What is the amount of heat to be supplied to prepare `128g` of `CaCO_(3)` by using `CaCO_(3)` and followed by reduction with carbon, Reactions are
`CaCO_(3)(s)=CaO(s)+CO_(2)(g),DeltaH^(@)=42.8Kcal`
`CaO(s)+3C(s)=CaC_(2)+CO(g),DeltaH^(@)=111Kcal`

Options

A. 102.6 kcals
B. 221.78 kcals
C. 307.6 kcals
D. 453.46 kcals

Answer

C

Explanation

`CaCO_(3)(s) = CaO(s) + CO_(2)(g), DeltaH^(@)=42.8` kcals
`CaO(s) + 3C(s) = CaC_(2)(s) =CaC_(2)(s) + CO(g), DeltaH^(@) = 111 kcals`
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`CaCO_(3)(s)+ 3C(s) =CaC_(2)(s) + CO(g) + CO_(2)(g) , DeltaH = 153.8` kcals
Thus, heat required to prepare 1 mole of `CaC_(2)` from `CaCO_(3) = 153.8` kcals
Molecular weight of `CaC_(2)` requires 153.8 kcals of heat
128 gm of `CaC_(2)` requires 307.6 kcals of heat