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What is the net ionic equation for the reaction between [tex]Na_2SO_4[/tex] and [tex]CaCl_2[/tex]?

A. [tex]SO_4^{2-}(aq) + Ca^{2+}(aq) \rightarrow CaSO_4(s)[/tex]

B. [tex]2 Na^{+}(aq) + SO_4^{2-}(aq) + Ca^{2+}(aq) + 2 Cl^{-}(aq) \rightarrow CaSO_4(s) + 2 Na^{+} + 2 Cl^{-}(aq)[/tex]

C. [tex]2 Na^{+}(aq) + 2 Cl^{-}(aq) \rightarrow 2 NaCl (aq)[/tex]

D. [tex]Na_2SO_4(aq) + CaCl_2(aq) \rightarrow CaSO_4(s) + 2 NaCl (aq)[/tex]

Sagot :

GinnyAnswer
To determine the net ionic equation for the reaction between [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] and [tex]\( \text{CaCl}_2 \)[/tex], we need to follow these steps:

1. Write the balanced molecular equation:
[tex]\[ \text{Na}_2\text{SO}_4(aq) + \text{CaCl}_2(aq) \rightarrow \text{CaSO}_4(s) + 2 \text{NaCl}(aq) \][/tex]

2. Write the complete ionic equation:
Since [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] and [tex]\( \text{CaCl}_2 \)[/tex] are soluble in water, they dissociate into their respective ions:
[tex]\[ 2 \text{Na}^+(aq) + \text{SO}_4^{2-}(aq) + \text{Ca}^{2+}(aq) + 2 \text{Cl}^-(aq) \rightarrow \text{CaSO}_4(s) + 2 \text{Na}^+(aq) + 2 \text{Cl}^-(aq) \][/tex]

3. Identify and remove the spectator ions:
Spectator ions are ions that appear on both sides of the ionic equation without undergoing any change. In this case, [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{Cl}^- \)[/tex] are the spectator ions.

4. Write the net ionic equation:
After removing the spectator ions, we are left with the ions that participate in forming the precipitate:
[tex]\[ \text{SO}_4^{2-}(aq) + \text{Ca}^{2+}(aq) \rightarrow \text{CaSO}_4(s) \][/tex]

The net ionic equation for the reaction between [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] and [tex]\( \text{CaCl}_2 \)[/tex] is:
[tex]\[ \text{SO}_4^{2-}(aq) + \text{Ca}^{2+}(aq) \rightarrow \text{CaSO}_4(s) \][/tex]

So, the correct answer is:
A. [tex]\( \text{SO}_4^{2-}(aq) + \text{Ca}^{2+}(aq) \rightarrow \text{CaSO}_4(s) \)[/tex]
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