H2-element
HCl-compound
Na-element
Al2O3-compound
H2O-compound
Answer: The enthalpy of formation of
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of 
The chemical equation for the combustion of propane follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of
is -396 kJ/mol
Dozen = 12,
ii. 1 score = 20
iii. 1 ream = 500
iv. 1 gross = 1.44
The answer is, All of the above