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Question

What is the amount of heat to be supplied to prepare 128 g of CaC2 by using CaCO3 and followed by reaction with carbon. Reactions are:
CaCO3(s)CaO (s)+CO2 (g); Ho=42 .8 kcal
CaO (s)+3C (s)CaC2+CO (g); Ho=111 kcal

A
102.6 kcal
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B
221.78 kcal
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C
307.6 kcal
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D
453.46 kcal
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Solution

The correct option is C 307.6 kcal
CaCO3 (s)CaO(s)+CO2 (g) H0=42.8 kcal
CaO (s)+3C (s)CaC2 (s)+CO (g) H0=111 kcal
___________________________________________________
CaCO3 (s)+3C (s)CaC2 (s)+CO (g)+CO2 (g) H=153.8 kcal

Thus heat required to prepare 1 mole of CaC2 from CaCO3
=153.8 kcal
Molecular weight of CaC2=40+24=64
64 g of CaC2 requires 153.8 kcal of heat
128 g of CaC2 requires 307.6 kcal of heat

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