A.
= (CaCO3 + H2O(g) ) - (Ca(OH)2 + (CO2)
= (-1207.6 -241.8) - (-985.2 -393.5)
= ( -1449.4 + 1378.7) KJ/mol
= - 70.7 KJ/mol.
B.
Moles of Ca(OH)2
= ( mass/molar mass)
= (3.77×1000/74)
= 50.94 mol.
Hence, change in enthalpy when 50.94 mol Calcium hydroxide reacts with excess carbon dioxide
= 50.94 (mol) × {-70.7 (KJ/mol)}
= - 3602 KJ.
Done The Romans used calclum oxide, CaO, to produce a strong mortar to build stone structures. Calclum oxide was mi...