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`
`CaCO(s)=3C(s)=CaC_(2)+CO(g),DeltaH^(@)=111Kcal`

Options

A. `102.6Kcal`
B. `221.78Kcal`
C. `307.6Kcal`
D. `453.46Kcal`

Answer

A

Explanation

`CaCO_(3)(s)=CaO(s)+O_(2)(g)DeltaH^(@)=42.8Kcal`
`(CaO(s)+3C(s)=CaC_(2)(s)+CO(g)DeltaH^(@)=111Kcal)/(CaCO_(3)(s)+3C(s)=CaC_(2)(s)+CO(g)+CO_(2)(g))`
`DeltaH=153.8Kcal`
Molecular weight of `CaC_(2)=40+24=64`
`64g` of `CaC_(@)` requires `307.6Kcal` of heat
`128g` of ` CaC_(2)` requires `307.6Kcal` of heat