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) $\to$ CaCO₃(s) + H₂O(g) $\Delta$ H = -69.1 kJ
What is the enthalpy change if 3.8 mol of calcium carbonate is formed?

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)

\to

CaCO₃(s) + H₂O(g)

\Delta

H = -69.1 kJ
What is the enthalpy change if 3.8 mol of calcium carbonate is formed?

Quizplus · Accepted Answer

The enthalpy change for the formation of 1 mole of calcium carbonate is -69.1 kJ. For 3.8 moles, the enthalpy change is 3.8 * (-69.1 kJ) = -262.58 kJ, which rounds to -260 kJ.

Calcium Hydroxide, Which Reacts with Carbon Dioxide to Form Calcium →\to→

Calcium Hydroxide, Which Reacts with Carbon Dioxide to Form Calcium $\to$