Calculate the enthalpy change when infinitely dilute solution of `CaCl_(2)` and `Na_(2)CO_(3)` are mixed. `Delta H_(f)^(0)` for `Ca^(+2)(aq)`.
`CO_(3)^(-2)(aq)` and `CaCO_(3)` are `-129.80,-161.7-288.50 kcal mol^(-1)` respectively.

Explanation

To calculate the enthalpy change when infinitely dilute solutions of \( \text{CaCl}_2 \) and \( \text{Na}_2\text{CO}_3 \) are mixed, we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
When \( \text{CaCl}_2 \) and \( \text{Na}_2\text{CO}_3 \) are mixed, the reaction can be written as:
\[ \text{CaCl}_2 (aq) + \text{Na}_2\text{CO}_3 (aq) \rightarrow \text{CaCO}_3 (s) + 2 \text{NaCl} (aq) \]

### Step 2: Identify the ions in the solution.
In an infinitely dilute solution, \( \text{CaCl}_2 \) dissociates into \( \text{Ca}^{2+} (aq) \) and \( 2 \text{Cl}^- (aq) \), and \( \text{Na}_2\text{CO}_3 \) dissociates into \( 2 \text{Na}^+ (aq) \) and \( \text{CO}_3^{2-} (aq) \). Thus, the ions present are:
\[ \text{Ca}^{2+} (aq), \text{Cl}^- (aq), \text{Na}^+ (aq), \text{CO}_3^{2-} (aq) \]

### Step 3: Write the simplified reaction considering the ions.
Since \( 2 \text{Na}^+ \) and \( 2 \text{Cl}^- \) are spectator ions, we can simplify the reaction to:
\[ \text{Ca}^{2+} (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{CaCO}_3 (s) \]

### Step 4: Use the enthalpy of formation values.
The enthalpy change for the reaction can be calculated using the enthalpy of formation values provided:
- \( \Delta H_f^0 (\text{Ca}^{2+} (aq)) = -129.80 \, \text{kcal/mol} \)
- \( \Delta H_f^0 (\text{CO}_3^{2-} (aq)) = -161.7 \, \text{kcal/mol} \)
- \( \Delta H_f^0 (\text{CaCO}_3 (s)) = -288.50 \, \text{kcal/mol} \)

### Step 5: Apply Hess's law.
According to Hess's law, the enthalpy change for the reaction can be calculated as:
\[ \Delta H = \Delta H_f^0 (\text{products}) - \Delta H_f^0 (\text{reactants}) \]
Substituting the values:
\[ \Delta H = \Delta H_f^0 (\text{CaCO}_3) - [\Delta H_f^0 (\text{Ca}^{2+}) + \Delta H_f^0 (\text{CO}_3^{2-})] \]
\[ \Delta H = (-288.50) - [(-129.80) + (-161.7)] \]
Calculating the right side:
\[ \Delta H = -288.50 - (-291.50) \]
\[ \Delta H = -288.50 + 291.50 \]
\[ \Delta H = 3.00 \, \text{kcal} \]

### Final Answer:
The enthalpy change when infinitely dilute solutions of \( \text{CaCl}_2 \) and \( \text{Na}_2\text{CO}_3 \) are mixed is \( 3.00 \, \text{kcal} \).

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