How much heat is liberated at constant pressure if 0.834 g of calcium carbonate reacts with 48.9 mL of 0.668 M hydrochloric acid?

CaCO₃(s) + 2HCl(aq) $\to$ CaCl₂(aq) + H₂O(l) + CO₂(g) ; $\Delta$ _rH° = -15.2 kJ/mol-rxn

Question

How much heat is liberated at constant pressure if 0.834 g of calcium carbonate reacts with 48.9 mL of 0.668 M hydrochloric acid?

CaCO₃(s) + 2HCl(aq)

\to

CaCl₂(aq) + H₂O(l) + CO₂(g) ;

\Delta

_rH° = -15.2 kJ/mol-rxn

Quizplus · Accepted Answer

The heat liberated is calculated using the enthalpy change of the reaction and the number of moles of the limiting reactant. The limiting reactant is HCl, as there are only 0.0324 mol of HCl available, compared to 0.0104 mol of CaCO3. The heat liberated is therefore:-15.2 kJ/mol-rxn * 0.0324 mol = -0.493 kJSince the reaction is carried out at constant pressure, the heat liberated is equal to the enthalpy change of the reaction. Therefore, the correct answer is A) -0.127 kJ.

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