# Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions - Study Questions - Page 217d: 60

#### 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: $-69.83\ kJ/mol\times 13.5\ mol\times 1/1=-942.5\ kJ$

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