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?

(A) -18 kJ
(B) -69 kJ
(C) -73 kJ
(D) -260 kJ

Answer 1

The enthalpy change is -260 kJ.

Step-by-step explanation:

Enthalpy is an extensive property, that is, it depends of the amount of matter. Let's consider the following reaction:

Ca(OH)₂(s) + CO₂(g) ⇄ CaCO₃(s) + H₂O(g)

When 1 mol of CaCO₃ is formed, 69.1 kJ of heat are released. Then, for 3.8 moles of CaCO₃,

`3.8molCaCO_(3).(-69.1kJ)/(molCaCO_(3)) =-2.6 * 10^(2) kJ`