Using the data (all values and in kcal/mol at 25^(@) C) given below: DeltaH_(l)^(@) CO_(2)(g) = -94.05 DeltaH_(t)^(@) CO(g) = -26.41 DeltaH_(l)^(@)H_(2)O(l) = -68.32 DeltaH_(l)^(@)H_(2)O(g) = -57.79 DeltaH_("combustion")^(@) (C_(6)H_(16)) = -1302.7 Mean molar heat capacities C_(p) (cal/mol) at 25^(@) C. CO(g) = 6.97 H_(2)O(g) = 5.92 CO_(2)(g) = 8.96 Calculate the no. of moles of CO and CO_(2) produced, when 0.1 mol of C_(8)H_(18) at 25^(@) C is completely burned at constant pressure in some oxygen gas at 25^(@) C yielding as products gaseous H_(2)O, CO and CO_(2) at 300^(@) C, the process yielded 90.20 kcal of heat to the surroundings.

Using the data (all values and in kcal/mol at 25^(@) C) given below:

1.	Using the data (all values and in kcal/mol at 25^(@) C) given below: DeltaH_(l)^(@) CO_(2)(g) = -94.05 DeltaH_(t)^(@) CO(g) = -26.41 DeltaH_(l)^(@)H_(2)O(l) = -68.32 DeltaH_(l)^(@)H_(2)O(g) = -57.79 DeltaH_("combustion")^(@) (C_(6)H_(16)) = -1302.7 Mean molar heat capacities C_(p) (cal/mol) at 25^(@) C. CO(g) = 6.97 H_(2)O(g) = 5.92 CO_(2)(g) = 8.96 Calculate the no. of moles of CO and CO_(2) produced, when 0.1 mol of C_(8)H_(18) at 25^(@) C is completely burned at constant pressure in some oxygen gas at 25^(@) C yielding as products gaseous H_(2)O, CO and CO_(2) at 300^(@) C, the process yielded 90.20 kcal of heat to the surroundings.
Answer» Solution :`C_(8)H_(18) + 25/2 O_(2)(g) to 8CO_(2)(g) + 9H_(2)O(L)` `DeltaH^(@) = 8DeltaH_(l)^(@) CO_(2)(g) + 9DeltaH_(l)^(@) H_(2)O(l) -DeltaH_(l)^(@)(C_(8)H_(18))` `DeltaH_(l)^(@)(C_(6)H_(18)) = -64.6` kcal REACTIONS under consideration re `C_(8)H_(18) + 25/2 O_(2)(g) to 8CO_(2)(g) + 9H_(2)O(g).DeltaH_(2)` `DeltaH_(1) = 8(-94.05) + 9(-57.79) -(-64.6) = -1207.9` kcal/mol. `DeltaH_(2) =8(-26.41) + 9(-57.79) -(-64.6) = -666.8` k cal/mol Let moles of CO PRODUCED =X `therefore` moles of `CO_(2)` produced = `0.8 -x` `(666.8 x)/8 + (0.8 -x)/8 1207.9 = 120.79-67.64`x. Heat utilized (in kcal): `90.20 + 275 [x(6.97) + (0.8 -x)8.96 + (0.9)(5.92)] xx 10^(-3)` `=93.64 - 0.547` Heat utilized = Heat generated `120.79 - 67.64 x = 93.64 - 0.547`x `x=0.405` mol CO `0.8-x =0.395` mol `CO_(2)`