When 50.0 mL of 0.400 M Ca(NO3)2 is added to 50.0 mL of 0.800 M NaF, CaF2 precipitates, as shown in the net equation below. The initial temperature of the solution is 30.0 °C. Assuming that the reaction goes to completion, and that the resulting solution has a mass of 100.0 g, what is the enthalpy change (ΔH) for the reaction if the final temperature of the solution is 25.0 °C?

• A. -1500 J
• B. -500 J
• C. -200 J
• D. -1000 J

Answer: (B)

Verified step by step guidance

1. Identify the reaction: The net ionic equation for the precipitation of calcium fluoride is: \( \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \rightarrow \text{CaF}_2 (s) \).
2. Calculate the moles of each reactant: Use the formula \( \text{moles} = \text{concentration} \times \text{volume} \). For \( \text{Ca(NO}_3)_2 \), \( \text{moles} = 0.400 \text{ M} \times 0.0500 \text{ L} \). For \( \text{NaF} \), \( \text{moles} = 0.800 \text{ M} \times 0.0500 \text{ L} \).
3. Determine the limiting reactant: Compare the stoichiometric ratio from the balanced equation with the moles calculated to find the limiting reactant.
4. Calculate the heat absorbed or released: Use the formula \( q = m \cdot c \cdot \Delta T \), where \( m \) is the mass of the solution, \( c \) is the specific heat capacity (assume \( 4.18 \text{ J/g°C} \) for water), and \( \Delta T \) is the change in temperature (final - initial).
5. Determine the enthalpy change (\( \Delta H \)): Since the reaction goes to completion, \( \Delta H = -q \) (because the system loses heat). Divide \( q \) by the moles of the limiting reactant to find \( \Delta H \) per mole of reaction.