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Consider the total ionic equation below.
[tex]\[ Ba^{2+} + 2 NO_3^{-} + 2 Na^{+} + CO_3^{2-} \longrightarrow BaCO_3 + 2 Na^{+} + 2 NO_3^{-} \][/tex]

Which is the net ionic equation for the reaction?

A. [tex]\[ Ba^{2+} + CO_3^{2-} \longrightarrow BaCO_3 \][/tex]

B. [tex]\[ 2 Na^{+} + CO_3^{2-} \longrightarrow Na_2CO_3 \][/tex]

C. [tex]\[ NO_3^{-} + Na^{+} \longrightarrow NaNO_3 \][/tex]

D. [tex]\[ Ba^{2+} + 2 NO_3^{-} \longrightarrow Ba(NO_3)_2 \][/tex]



Answer :

GinnyAnswer
To determine the net ionic equation from the given total ionic equation, follow these steps:

1. Write the total ionic equation:

[tex]\[ \text{Ba}^{2+} + 2 \text{NO}_3^- + 2 \text{Na}^+ + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 + 2 \text{Na}^+ + 2 \text{NO}_3^- \][/tex]

2. Identify the spectator ions:

Spectator ions are ions that do not participate in the actual reaction and remain unchanged on both sides of the equation. In this equation, the spectator ions are:
- [tex]\( \text{NO}_3^- \)[/tex]: nitrate ion appears unchanged on both sides.
- [tex]\( \text{Na}^+ \)[/tex]: sodium ion appears unchanged on both sides.

3. Remove the spectator ions from the equation:

By canceling the common ions on both the reactant and product sides:

[tex]\[ \text{Ba}^{2+} + 2 \text{NO}_3^- + 2 \text{Na}^+ + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 + 2 \text{Na}^+ + 2 \text{NO}_3^- \][/tex]
After removing 2 [tex]\( \text{NO}_3^- \)[/tex] and 2 [tex]\( \text{Na}^+ \)[/tex]:

[tex]\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \][/tex]

4. Write the net ionic equation:

The remaining ions form the net ionic equation:

[tex]\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \][/tex]

Thus, the correct net ionic equation is:

[tex]\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \][/tex]

So, the correct option for the net ionic equation is:
[tex]\( \text{Ba}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{BaCO}_3 \)[/tex].

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