Calculate the heat of formation of anhydrous `Al_(2)Cl_(6)` from the following data :
(i) `2Al(s)+6HCl(aq.) to Al_(2)Cl_(6)(aq.)+3H_(2),DeltaH=-244 kcal`
(ii) `H_(2)(g)+Cl_(2)(g) to 2HCl (g),DeltaH=-44 kcal`
(iii) `HCl(g)+aq. to HCl(aq),DeltaH=-17.5 kcal`
(iv) `Al_(2)Cl_(6)(s)+aq. to Al_(2)Cl_(6)(aq.),DeltaH=-153.7 kcal`

Explanation

we have to calculate `DeltaH` of the equation,
(v) `2Al(s)+3Cl_(2)(g) to Al_(2)Cl_(6)(s),DeltaH=?`
in the given equation we see that Equation (iii) does not contain any species Eqn. (v). Let us thus first consider Eqn. (i), (ii) and (iv) only and apply,
`[Eqn. (i)+3xxEqn. (ii)-Eqn.(iv)]`, we get,
`2Al(s)+6HCl(aq.)+3H_(2)(g)+3Cl_(2)(g)-Al_(2)Cl_(6)(s)-aq. to`
`Al_(2)Cl_(6)(aq.)+3H_(2)(g)+6HCl(g)-Al_(2)Cl_(6)(aq.),`
`DeltaH=-244+3xx(-44)-(-153.7)`
or `2Al(s)+3Cl_(2)(g)+6HCl(aq)-aq=Al_(2)Cl_(6)(s)+6HCl(g),`
`DeltaH=-222.3kcal.`
Now multiplying Eqn. (iii) by `6` and adding it to the equation just above we get,
`6HCl(g)+aq+2Al(s)+3Cl_(2)(g)+6HCl(aq)-aq to `
`6HCl(aq)+Al_(2)Cl_(6)(s)+6HCl(g),`
`DeltaH=6xx(-17.5)+(-222.3)`
or `2Al(s)+3Cl_(2)(g) to Al_(2)Cl_(6)(s),DeltaH=-327.3kcal.`