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=42.8kcal CaO(s)+3C(s)→CaC2+CO(g);△Ho=111kcal
A
102.6kcal
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B
221.78kcal
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C
307.6kcal
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D
453.46kcal
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Solution
The correct option is C307.6kcal CaCO3(s)→CaO(s)+CO2(g)△H0=42.8kcal CaO(s)+3C(s)→CaC2(s)+CO(g)△H0=111kcal ___________________________________________________ CaCO3(s)+3C(s)→CaC2(s)+CO(g)+CO2(g)△H=153.8kcal
Thus heat required to prepare 1 mole of CaC2 from CaCO3 =153.8kcal 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