Answer
See answer below.
Work Step by Step
a) Enthalpies of formation (kJ/mol):
$Ca(OH)_2(s): -986.09, CO_2(g): -393.509 , CaCO_3(s):-1207.6 ,H_2O(g): -241.83$
Enthalpy of reaction:
$-1207.6-241.83+393.509+986.09=-69.83\ kJ/mol$
b) Number of moles of hydroxide:
$1000\ g \div 74.09\ g/mol=13.5\ mol$
Enthalpy:
$-69.83\ kJ/mol\times 13.5\ mol\times 1/1=-942.5\ kJ$
