What is the amount of heat to be supplied to prepare 28 g of CaC2 by using CaCO3 and followed by reduction with carbon. Reactions areCaCO3(s)=CaO(s)+CO2(g) ΔH∘=42.8kcalCaO(s)+3C(s)=CaC2+CO(g) ΔH∘=111kcal

CaCO3(s)=CaO(s)+CO2(g);    ΔH∘=42.8kcalCaO(s)+3C(s)=CaC2(s)+CO(g);    ΔH∘=111kcal ---------------------------------------------------------------------------------------CaCO3(s)+3C(s)=CaC2(s)+CO(g)+CO2(g)  ΔH=153.8kcalThus heat required to prepare I mol of CaC2 from CaCO3 = 153.8 kcalMolecular weight of CaC2 = 40 + 24 =  6464 g of CaC2 requires 153.8 kcal of heat128 g of CaC2 requires 307 .6 kcal of heat