Question

Calcium hydroxide, which reacts with carbon dioxide to form calcium carbonate, was used by the ancient Romans as mortar in stone structures. The reaction for this process is Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(g) ΔH = –69.1 kJ. What is the enthalpy change if 3.8 mol of calcium carbonate is formed?

Answer 1

Answer:

= - 262.58 kJ

Explanation:

From the reaction:

Ca(OH)₂(s) + CO₂(g) → CaCO₃(s) + H₂O(g)   ΔH = –69.1 kJ.

For 1 mole of CaCO₃ to be formed, 69.1 kilo joules energy will be released.

Therefore; for 3.8 moles of CaCO₃ to be formed, the amount of energy released will be;

= 3.8 moles × -69.1 kJ

= -262.58 kJ

The negative sign shows that the reaction is exothermic and energy is released to the surrounding. Therefore, for 3.8 moles of CaCO₃ to be formed an energy of 262.58 kJ will be released.