Calculate the heat required to make 6.4 kg of `CaC_(2)` from CaO(s) and C(s) from the reaction :
`CaO(s) + 3 C(s) rarr CaCl_(2)(s) + CO(g)` given that `Delta_(f)H^(theta) (CaO) = - 151.6 kcal`. `Delta_(f)H^(theta) (CaC_(2)) = - 14.2 kcal`. `Delta_(f)H^(theta) (CO) = - 26.4 kcal`

Options

A. 5624 kcal
B. `1.1 xx 10^(4) kcal`
C. `86.24 xx 10^(3)`
D. 1100 kcal

Answer

B

Explanation

`Delta H` for the reaction,
`CaO (s)+ 3C (s) rarr CaC_(2)(s)+CO (g)`
`Delta H=[Delta_(f) H^(theta)(CaC_(2)) +Delta_(f) H^(theta) (CO)]- [Delta_(f)H^(theta) (Cao)+3 Delta _(f) H^(theta) (C ) ]`
`= -14.2-(-26.4) -(-151.6+3 xx 0)`
`=111.0` kcal
Heat required to prepare 64g of `CaC_(2)`
`=111.0`kcal
Heat required to prepare of 6.4 kg of `CaC_(2)`
`=(111.0)/(64) xx 6400 = 1.1 xx 10^(4)` kcal