Saved Bookmarks
| 1. |
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 |
|
Answer» 5624 kcal `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 |
|