Determine the enthalpy of formation of anhyd. Al2Cl6(s) Given2Al(s)+6HCl(aq)→Al2Cl6(aq)+3H2(g)+239760calΔHf(HCl(g))=−22kcal/molMaximum amount of heat of solution of HCl(g) in water = - 17315 cal / molMaximum amount of heat of solution of Al2Cl6(s) in water = - 153.69 kcal / mol.

For 2Al(s)+3Cl2(g)→Al2Cl6(s) enthalpy change is ΔH which  isΔHfAl2Cl6Now, (i) 2Al(s)+6Hcl(aq)→Al2C6(aq)+3H2(g)ΔH1=−239760cal (ii)     H2(g)+Cl2(g)→2HCl(g)    ΔH2=−44000cal (iii)     HCl(g)+aq→HCl(aq)    ΔH3=−17315cal (iv)     Al2Cl6(s)+aq→Al2Cl6(aq)    ΔH4=−153690cal    Apply(i)+3×(ii)−(iv)+6×(iii)    ΔHf=−239760+3(−44000)−(−153690)+6(−17315)    =−321960cal/mol