Please help me with chemistry about Enthalpy

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 shown below.

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.-260 kJ

B.-69 kJ    

C.  -18 kJ

D.-73 kJ

E. none of these choices is correct

Answer 1

We are told that 69.1 kJ of heat energy is released when ONE mol of calcium carbonate (CaCO3) is formed.

So, for 3.8 mols of CaCO3, we have 3.8 mols x 69.1 kJ/mol = 262.58 kJ and rounding to 2 sig. figs, we have

260 kJ. Since heat is RELEASED, there would be a negative sign in front of this value signifying it is an exothermic reaction. So, the correct answer would be (A) -260 kJ