Find `Delta H_(f)^(@)` of `Fe_(2)O_(3)(s)` if the standard heat reaction `Fe_(2)O_(3)(s)+3CO(g)rarr 2Fe(s)+3CO_(2)(g)` is -6.6 kcal. Given `Delta H_(f)^(@)CO(g)=-26.4 kcal. Delta H_(f)^(@)CO_(2)(g)=-94 kcal`.

Explanation

For following thermochemical equation
`Fe_(2)O_(3)(s)+3CO(g)rarr 2Fe(s)+3CO_(2)(g), Delta H=-6.6 kcal`
Thus, `Delta H_("of reaction") = [Sigma Delta H_(f)^(@)" of " Fe(s)+3 Delta H_(f)^(@)" of " CO_(2)(g)-(Delta H_(f)^(@)" of " Fe_(2)O_(3)(s)+3xxDelta H_(f)^(@)" of " CO(g))]`
`-6.6 = [(2xx0)+(3xx-94)-(Delta H_(f)^(@)" of " Fe_(2)O_(3)(s)+3xx-26.4)]kcal`
`-6.6=[0-282-Delta H_(f)^(@) " of " Fe_(2)O_(3)" of " Fe_(2)O_(3)(s)+79.2]`
`therefore Delta H_(f)^(@)" of " Fe_(2)O_(3)(s)=-196.2 kcal // mol`