

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

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


Sagot :

When dealing with ionic reactions, it is often useful to identify and eliminate the spectator ions, which are the ions that do not participate directly in the reaction. Here’s the step-by-step method to find the net ionic equation:

1. Write the Complete Ionic Equation:

The given total ionic equation is:
[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 the Spectator Ions:

Spectator ions are those that do not change during the reaction and appear unchanged on both sides of the equation. In this case, the spectator ions are:
[tex]\[ \text{Na} ^{+} \text{ and } \text{Cl}^{-} \][/tex]
These ions appear in the same form on both the reactants and products side.

3. Eliminate the Spectator Ions:

Removing [tex]\(\text{Na} ^{+}\)[/tex] and [tex]\(\text{Cl}^{-}\)[/tex] from both sides of the equation leaves us with the ions that actually participate in the reaction:
[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:

After removing the spectator ions, the net ionic equation is:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]

Therefore, 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]