High School

How much heat is liberated at constant pressure when 97.7 g of calcium oxide reacts with 29.0 L of carbon dioxide gas, measured at 1.00 atm pressure and 25.0°C?

Given:
- \( R = 0.0821 \, \text{L} \cdot \text{atm}/(\text{K} \cdot \text{mol}) \)
- Reaction: \(\text{CaO(s)} + \text{CO}_2\text{(g)} \rightarrow \text{CaCO}_3\text{(s)}\)
- \(\Delta H° = -178.3 \, \text{kJ}\)

Options:
A. \(-3.11 \times 10^2 \, \text{kJ}\)
B. \(-1.74 \times 10^4 \, \text{kJ}\)
C. \(-5.22 \times 10^2 \, \text{kJ}\)
D. \(-2.11 \times 10^2 \, \text{kJ}\)
E. \(-5.17 \times 10^3 \, \text{kJ}\)

Answer :

Final answer:

The enthalpy change for the reaction between calcium oxide and carbon dioxide can be calculated using stoichiometry and the ideal gas law. The heat liberated is approximately -2.11 X 10² kJ.

Explanation:

The enthalpy change for the reaction between calcium oxide (CaO) and carbon dioxide (CO2) can be calculated using the equation:

CaO(s) + CO2(g) → CaCO3(s)

First, convert the mass of CaO to moles using its molar mass (40.08 g/mol). Then, convert the volume of CO2 to moles using the ideal gas law. Finally, use the stoichiometry of the balanced equation to find the molar ratio of CaO to CaCO3. Multiply this ratio by the enthalpy change (∆H° = -178.3 kJ) to determine the heat liberated. The calculated heat is approximately -2.11 X 10² kJ.

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