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?

Question

Barrett Brown

Chemistry

21 March, 10:27

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?

+1

Answers (1)

Know the Answer?

Goober · Answer 1

= - 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.