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

[tex]\[
\begin{array}{l}
Na^{+} + 2PO_4^{3-} + 3Ca^{2+} + 6Cl^{-} \longrightarrow 6Na^{+} + 6Cl^{-} + Ca_3(PO_4)_2 \\
2Na_3PO_4 + 3CaCl_2 \longrightarrow 6NaCl + Ca_3(PO_4)_2 \\
2PO_4^{3-} + 3Ca^{2+} \longrightarrow Ca_3(PO_4)_2 \\
\end{array}
\][/tex]

Sagot :

GinnyAnswer
To find the net ionic equation for the given reaction, we need to follow these steps:

1. Write down the total ionic equation:
The given equation is:
[tex]\[ \text{Na}^+ + 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} + 6 \text{Cl}^- \longrightarrow 6 \text{Na}^+ + 6 \text{Cl}^- + \text{Ca}_3(\text{PO}_4)_2 \][/tex]

2. Identify and cancel spectator ions:
Spectator ions are ions that do not participate in the actual reaction and appear unchanged on both sides of the equation. In this case, Na^+ and Cl^- are spectator ions because they appear on both sides of the equation.

3. Remove spectator ions to obtain the net ionic equation:
After removing the spectator ions (Na^+ and Cl^-), we are left with:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]

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