Thermochemistry

Question

How much heat (in kJ) is liberated at constant pressure if 0.834 g of calcium carbonate reacts with excess hydrochloric acid?

CaCO3(s) + 2HCl(aq) --> CaCl2(aq) + H2O (l) + CO2(g) ΔH = -15.2 kJ/mol-rxn

a. -10.2 kJ

b. -0.375 kJ

c. -0.127 kJ

d. -12.7 kJ

e. -0.248 kJ

Answer 1

Hydrochloric acid is used in excess. This means there is plenty of acid for the calcium carbonate to be fully used up in the reaction. It also makes CaCO3 the limiting reagent, so the key to this problem is finding how many moles of CaCO3 were used.

We know from ΔH that, for every mole of CaCO3 we react with HCl, we'll release 15.2 kJ of energy. By finding the moles of CaCO3 used, we can find the amount of energy released using the equation:

moles of CaCO3 * ΔH = energy released

The molecular weight of CaCO3 is 100.0869 grams/mol. This gives us the equation:

(0.834g)/(100.0869g/mol) = 8.33*10^-3 moles of CaCO3

Plugging this in to the energy equation, we get:

8.33*10^-3mol*(-15.2 kJ/mol) = -0.127kJ.

c is the right answer.