Answer:
Option A
Explanation:
Given,
(a) 2C+3H2→C2H6;△H=−21.1kcal
(b) C+O2→CO2;△H=−94.1kcal
(c) H2+12O2→H2O;△H=−68.3kcal
Now, eqs, 2×(b)+3×(c)−(a)
C2H6+32O2→2CO2+3H2O
Heat of combustion of ethane △x=2(−94.1)
+3(-68.3)-(-21.1)
=(−188.2)+(−204.9)−21.1
=−372kcal