Let's start by analyzing the given total ionic equation:
[tex]\[ Ba^{2+} + 2 NO_3^- + 2 Na^+ + CO_3^{2-} \rightarrow BaCO_3 + 2 Na^+ + 2 NO_3^- \][/tex]
1. Identify the spectator ions:
Spectator ions are ions that do not participate in the actual chemical reaction. They remain unchanged on both the reactant and product sides of the equation. In this case:
- [tex]\( NO_3^- \)[/tex] (nitrate ion) appears on both sides of the equation.
- [tex]\( Na^+ \)[/tex] (sodium ion) also appears on both sides of the equation.
2. Remove the spectator ions:
By removing the spectator ions [tex]\( 2 NO_3^- \)[/tex] and [tex]\( 2 Na^+ \)[/tex], we are left with the ions that participate in the reaction:
[tex]\[ Ba^{2+} + CO_3^{2-} \rightarrow BaCO_3 \][/tex]
This equation shows the precipitation of barium carbonate (BaCO_3) from barium ions (Ba^{2+}) and carbonate ions (CO_3^{2-}).
Therefore, the net ionic equation for this reaction is:
[tex]\[ Ba^{2+} + CO_3^{2-} \rightarrow BaCO_3 \][/tex]
Now, let's match this with the given options:
1. [tex]\[ Ba^{2+} + CO_3^{2-} \rightarrow BaCO_3 \][/tex]
2. [tex]\[ 2 Na^+ + CO_3^{2-} \rightarrow Na_2CO_3 \][/tex]
3. [tex]\[ NO_3^- + Na^+ \rightarrow NaNO_3 \][/tex]
4. [tex]\[ Ba^{2+} + 2 NO_3^- \rightarrow Ba(NO_3)_2 \][/tex]
Upon comparison, the correct net ionic equation is:
[tex]\[ Ba^{2+} + CO_3^{2-} \rightarrow BaCO_3 \][/tex]
Therefore, the correct answer is:
[tex]\[ Ba^{2+} + CO_3^{2-} \rightarrow BaCO_3 \][/tex]
