Question

Romans used calcium oxide, CaO, to produce a strong mortar to build stone structures. Calcium oxide was mixed wit ch reacted slowly with CO2 in the air to give CaCO3.
Ca(OH)2(s) +CO2(g) → CaCO3(s)+H20(g)

Required:
a. Calculate the standard enthalpy change for this reaction.
b. How much energy is evolved or absorbed as heat if 7.50 kg of Ca(OH)2 reacts with a stoichiometric amount of CO2.

Answer 1

Answer:

The given reaction is:  

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

The ΔH°f of Ca(OH)₂ (s) is -986.09 kJ/mole, the ΔH°f of CO₂ (g) is -393.509 kJ/mol, the ΔH°f of CaCO₃ (s) is -1207.6 kJ/mol, and the ΔH°f of H₂O (g) is -241.83 kJ/mol.  

ΔH°rxn = 1 × ΔH°f of CaCO₃ (s) + 1 × ΔH°f of H₂O (g) - 1 × ΔH°f of Ca(OH)₂ (s) - 1 × ΔH°f of CO₂ (g)

ΔH°rxn = 1 (-1207.6) + 1(-241.83) - 1 (-986.09) - 1 (-393.509)

ΔH°rxn = -69.831 kJ

b) The molecular mass of calcium hydroxide is 74.096 gram per mole.  

The mass of calcium hydroxide given is 7.50 Kg or 7500 grams.  

The number of moles of calcium hydroxide is,  

n = Mass of Ca(OH)₂ / Molecular mass of Ca(OH)₂

n = 7500 / 74.1

n = 101.21 moles

As ΔH is negative, therefore, release of heat is taking place. Thus, when one mole of calcium hydroxide reacts, the heat released is -69.831 kJ. Therefore, 101.21 moles of calcium hydroxide will release the heat,  

= 101.21 × 69.831 kJ

= 7.067 × 10³ kJ